Download TEMPERATURE HEAT

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C H A P T E R 12
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TEMPERATURE
AND HEAT
It is winter, and these Macaque
monkeys are enjoying a dip in the
Jigokudani hot springs in Japan. The
water in the photograph exists in
three forms or phases , solid (snow),
liquid , and gas (water vapor). Water
can change from one phase to
another, and heat plays a role in the
change , as we will see in this chapter.
(@ Shusuke Sezai/Corbis)
COMMON TEMPERATURE SCAlES
Fahrenheit
scale
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To measure temperature we use a thermometer. Many thermometers make use of
the fact that materials usually expand with increasing temp巳rature. For example , Figure
12.1 shows th巳 common mercury-in-glass th巳rmom巳ter， which consists of a mercury-filled
glass bulb conn巳cted to a capillary tub巳 . When the mercury is heated , it expands into the
capillary tube , the amount of expansion being proportional to the chang巳 in t巳 mperature.
The outside of th巳 glass is marked with an appropriat巳 scale for reading the t巳 mp巳 rature.
A number of different temp巳raωre scales have be巳 nd巳vised ， two popular choic巳 s being the Celsius (formerly, centigrad巳) and Fahrenheit scales. Figure 12.1 illustrates these
scales. Historically,* both scales were d巳自 ned by assigning two temperature points on the
scale and then dividing the distance betw巳en them into a number of equally spaced intervals. One point was chosen to b巳 the temperature at which ic巳 melts under one atmosphere
of pressure (the "ice point") , and the other was th巳 temperature at which water boils under
one atmosphere of pressur巳 (the "steam point"). On the Celsius scale , an ice point of 0 oC
(0 degrees Celsius) and a st巳am point of 100 oC wer巳 s巳lected. On the Fahrenheit scale , an
ice point of 32 OF (32 degrees Fahrenheit) and a st巳 am point of 212 oF w 巳r巳 chosen. The
Celsius scale is used worldwide , while th巳 Fahrenheit scale is used mostly in the United
States , often in home medical thermometers.
There is a subtle differenc巳 in the way the temperature of an object is reported , as compared to a change in its temperature. For example，由e temperature of th巳 human body is about
37 oC , where 由巳 symbol oC stands for "degr，巳巳s Celsius." Howev町，由e change between two
temperaωres is specified in
*Today, the Celsius and Fahrenh巳 it scales are defined in terms of the Kelvin temperature scale; Section 12.2
discllsses 由e Kelvin scale
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Example 1
Ah巳 althy p巳rson
Converting from a Fahrenheit to a Celsius Temperature
has an oral
temperatur巳 of
98.6 oF. What wou ld this reading be on
th巳 Cel s iu s
sca l 巳?
Reasoning and Solution A
po 川 of 32.0 oF. 缸
S mc
怡t 巳mp巳ratωur陀巳
of 98.6 OF
比
is
66.6
F址抽11 巳 nh巳
eit d巳gr巳es abov巳 t由
h巳 l陀
c巳
( 1C \
(66.6 F O) l ; '::0 ! = 37.0 C 。
、
3
‘
'
Thus , the person 's temperature is 37.0 Celsius degrees above th巳 ic巳 poin t. Adding 37.0
Celsius d巳grees to th巳 ice point of 0 oc on th巳 Celsiu s scale giv巳 s a Celsius t巳 mperature of
o
137.0 c l
。
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Atim巳 and 怡t 巳mp巳ratωure
毡 sign
on a bank indi比cat优
:es 由
t ha
创t 由
t h巳 outdoor t巳 mperature is - 20.0 oc. Find
on the Fahr巳 nheit scale
th 巳 corresponding temp巳rature
Reasoning and Solution The temperature of -20.0 oc is 20.0 Celsius
point of 0 oC. This numb巳r of Celsius degre巳s corr巳 sponds to
(20.0
C) !飞 主二)
=
1 CJ
36.0
degr巳巳 s
below
th巳 ice
F。
The t巳 mp巳 ratur毡， then , is 36.0 Fahrenheit d巳grees b巳low th 巳 lC巳 poin t. Subtracting 36.0
Fahrenheit d巳gr些 s from the ic巳 point of 32.0 oF on the Fahrenheit scale gives a Fahrenheit
temperature of 卜 4.0 。 叫
。
The reasoning strategy us巳d in Examples 1 and 2 for converting
is summariz巳 d below.
betwe巳 n differ巳 nt
t巳 mp巳ratur巳 scal巳 s
REASONING STRATEGY
Converting Between Different Temperature Scales
1. Determine the magnitude of the difference between
the ice point on the initial scale.
th巳 stated temp巳rature
and
2. Convert this number of degrees from one scale to the other scal巳 by using the appropriat巳 conv巳rsion factor. For conversion betw巳en th 巳 Celsius and Fahrenheit
scales , the factor is bas巳d on the fact 阳t 1 C。二 ; F。
3. Add or subtract the
on th巳 n巳w scal巳.
numb巳r
of degrees on the new scale to or from the ice point
~ CHECK YOUR UNDERSTANDING
(The answer is given at the end of the book.)
1. On a new temperature scale the steam point is 348 oX , and the ice point is 112 oX. What
is the temperature on this scale that corresponds to 28.0 oC?
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body temperature rises to 39 oC , the change in temperature is 2 Celsills d 巳gre巳 s or 2 C O,
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not 2 oc.
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As Figure 12.1 indicates , the separation b巳tween the ice and steam points on th巳
Celsius scal巳 is divided into 100 Celsills degr巳白， while on the Fahrenheit scale the s巳pa­
ration is divided into 180 Fahr巳 nheit degrees. Therefore , th 巳 size of th巳 Celsius degree is
larger than that of the Fahrenheit d巳gree by a factor of f击， or ~ . Examples 1 and 2 illustrate how to convert betwe巳 n the Celsills and Fahrenheit scales using this factor.
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CHAPTER 12 TEMPERATURE AND HEAT
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Steam
point
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14
d巳grees
(1 2.1)
273.15
The number 273 .1 5 in Equation 12.1 is an exp巳rimental result , obtained in studies that utilize
a gas-bas巳d thermom巳ter.
When a gas confined to a fixed volum巳 is heated , its pressure increases. Conversely,
wh巳n the gas is cool 巳d ， its pr巳 ssur巳 decr巳 as巳s. For 巳xample ， the air pressur巳 in automobil 巳
tires can rise by as mllch as 20% after the car has been driven and the tires have become
wann. Th巳 chang巳 in gas pressure with temperature is th巳 basis for the constant-volume
gas thermometer.
A constant-volum巳 gas thermometer consists of a gas-fi lI巳d bulb to which a pressure
gallge is attached , as in Figur巳 12.3. The gas is oft巳 n hydrogen or h巳 l illm at a low density ,
and th巳 pr 巳 ssur巳 gaug巳 can be a U-tllb巳 manometer filled with m巳rcury. The blllb is placed
in thermal contact with th巳 sllbstance whos巳 temp巳rature is b巳II1 g m 巳 asured . The volume
of the gas is held constant by raising or lowering the right column of the U-tllbe manometer in order to keep the mercury level in the left column at th巳 same r巳f巳 rence leve l. Th 巳
absolllt巳 pressllr巳 ofth巳 gas is proportional to the height h of the mercury on the righ t. As
th巳 t巳mp巳rature chang巳s ， the pressllr巳 chang巳s and ca l1 b巳 used to indicate the temperature ,
onc巳 th巳 cO l1 stant-volllme gas thermomet巳r has b巳巳 n calibrated.
Sllppose that th巳 absolllt巳 pr巳ssure ofth巳 gas in Figur巳 12.3 is measur巳d at different temperatures. If the results are plott巳d on a pressure-versus-t巳mperature graph , a straight line is
obtained , as in Figure 12 .4. If th巳 straight lin巳 is extended or 巳xtrapolated to lower and lower
t巳mperatw白， the Li n巳 crosses the temperature axis at -273.15 o
In reality , 110 gas can be
cooled to this t巳 mperature ， becaus巳 all gases liqllify before reaching i t. However, helium and
hydrogen liquify at such low temperatures that they are often llsed in th巳 thermometer. This
kind of graph can be obtained for differ巳nt amounts and types of low-density gases. In all
cases , a straight line is found that 巳xtrapolates to -273.15 0 C on th巳 temperature axis , which
suggests that th巳 valll巳 of -273 .1 50 C has fundam巳ntal significance. The significance of this
nllmber is that it is th巳 absolute zero point for t巳 mperatllre measuremen t. The phras巳 " ab­
SOlllt巳 zero" means that temperatures lower than - 273.15 oC cannot be reach巳d by continllally cooling a gas or any oth巳r substanc巳. If lower t巳 mp巳ratures could b巳 r巳 ach巳d ， th巳n
fllrther extrapolation of th巳 straight line in Figur巳 12 .4 wOllld sugg巳st that negativ巳 absolute
gas pressures could 巳Xl
c.
Absolute
-.J I All thermom 巳ters make u
ature. A property that changes with temperature is called a thermometric proper，砂.FOl巳X
ampl巳， the thermometric prop巳rty ofthe m巳 rcury tbermometer is th 巳 l 巳 ngth of the mercury
7J
Figure 12.2 A comparison of the
and Celsius temperature scales.
K巳 l vin
」一一一γ一....-/
U-tube
Substance whose
manometer
temperature is
bei ng measured
Figure 12.3 A constant-volum巳 gas
thermomete r.
Absolute
pressure
/
~I THERMOMETERS
nu
zero
民d
T=]飞+
One kelvi 门
equals one
Celsius degree
m/
』
kelvin ." Th 巳 kelvin is the SI base unit for t巳 mperature.
Figure 12.2 comp创r巳s th巳 Kelvin and Celsius scales. The siz巳 ofon 巳 kelvin is id巳 ntical t。
由e size of one C冶 lsius d巳gree because there are one hundred divisions betwe巳n the ice and
steam points on both sc aJ es. As we will discuss shortly，巳xp巳 rim巳 nts have shown that thel它 ex­
ists a lowest possible t巳 mp巳ratur毡， below which no substance can be cool巳d. This low巳 st temp巳rature is defined to be the zero point on the Kelvin scale and is referred to as absolute zero.
The ic巳 point (0 oc) occurs at 273. j 5 K on the Kelvin scale. Thus , th巳 Kelvin temperature T and the Celsius temperatur巳 Tc are related by
Ice
point
qζ
11 Although the Celsius and Fahrenheit scales are widely used , the Kelvin temperature scale has greater scientific significance. It was introduced by the Scottish physicist
William Thompson (Lord Kelvin , 1824-1907) , and in his honor 巳ach degree on the scale
is call巳d a kelvin (K). By international agreement , the symbol K is not written with a degre巳 sign (0) , nor is the word "degrees" us巳d when quoting temperatures. For exampl 巳， a
t巳 mperature of 300 K (not 300 OK) is read as "three hundred kelvins ," not "three hundr 巳 d
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373.15 一一一一一 100.00
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Celsius , oc
Kelvin , K
NII THE KELVIN TEMPERATURE SCALE
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12.3 THERMOMETERS
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-200 - 100
+100 +200
Temperature , oc
negative-pressure
reglon
Figure 12.4 A plot of absolute pressure
V巳rsus temperature for a low-density
gas at constant volume. The graph is a
straight line and , when 巳xtrapolated
(dash巳d line) , crosses the temperature
axis at -273.15 oC
TEMPERATURE AND HEAT
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Hot junction
Constantan
Reference junction
Ice-point bath , 0 oc
(a)
Figure 12.5 (α) A thermocouple is
made from two diffl巳rent types of wires ,
copper and constantan in thjs case.
(b) A thermocouple junction between two
different wires. (b , @ Omega Engineering ,
Inc. AlI rights reserved. Reproduced with
permission of Omega Engineering , lnc. ,
Stamford , CT, www. om巳ga.com.)
Figure 12.6 Invasive carcinoma
(cancer) of th巳 bre a st registers colors
from red to yellow/white in this thermogram , indicating markedly el巳vated
temperatures. (Science Photo Library/
Photo Researchers, lnc.)
The physics of
thermography.
Figure 12.7 Thermogram showing a
smoker's forearms before (left) and
5 minutes after (right) he has smoked a
cigarette. Temperatures range from
over 34 oC (white) to about 28 oC
(blue). (Dr. Arthur Tuc k巳 r/S c i en ce Photo
Library/Photo Researchers, Inc.)
(b)
column , while in the constant-volume gas th巳rmometer it is th巳 pressure of the gas. Several
oth巳r important thermometers and their th 巳rmometnc prop巳rties will now b巳 discussed.
Th巳 thermocouple is a thermometer used 巳xt巳 nsively in scientific laboratori 巳 s. It consists of thin wires of different metals , welded together at the ends to form two junctions , as
Figure 12.5 illustrates. Often th巳 m巳tals ar巳 copper and constantan (a copper-nickel alloy)
On巳 of the junctions , called the "hot" junction , is placed in thermal contact with th巳 0均巳 ct
whose temp巳rature is being measured. Th巳 other junction ，也rm巳d the "reference" jllnction ,
is k巳pt at a known constant temp巳rature (usually an ic巳-water mixture at 0 oC). The th巳 r­
mocouple g巳n巳rates a voltage that depends on th巳 difJ社 rence in temperature betwe巳n th巳
two junctions. This voltag巳 is the thermom巳 tric prop巳rty and is measur巳 d by a voltmet町， as
the drawing indicates. With the aid of calibration tables , the temperature of the hot junction
can b巳 obtained from th巳 voltage. Thermocouples ar巳 used to measure t巳 mperatures as high
as 2300 oC or as low as - 270 。巳
Most substances offer resistanc巳 to th巳 flow of electricity, and this resistanc巳 changes
with temp巳rature . As a r巳 sult， 巳lectrical resistance provides another th巳rmometric property.
Electrical resistance thermometers are often made from platinum wire , b巳caus巳 platinum
has excellent mechanical and electrical properties in th巳 temperatur巳 range from -270 oc
to + 700 o Th巳 resistance of platinum wire is known as a function of temperature. Thus ,
the temperatur巳 of a substance can be det巳rmined by placing th巳 resistanc巳 thermometer
in th巳rmal contact with the substance and measuring the resistance of the wir巳Radiation emitt巳d by an object can also be used to indicate temperature. At low to
moderate t巳mperatur町， th巳 pr巳dominant radiation emitted is infrared . As the temp巳rature
lS ra1 S巳d ， th巳 mt巳 nsity of th巳 radiation incl 巳 ases substantially. 1n on 巳 interesting application , an infrared camera regist巳rs the intensity of the infr缸.ed radiation produced at different locations on the human body. The cam巳ra is connected to a color monitor that displays
the diff，巳r巳 nt infrared intensities as di 征"erent colors. This "thermal painting" is called a
thermograph or thermogram. Thermography is an important diagnostic tool in medicine.
For example , br巳 ast cancer is indicat巳d in th巳 thermogram in Fig
c.
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NEAR THERMAL EXPANS ION 361
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Ll NEAR THERMAL EXPANSION
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12 .4 Ll
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NORMAL SOllDS
Have you ever found the metallid on a glass jar too tight to open? On巳 solution is
to run hot water over th巳 lid ， which loosens it becaus巳 the metal 巳xpands more than the
glass does. To varying 巳xt巳nts ， most materials expand when heat巳d and contract when
cooled. Th巳 mcr巳 ase in any one dimension of a solid is called linear expansion, linear in
the sense that the 巳xpansion occurs along a line. Figure 12.9 illustrates the linear expan
sion of a rod whose length is Lo when th巳 t巳mperature is T o. When the temperature of the
rod increases to T o + !1 T , the length becomes Lo + !1 L , wh巳re !1 T and !1 L are the changes
in temp巳rature and length , respectively. Conv巳rsely， when th巳 t巳mperature decreases to
To - !1 T , the length d巳creas巳s to Lo - !1L.
For modest temperature changes , exp巳rim巳nts show that the change in length is directly proportional to the change in temp巳rature (!1 L IX !J.T) . In addition , the change in
l巳ngth is proportional to the initiallength of 出巳 rod ， a fact that can be und巳rstood with th巳
aid ofFigur巳 12 . 10. Part a of th巳 drawing shows two identical rods. Each rod has a 1巳ngth
Lo and expands by !1 L wh巳n the temperature incr巳 as巳 s by !1T. Part b shows th巳 twoh巳 ated
rods combined into a single rod , for which the total expansion is th巳 sum of the expansions
of 巳ach part-namely, !1 L + !1 L = 2 !1L. Clearly, th巳 amount of expansion doubles if th巳
rod is twice as long to begin with. In other words , the change in length is dir巳ctly proportional to th巳 originall巳ngth (!1 L IX Lo). Equation 12.2 expr巳sses th巳 fact that !1 L is proportional to both Lo and !1 T (!1 L IX Lo !1 T) by using a proportionality constant α， which is
called the coeJ.声cient 01 linear expansion.
Figure 12.8 A thermogram of the
1997/98 El Niño (r巳d) ， a large region of
abnormally high temperatures in the
Pacifìc Ocean. (Courtesy NOAA)
Ll NEAR THERMAL EXPANSION OF A SO Ll D
The length Lo of an object changes by an amount !1 L when its
an amount !1 T:
t巳 mperatur巳 changes
!1 L= αLo !1 T
whereαis th巳 coefficient
by
(1 2.2)
Temperature = To
lTempe
of linear expansion.
ωm仰 Uni，咖 the Coφcie叫 Line川xpans毗古=阶 l
LO
Solving Equation 12.2 for αshows that α = !1L1 (Lo !1 T). Sinc巳 th巳 l巳ngth units of !1 L
and Lo algebraically cancel , the co巳fficient of linear expansionαhas th巳 unit of (CO)- l
when th巳 temperature difference !1 T is expressed in Celsius degrees (C O). Different
materials with the same initial length expand and contract by diff，巳 rent amounts as the
/
2 ð. L
Figure 12.10 (α) Each of two identical rods
expands by ßL when heated. (b) When the rods
ar巳 combined into a single rod of length 2L口， the
"combined" rod expands by 2 ß L.
ð. L
Fig ure 12.9 When the t巳 mperature of a
rod is raised by ßT, th巳 length of the
rod increases by ß L.
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TEMPERATURE AND H EAT
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Coeffìcient
ofTh巳rmal Expansion (CO) - I
S lI bstanc巳
Linear
(α)
Volume
.d o
(β)
Solids
23 X 10- 6
A111minllm
10- 6
19 X
12 X 10- 6
17 X 10- 6
Brass
3.00000 m
(a)
Concr巳 te
Copper
Glass (common)
(1))
Figure 12.11 (α) Two concl 巳t巳 s l abs
completely fill the space between the
buildings. (b) When the te mperatUl巳
incr巳a ses ， each slab expands , causing
the sidewalk to buckle.
Lead
Nickel
QlI artz (fllsed)
10- 6
10- 6
10- 6
10- 6
26 X 10- 6
9.9 X 10- 6
42 X 10- 6
13 X 10- 6
39 X 10- 6
1.5 X 10- 6
X
X
X
X
X
0.50 X 10- 6
19 X 1O- 6
Silver
X
X
X
X
10- 6
10- 6
10- 6
10- 6
10- 6
8.5
3.3
14
12
29
Glass (Pyrex)
G01d
lron or steel
69
57
36
51
36 X 10- 6
87 X 10- 6
57 X 10- 6
Liquids b
1240 X 10- 6
1240 X 10- 6
1120 X 10- 6
Benz巳n 巳
Carbon tetrachlorid巳
Ethyl alcohol
Gasolin巳
950
182
1200
207
M巳rcury
Methyl alcohol
Water
•
x 10
X 10- 6
X 10- 6
X 10- 6
"The valu巳sforαand βp 巳 rta ll1 10 a 1巳 mpe rature near 20 oC
bSince liquids do not have fìxed shapes , the coeffì ci巳 nt of linear expansion is not defìned for them
temperature changes , so th巳 v al l1巳 of αdep巳 nds on th巳 nature of th 巳 material. Tabl巳 12 . 1
shows some typical valll巳 s . Coefficients of linear expansion also vary somewhat dep巳 nd­
ing on the rang巳 of t巳 mperatur巳 s involved , bllt the values in Table 12.1 ar巳 adequate approximations . Exampl 巳 3d巳 al s with a sitllation in which a dramatic effect due to thermal
expansion can be observed , even though the change in temp巳rature is small.
。 Example 3
Buckling 们 S i dewalk
A concrete s id巳walk is constrllcted b etw巳巳 n two bllildings on a day when the , temp巳rature is
25 oc . Th巳 sidewalk consists of two slabs , each thre巳 meters in length and of negligible thickness (Figllre 12.11a) . As the t巳 mp巳rature rises to 38 oc , the slabs expand , bllt no s pac巳 is provided for thermal 巳x pan s ion. The buildings do not move , so the slabs buckl 巳 upw创'd.
D巳termm巳 the vertical di s tanc巳 y in part b of th 巳 draw ing
Reasoning Th巳巳x pand巳 d length of 巳 ach slab is eqllal to its originallength pllls the chang巳 In
length I:::. L dll巳 to th巳 ri se in temp巳rature. We know the original length , and Eqllation 12.2 can
be used to fì nd th 巳 chan g巳 in length. Once th 巳 ex pa nd 巳d length has be巳 n determined , th 巳
Pythagorean theorem can b巳巳 mployed to fìnd the vertical distanc巳 y in Figur在 12.11b.
Solution The change in temp巳 rature is I:::. T = 38 oc - 25 oc = 13 C O, and the co巳f白 c i e nt of
linear expansion for concrete is g iv巳 n in Table 12. 1. The change in 1 巳 ngth of each slab associated with this temperature change is
O
6
I:::. L = α Lo l:::. T = [1 2 x 10- (CO) - I](3 .0 m) (1 3 C ) = 0.00047 m
(1 2.2)
Th巳 expanded
by applying
Figure 12.12 An expansionjoint in a
bridge. (Richard Choy/Peter Arnold , Inc.)
length
of 巳ach
slab is , thus , 3.00047 m. Th 巳 V巳rtical distanc巳
to the right triangl 巳 in Figure 12.11b :
y
can be
obtain 巳d
th巳 Pythago rea n theor巳 m
y= 的 00047 m)2 一 (3 . 00000 m) 2 = I 0.053 m I
。
m
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Table 12.1 Coefficients of Thermal Expansion for Solids a nd Liquids a
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Actuator
spri ng
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Movable
plunger
to
The buckling of a sidewalk is on巳 cons巳 qu巳 nce of not providing s u ffici 巳 nt room for
therm
a1 expansion. To eliminate such probl巳 ms ， engine巳rs incorporat巳巳xpan s ion joints or
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spaces at intervals along bridge roadbeds , as Figure 12.12 shows
ÞJ" The physics 01 an antiscalding device. Although Example 3 shows how thermal
expansion can cause problems , ther 巳 are also times when it can be usefu l. For instance , each year thollsands of children ar巳 taken to em巳rg巳ncy rooms sllffering
from burns caus巳d by scalding tap water. Such accid巳 nts can be redllced with the aid of the
antiscalding devic巳 shown in Figure 12.13. This devic巳 scr巳ws onto the 巳 nd of a faucet and
qllickly shuts off the ftow of water when it becomes too hot. As th巳 wat巳r temperature rises ,
the actuator spring expands and pllshes th巳 plllng巳r forward , shlltting off th巳 ftow. Wh en the
water cools, the spring contracts and the water ftow reSllmes.
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Water
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Figure 12.13 An antiscalding devic巳.
THERMAL STRESS
If the concrete slabs in Figure 12.11 had not bllckled llpward , they would have be巳n
to imm巳nse forc巳s from the buildings. Th巳 forc巳 s n 巳巳d巳d to prev巳 nt a solid ob
j 巳ct from expanding mllst be strong enough to COllnt巳ract any cha吨巳 in length that wOllld
occur du巳 to a change in temperature. A1 though th巳 change in t巳 mperature may be small ,
th巳 forces-and hence the stresses-can b巳 enormou s. Th 巳y can , in fact , lead to serious
structllral damag巳. Example 4 illllstrates jllSt how large th巳 str巳 sses can b巳 .
Sll均 ect巳d
A[~lil l'.. . \'._\~.I.II. :I .4 ' ''-IIi'I E P T P R 1.:'. :fI \'J"~
Example 4
The Stress on a Steel Beam
A steel beam is used in th巳 roadb巳d of a bridge. The beam is mo unt巳db巳tween two concrete supports wh巳n the temp巳rature
is 23 oC, with no room provided for thermal expansion (see Figure 12 .1 4). What compressional stress must the concr巳t巳 sup ­
ports apply to each 巳 nd of the beam , if they ar巳 to k，巳巳p th巳 b巳 am from 巳xpanding wh 巳 n th巳 temperature rises to 42 OC?
Assume that th巳 distanc巳 between th巳 concret巳 supports does not change as the temperature rises.
Reasoning When th巳 temperatllre ns巳s by an amollnt ð. T, the natural tendency of the
beam is to 巳xpand. If the b巳am were fr巳e to 巳xpand ， it would lengthen by an amount
ð. L = αLo ð. T (Eqllation 12.2). How巳V巳r， the concrete sllpports prevent this 巳xpansion from
occllrring by exerting a compressional force on 巳 ach 巳 nd of the beam. The magnitude F of
this forc巳 d巳pends on ð. L throllgh the relation F = YA( ð.L/Lo) (Equation 10.1 7) , where Y is
Young's modulus for steel and A is the cross-sectional area of the beam. Accord ing to the
discussion in Section 10.8 , th巳 compressional str巳 ss is eqllal to the magnitude of the compr巳 ssional force divided by th巳 cross 四 sectional area , or Str巳 ss = F/A.
Knowns and Unknowns The data for this problem are listed in the
Description
tabl巳:
Value
23
Concrete
support
oc
m
r-a
-HU
L
mα
e
V且，同 U
n
iv A
ν
且
，A
町
AmmM
巳咀
St
『
··
气J·
ny
m
…
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ffρLVL l
4EE
巳
7
a--nur
-md
Z dnd
dph
Ltf
tE
t
t idDA
EJU
lejO
IHWO
i
l U}I
dm c
-
Frhur
Unknown Variable
Stress
supp口时
m川m
42 0 C
Concrete
L
山刀
To
T
m附
lnitial temperature
Final temp巳rature
Symbol
The physics 01
thermal stress.
L
Modeling the Problem
E圃 S…
compr'陀巳s岱sional forc巳 div
叮id巳d
by 由
th巳 cross-s巳ctional area A of 由
th巳 be缸丑
(s巳巳 Section 10.8
创) ， or
S仕ess =王
A
Continued
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( ð. L \
YAI 一一-1
\ Lo /
Y is Young's modulus , ð.L is the change in length , and La is the originallength of
tress
c;r:
-i..
‘
Substituting this expr臼 sion for F into the d巳finition of stress giv巳s Equation 1 in
由巳 right colurnn. Young's modulus Y for st巴巴1 is k:nown (see Table 10 .1), but we do not
k:n ow 巳ither ð.L or La. However, M is proportionalω 勾， so we will focus on ð.L in Step 2.
、
出巳 beam.
/ ·飞
where
.d o
,
,,
E固z
岛缸
Linear The
忧阳
e创rn
巳xpand巳d
、
J
‘ E
aLo(T
rsE飞
|ð.L =
唱 EA
by an amount M =αL
乌o ð. T (Equation 12.2) , where αis the coefficient of linear
expansion. The change in temperature is the final temperature T rninus the initial temperature To, or ð. T = T - To. Thus , the beam would have expand巳d by an amount
ω |
expression the variabl巳s T and To are k:nown , and αis available in Table 12. 1. We
substitute this relation for ð.L into Equation 1, as indicat巳d in th巳 right colurnn.
1n 由is
Solution Algebraically combining the results of each step , we have
Stress
,_
ð. L
Y 丁一
"'0
，
Y
α与(T 一 To )
y，α (T
- To)
"'0
Note that 由巳 original length Lo of the beam is elirninated alg巳braically from this resul t.
Taking the value of Y = 2.0 X 10 11 N/m 2 from Table 10.1 and α= 12 X 10- 6 (C O ) - 1
from Table 12.1 , we find that
Stress =
Yα (T
- To)
= (2.0 X 10 11 N/的 [12 X 10- 6 (C O )- I](42 oC - 23 0 C) = I4.6 X 10 7 阳时|
This is a large stress , equivalent to
ne缸忖 on 巳 rnillion
Related Homework: Problems 20,
pounds per square foo t.
2 又 98
THE BIMETAL Ll CSTRIP
A bimetallic strip is made from two thin strips of metal that have different
co巳ffi­
Cl巳nts of linear expansion , as Figure 12 . 15αshows. Often brass [α= 19 X 10- 6 (C )- I]
O
(a)
(b) Heated
(c ) Cooled
Figure 12.15 (a)A bim巳tallic strip and
how it behaves when (b) heated and
(c) cooled
O
and steel [α= 12 X 10- (C ) 一 1] are selected. The two pieces ar巳 welded or riveted togeth巳r. When the bimetallic strip is heated , th巳 brass ， having th巳 larger value of α ， expands
more than th巳 ste巳1. Since the two metals 缸 e bonded together, the bimetallic strip b巳 nds into
an arc as in part b , with the longer brass piece having a larger radius than the steel piec巳­
When the strip is cooled , the bimetallic strip bends in the opposite direction , as in part c.
The physics of an automatic coffee maker. Bim巳 tallic strips are frequ 巳 nt ly us巳 d as adjustable automatic switches in el 巳ctrical appliances. Figure 12.16 shows an automatic coffee m丛cer that turns off when the coffee is brewed to the selected str巳 ngth. 1n part a , while
the brewing cy c1 e is on ，巳 lectricity passes through the heating coil that h巳 ats th巳 water. The
巳lectricity can f1 0w becaus巳 th巳 contact mount巳d on the bimetallic strip touches th巳 con­
tact mount巳d on the "strength" adjustm巳 nt knob , thus providing a con tÎ nuolls path for the
巳lectricity. When the bimetallic strip gets hot enough to bend away , as in part b of th巳
drawing , the contacts separate. The electricity stops becaus巳 it no longer has a continuolls
path along which to f1 ow, and th巳 brewing cy c1 e is shut off. Tllrning the "str巳 ngth" knob
adjusts th巳 brewing tim巳 by adjusting th巳 distanc巳 through which the bimetallic strip must
bend for the contact points to separat巳.
6
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F=
C
According to Equation 10.17 , the magnitude of the compressional force that the concrete
巳xert on each end of th巳 st巳el beam is given by
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CHAPTER 12 TEMPERATURE AND HEAT
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Bimetallic strip (cold) I
Contacts closed
(0)
Coffee
,
I辈革
pot 、 n "
\
(b)
THE EXPANSION OF HOlES
An interesting example of linear expansion occurs wh巳n there is a hole in a piec巳
of solid material. We know that the mat，巳rial itself expands when heat巳d ， but what about
the hole? Do巳 s it expand , contract, or remain the sam巳? Conceptual Example 5 provides
som巳 insight into the answer to this question.
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Bimetallic strip (ho t)
Coffee pot "o ff"
Figure 12.16 A bimetallic strip controls
whether this ∞仔ée pot is (α) "on"
(strip cold , straight) or (b) "off" (strip
hot, bent).
Co nceptual Example 5
Do Holes Expand or Contract When the Temperature Increases?
Figure 12.17a shows eight square tiles that are attach巳 d tog巳ther and arrang巳 d to form a square
pattern with a hol巳 in th巳 center. lf the tiles ar巳 heated ， does the size of th巳 hole (a) decrease or
(b) incr巳 ase?
Reasoning We can analyze this problem by disass巳 mbling the pattern into separate tiles , heating them , and then reassembling the pattern. What happ巳 ns to each of the individual tiles can
b巳 explained using what w巳 know about linear expansion.
Answer (a) is incorrect. When a tile is heated both its length and width expand. It is tempting
to think , therefor巳 ， that the hole in the pattern decreases as th巳 surrounding tiles 巳xpand into it.
However, this is not corr巳 ct ， b巳cause any on巳 tile is prev巳nted from expanding into the hole by
the expansion of the tiles next to it.
Answer (b) is correct. Sinc巳巳ach tile expands upon heati吨， th巳 pattern also expands , and 由ehole
along with it , as shown in Figure 12.17b. In fact , if w巳 had a ninth tile that was id巳ntical to the
oth巳rs and heat巳dittoth巳 same extent , it would fit exactly into the hol 巳， as Figure 12.7c indicates.
Thus , not only does the hole expand , it does so exactly as 巳 ach of th巳 tiles does. Since the ninth
tile is made of the same materi aJ as th巳。由巳rs ， w巳 S巳巳 that the hole expands just as if it were mad巳
of th巳 mat巳rial of th巳 sUlTounding tiles. The thermal expansion of 由 e hole and the surrounding
material is an aJ ogous to a photographic enlarg巳ment:巳verything is enlarged , including holes.
Related Homework: Problems 13, 23
。
Instead of the separat巳 tiles in Example 5 , w巳 could hav巳 used a squar巳 plate with a
square hole in the c巳 nter. Th巳 hole in the plate would hav巳巳xpand巳d jllSt lik巳 the hole in
the pattern of tiles. Fllrth巳 rmore ， the same conclllsion applies to a hole of any shap巳. Thus ,
[Jil〔巴山川 」 副「口
MV- 口 | 川 山 丁 口
口口已D川
(a)
Unheated
(b )
Heated
二口口
(c)
Figure 12.17 (α) The tiles are arranged
to form a square pattern with a hole in
the center. (b) Wh en the tiles are
heated , the pattern , including th巳 hole
in the center，巳xpands. (c) The
巳xpanded hole is the same size as a
heated tile.
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Example 6
A Heated Engagement Ring
A gold 巳n gagem巳nt ring has an inner diameter of 1.5 x 10- 2 m and a t巳 mp巳rature of 27 oc.
The ring falls into a sink of hot water whose te mp巳 rat ure is 49 oc. What is the change in th 巳
diameter of th巳 h o l e in the ring?
Reasoning The hole expands as if it were 白 ll ed with gold , so th巳 change in th巳 diameter is
given by !::.L = αLo!::.T， whereα= 14 X 10 - 6 (C O)- I is the coefficient oflin巳ar expansion for
gold (Table 12.1) , Lo is the ori ginal di am巳ter， and !::.T is the c han g巳 10 temp巳ratur巳.
Solution The change in
th巳 ring 's
diameter is
!::.L= αLo !::.T = [1 4 X 10- 6 (C O)- I] (l .5 X 10- 2 m)(49 oC - 27 0 C) = I4.6 X 10- 6 ml
The previous two examples illustrate that ho les expand lik巳 th e surrounding mat巳rial
when h巳 ated . Therefore , holes in materials with larger coefficients of linear expansio n 巳x ­
pand more than those in materials with smaller coefficients of linear expansion. Conceptual
Example 7 巳xplores this asp巳ct of thermaJ expansio n.
。 Concep叫 Example 7
Ex阳 nding C州的
Figure 12.18 shows a cross-sectional view of three cylind巳rs ， A ， B , and 巳 One is made from lead ,
one from brass , and one fro m stee1. AU three have the same tem p巳ratu re， and they barely fìt in
side each other. As the cy Jj nd ers 缸它 h eated to the sam巳 higher temp巳 rature ， C falls off, whi le A
becomes tightly wedged agai nst B. Which cylinder is made fro m which material ? (a) A is brass,
B is lead , C is steel (b) A is lead , B is brass , C is steel (c) A is lead , B is steel , C is brass (d) A is
brass , B is st叫， C is lead (e) A is steel , B is brass , C is lead (f) A is steel , B is lead , C is brass
(α)
Re asoning We will consider how the outer and inner diameters of each cy lind巳 r change as th巳
temp巳ratu re is raised. In particular, with regard to th巳 mn巳r diameter we note that a hole expands
as if it were fìlled with the surrounding materia l. According to Table 12.1 , lead has the gr巳atest
coefficient of Jj near expansion, followed by brass , and then by steel. Thus , th巳 outer and inner
diam巳ters of the lead cy Jj nder change the most , while those of th巳 ste巳 I cy lind巳r change the leas t.
Answers (a) , (b) , (e) , and (1) are incorrect. Since the s也巳 I cylinder expands the least, it cannot
b巳 th巳 o uter one , for if it were , th巳 greater 巳xpan s i o n of the middle cy Jj nder would prevent the
steel cy lind er 仕om fa lJ ing 0旺" as outer cylinder C actuaUy d o巳s. The ste巳 I cylinder also cannot be
the inner one , because then 由e gr飞eater 巳xpa n si o n of th巳 middle cyLinder would allow th巳 steel
cylind巳r to fall out , contrary to what is observed for inner cy linder A
(b)
Figure 12.18 ConceptuaJ Example 7
discusses the arrangements of the thre巳
cylinders shown in cutaway views in
parts a and b.
Answers (c) and (d) are correct. S in ce 出e steel cylinder cannot be on th巳 o utside or on 出巳 in s ide，
it must be the middle cy Jj nder B. Figure 12.18a shows th巳 lead cy lind巳r as 由巳 o u ter cylinder C. It
wiU fall off as the temperaωre is rai s时 ， since lead expands more than steel. The brass inner cyLinder A expands more than the s te巳I cy lind巳 r that surrounds it and b巳co m es tightly wedged , as observed. Similar reasorung applies aJ so to Figure l 2.18b , which s hows 由巳 brass cy hnd巳r as the o ut巳r
cylinder and 出巳 lead cy Jj nder as the inn巳ron巳， since bo由 brass and l ead 臼 pand more 由 an steel
yl' CHECK
YOUR UNDERSTANDING
(The answers are given at the end of the book.)
Aluminum frame
2. A rod is hung from an aluminum frame , as the drawing
shows. The rod and the frame have the same temperat ure ,
and there is a small gap between the rod and the floor. The
frame and rod are th en heated uniformly. Will the rod ever
touch the floor if the rod is made from (a) aluminum ,
(b) lead , (c) brass?
3. A simple pendulum is made using a long , thin meta l
wire. When the temperature drops , does th e period of the
pendulum increase , dec rease , or remain the sa me?
←一一-Smal l
gap
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it follows that a hole in a piece of solid material expands when heated and contracts
when cooled, just as 扩 it were filled with the material that surrounds it. If the hole is cirw
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c ular, the equation ð. L = αLo ð. Tca n b巳 us巳d to find the change in any linear dimensio n of the hol 巳， such as its radius or diameter. Examp l 巳 6 illustrates this type of linear expansion .
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Problem-solving insig ht
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crete
(concrete that is reinforced with embedded steel rods). Table 12.1 shows that the coefficient of linear expansionαfor concrete is the same as that for stee l. Why is it important
that these two coefficients be the same?
5. One type of cooking pot is made from stainless steel and has a copper coating over the
outside of the bottom. At room temperature the bottom of this pot is flat , but when heated
the bottom is not flat. When the bottom of this pot is heated , is it bowed outward or inward?
6. A metal ball has a diameter that is slightly greaterthan the diameter of a hole that has
been cut into a metal plate. The coefficient of linear expansion for the metal from which
the ball is made is greaterthan that for the metal of the plate. Which one or more of the
following procedures can be used to make the ball pass through the hole? (a) Raise the
temperatures of the ball and the plate by the same amount. (b) Lower the temperatures
of the ball and the plate by the same amount. (c) Heat the ball and cool the plate
(d) Cool the ball and heat the plate .
在 A hole is cut through an aluminum plate. A brass ball has a diameter that is slightly
smallerthan the diameter of the hole . The plate and the ball have the same temperature at
all times. Should the plate and ball both be heated or both be cooled to preventthe ball
from falling through the hole?
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4. For added strength , many highways and buildings are constructed with reinforced con-
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VOlUME THERMAl EXPANSION
Th巳 volum巳 of a normal material increases as the temp巳rature incr巳 ases. Most
solids and liquids behave in this fas hio n. By analogy with lin巳 ar thermal expansion , the
change in volume !1 V is proportional to th巳 change in temp巳rature !1 T and to the initial volum巳 Vo ， provided th巳 change in temperatur巳 is not too larg巳 Th ese two proportionalities
can b巳 conv巳rted into Equation 12.3 with the aid of a proportionality constant ß, known as
th巳 coefficient ofvolume expansion. The algebraic form of this 巳quation is similar to that
fo r linear expansion , !1 L = αLo !1 T
VOLUME THERMAL EXPANSION
The volum巳 Vo ofan obj 巳ctchang巳s by an amount !1 V when its temperature changes by
an amount !1 T:
!1 V = ßVo!1 T
where 卢 is
the
co巳fficient
(12.3)
of volume expansion .
Coolant reservoi r
O
Common Unit for the Coefficient of 协lume Expansion: (C )一 l
The unit for ß, like that forα， is (c o )- I. Values for ß depend on the natur巳 of the
material , and Table 12.1 lists some examples measured near 20 。巳 Th 巳 values of ß for
Ji quids are substantially larger than those for solids , becaus巳Ii quids typica lly expand
more than solids , giv巳 n the same initial volumes and temperature changes. Table 12.1
also shows that , for most solids , the co巳fficient of volum巳巳 xpansion is thre巳 tim巳 s as
much as th巳 co巳伍 ci 巳nt of lin 巳 ar expans io n: 卢 =3α.
If a cavity exists within a solid object , the volume of the cavity increases wh巳n the object expands , just as if the cavity were filled with the surrounding material. Th巳 expans lO n
of the cavity is analogous to the expansion of a hole in a s h e巳t of m at巳rial. Accordingly,
the chang巳 in volum巳 of a cavity can be found using the r巳lation !1 V = ßVo!1 T , wh巳re ß is
the coef自 cient of volume 巳xpa nsion of th巳 mat巳rial that surrounds th巳 c a vity. Example 8
illustrates this poin t.
Figure 12.19 An automobile radiator
and a coolant reservoir for catching the
overflow fro111 the radi atoI二
。 …… 川ut叫e 叫主tor
A small plastic contain巳 r， called the coolant 1 巳s巳rvoir， catches the radiator f1 uid that overflows
whe n an a utomobi l 巳 en gi n e becomes hot (see Figure 12. 19). The radiator is made of copper,
and th巳 coolant has a coefficient of volume ex pansion of ß = 4 .10 X 10- 4 (C O )- I. If the radiator is fìlled to its 15-quart capacity wh巳 n the e n gi n巳 i s cold (6.0 oc) , how much overflo w wiU
spill into the r巳 servoir when th 巳 coolant r巳 ac h es its operating temperature of 92 OC?
The physics of
the overflow of an automobile
radiato r.
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Solution When the temperature increases by 86 C O, the coolant expands by an amount
ð. V = ßVoð. T = [4.10 x 10- 4 (C ) 一 1] (1 5 quarts)(86 C O) = 0.53 quarts
(12.3)
O
The radiator cavity expands as if it w巳re fìlled with copper [β = 51 X 10- 6 (CO) - I; se巳 Tabl e
12.1] . Th巳巳x pan s ion of the radiator cavity is
ð. V = ßVoð. T = [51 X 10- 6 (C O)- I] ( 15 qu缸 ts)(86 CO) = 0.066 quarts
The 0阳日 ow volume is 0.53 quarts - 0.066 quarts = I0 .46q川s l
Although most substances expand
wh巳nh巳ated ，
a few do no t. For instance , if water at
o oC is heated , its volume decreases until the temperature reaches 4 oc. Abov巳 4 oC wa-
The physics of
ice formation and the
survival 0' aquatic life.
ter behaves normally, and its volume increases as the temperature increases. Because a
given mass of water has a minimum volume at 4 0 C , the d巳 n sity (mass per unit volume) of
water is greatest at 4 oC , as Figure 12.20 shows .
The fact that water has its greatest density at 4 oC , rath巳r than at 0 oC , has important
consequences for the way in which a lak巳 fr巳ez巳s . When the air t巳mperature drops , the surface layer of water is chilled. As the temperature of the surface layer drops toward 4 oC , this
layer b巳comes more d巳nse than th巳 warmer water b巳low. The denser water sinks and pushes
up the deeper and warm巳r water, which in turn is chilled at the s urfac巳 . This proc巳ss continues until the t巳mperature of the entire lake reaches 4 o Further cooling of the s urfac巳
water below 4 oC makes it l ess dense than the d巳eper layers; consequently, the surface layer
does not sink but stays on top. Continu巳d cooling of the top layer to 0 oC leads to the formation of ice that fioats on th巳 water， because ice has a small巳rd巳 nsity than water at any
temperature. Below the ice , h owev巳r， the water temperature remains above 0 o Th 巳 s h巳 et
of ice acts as an insulator that reduc巳 s the loss of heat from 由巳 lake ， esp巳 cially if th巳 ice is
cov巳red with a blanket of snow , which is also an insulator. As a result , lakes usually do not
freeze solid , even during prolonged cold spells , so fì sh and other aquatic life can survive.
The physics 0' bursting water pipes. Th巳 fact that the density of ic巳 is smaller than the
d巳nsity of water has an important consequenc巳 for homeown巳rs ， who have to contend with
由e possibility of bursting wat巳r plp巳s during severe winters. Water often fre巳zes in a section
of pipe exposed to unusually cold temperatures. The ice can form an immovable plug that
prevents th巳 subsequent fiow of wat巳r， as Figure 12.21 illustrates. wh巳n water (larger density) turns to ice (smaller d巳nsity) ， its volume expands by 8.3%. Th巳refore ， when more wat巳r fre巳zes at th巳 left sid巳 of 出e plug , the 巳xpanding ice pushes liquid back into the pipe
leading to 由巳 stre巳t connection , and no damage is don巳 How巳ver， when ice forms on 出e
right side of 由e plug , the 巳xpanding ice pushes liquid to th巳 right-but it has nowhere to go
if the faucet is closed. As ice continues to form and expand ， 出e water pres s ur巳 b巳tween th巳
plug and faucet rises
c.
c.
1000.0
m
E
bEEO
的 ha
QdQJQJ
QJau7'
nyQJnwd
nynuJqd
999.6
4
8
10
Temperature , oC
Figure 12.20 The density of water in
the temp巳rature range from 0 to 10 o
At 4 oC water has a maximum density
of 999.973 kg/m3 . (This value is
equivalent to the often-quoted density
of 1.000 00 grams per milliliter.)
O
2
c.
~ CHECK Y 。
ωUR 川
U川
ND
、唱aωDERS τ ANDI川川
N唱
Closed
pressure
pressure
faucet
Figure 12.21 As water freezes and
expands , enormous pressure is applied
to the liquid water between the ice and
the closed fauce t.
(The answers are given at the end of the book.)
8_ Suppose th at liquid mercury and glass both had the same coefficient of volume expan-
sion. Would a mercury-in-glass thermometer still work?
9. Is the buoyant force provided by warmer water (above 4 OC) greater than , less than , or
equal to the buoyant force provided by cooler water (also above 4 C)?
0
o
m
lic
C
m
o
The
way. c in which the level 01 a liquid in a
c u -tr a c k
container changes with temperature depends
on the change in volume 01 both the liquid
and the containe r.
w
w
.d o
Reasoning When the temperature increas町， both th巳 coolant and th 巳 radiator expand. If th 巳y
w
were to expand by the same amount, there would be no overflow. How巳ver， the liquid coolant
.c
.d o
c u -tr a c k
expands more than th巳 radiator， and the overflow volume is the amount of coolant expansion
minus the amount of the radiator cavity expansion
w
w
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k
Problem-solving insight
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TEMPERATURE AND HEAT
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An object with a high temperature is said to be hot, and th巳 word "hot" brings to
mind the word "heat." Heat f1 0ws from a hotter object to a cooler 0时 ect wh 巳 n the two are
placed in contac t. It is for this reason that a cup of hot coffee feels hot to the touch , while
a glass of ic巳 water fe巳 Is cold. When th巳 p巳rson in Figure 12.22a touches the coffee cup ,
heat f1 0ws from the hotter cup into th巳 cooler hand. When th巳 person touches the glass in
part b of the drawing , heat again f1 0ws from hot to cold , in this case from th巳 warmer hand
into th巳 colder glass. Th巳 r巳spons巳 of the n巳rves in th巳 hand to th巳 arrival or departure of
heat prompts the brain to identify the coffe巳 cup as being hot and th巳 glass as being cold.
What exactly is heat? As the following defìnition indicat巳 s ， h 巳at is a form of en巳rgy，
energy in transit from hot to cold.
.d o
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m
C
m
HEAT AND INTERNAL ENERGV
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12.7 HEAT AND TEMPERATURE CHANGE : SPECIFIC HEAT CAPACITY
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(α)
DEFINITION OF HEAT
Heat is energy that f1 0ws from a higher-temperature
obj 巳ct b巳cause of th巳 differenc巳 in temperatures.
obj 巳ct
to a lower-temperature
SI Unit of Heat: joule (J)
Being a kind of energy, heat is measured in the same units used for work , kin巳 tic energy,
and potential energy. Thus , the SI unit for h巳 at is th巳 joule.
The heat that f1 0ws from hot to cold in Figure 12.22 originates in the internal energy
ofth巳 hot substanc巳 . The internal energy of a substanc巳 is th巳 sum of the molecular kin 巳 tic
energy (due to random motion of th巳 molecules) ， the molecular pot巳 ntial energy (due to
forces that act betw巳巳n the atoms of a molecul 巳 and between molecules) , and other kinds
ofmol巳cular energy. When heat f1 0ws in circumstances wh巳r巳 no work is done , the internal en巳rgy of th巳 hot substanc巳 decreases and th巳 internal energy of the cold substance increas巳 s. Althollgh h巳 at may O!毡 inate in the internal energy supply of a sllbstance , it is not
correct to say that a substance contains heat. The substance has internal 巳 nergy， not hea t.
The word "heat" only refers to the 巳 nergy actually in transit from hot to cold.
The next two sections consider som巴巴ffl巳cts of hea t. For instance , when preparing
spaghetti , the fìrst thing that a cook does is to heat the water. Heat from the stove callses the
internal energy of the wat巳 r to incr巳 ase . Associat巳d with this increase is a rise in t巳mpera­
ture. Aft巳r a while , the t巳mperature reach巳 s 100 oC , and th巳 water b巳gins to boil. During
boiling , th巳 add巳d heat causes the water to change from a liquid to a vapor phase (steam)
Th巳 n巳xt section inv巳 stigates how the addition (or removal) of h巳 at causes th巳 temperature
of a substance to chang巳. Then , Section 12.8 discusses th巳 I 巳lationship betw巳en h巳 at and
phas巳 chang巳， such as that which occurs wh巳 n water boils.
HEAT AND TEMPERATURE CHANGE: SPECIFIC HEAT CAPACITV
SOllOS ANO lI QUIDS
Greater amounts of heat ar巳 needed to raise th巳 temperature of solids or liqllids to
higher values. A great巳r amount of heat is also required to rais巳 th巳 temp巳 ratur巳 of a
greater mass of mat巳 rial. Similar comments apply when th巳 t巳 mperature is low巳 red ，巳xcept
that h巳 at must b巳 r巳 mov巳d. For limited temperature rang巳 s ， experiment shows that the heat
Q is directly proportional to the chang巳 in t巳 mperature I1 T and to th巳 mass m. These two
proportionalities are expressed below in Eqllation 12 .4, with the help of a proportionality
constant c that is referred to as the specifìc heat capacity of the m at巳 ria l.
HEAT SUPP Ll ED OR REMOVED IN CHANGING THE TEMPERATURE OF A SUBSTANCE
The heat Q that must be supplied or removed to chang巳
of mass m by an amount I1 T is
Q = cm l1 T
where c is the
sp巳cific
heat capacity of the
substanc巳­
Common Unit for Specifìc Heat Capacity: J/(kg. C O )
the temperatur巳 of
a
s llbstanc巳
(12 .4)
(b)
Figure 12.22 Heat is energy in transit
from hot to cold. (α) Heat f1 0ws from
the hotter coffe巳 cup to the colder hand.
(b) Heat f1 0ws from the warmer hand to
the colder glass of ice wate r.
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TEMPERATURE AND HE AT
y
bu
Capacity, c
J/(kg' C O )
S lI b sta n c巳
Solids
AI lI lllinlllll
Coppel
Gl ass
H lI lllan body
(37 oC , average)
Ice (- 15 oc)
lron
0 1'
steel
Lead
Sil ver
Liquids
Benzene
Ethyl alcoho l
Glycerin
Mercury
Water ( 15 oc)
9.00 X 10 2
387
840
to
k
O Exam回e 9
A Hot Jogger
In a ha lf hour, a 65-kg jogger can g巳nerat巳 8.0 x 10s J of hea t. This h巳at is removed frolll the
j ogger's body by a variety of Ill eans , inclllding th 巳 body 's own telllp巳 ratllre- regul at in g Ill echani sllls. If the h巳at were not rellloved , how Ill llch wOllld the j ogger's body t巳Ill peratllre incl 巳 as巳?
2.00 X 10 3
Reasoning The increase in body telll peratllre dep巳 nd s on th 巳 a mount of heat Q generated by
the jo gg町， her Ill ass m , and the s p巳 c ifì c heat capacity c of the hllman body. Sinc巳 nlllll e ri ca l valU 巳 s are known for these thr巳:e variables , we can deterllline the potenti al ris巳 in t巳Ill perature by
using Eg llation 12 .4.
452
128
Solution Tab le 12.2 gives the av巳 rag巳 s p巳c ifìc heat capacity of the human body as
3500 J/(kg' C O ) . With thi s va llle, Egll ati on 12 .4 shows that
3500
235
Q
cm
1740
2450
2410
139
4186
"Except as noted, the values are for 25 oc
and 1 alm of press Ul 巳
8.0 x 10 s J
[3500J/(kg'C )](65kg)
厅7士~
ð. T = 一一-------- = 13.5 C 门
O
'-一一一」
An increase in body telllperatllre of 3.5 oC cOllld b巳 life-threatening. 0ne way in which the
it fro lll occllrring is to relllov巳 excess h 巳 at by perspiring. ln contras t,
cats, sllch as the on 巳 in F igllre 12.23 , do not perspire bllt often pant to relllove 巳xcess hea t.
j ogge l、 body pr 巳V巳 nts
。
。 Exam帅 10 Taking a Hot Shower
Co ld water at a telllperatllre of 15 oC enters a h 巳 ate r， and the 1 巳 s lll ting hot water has a telllpe ratllre of 6 1 o A person lI ses 120 kg of hot wat巳 r in taking a shower. (a) F ind th巳巳n e rgy n 巳ed ed
to h巳 a t th巳 wat巳 r. (b) Ass urIlÍ ng that the lI tility cOlllpany charges $0.10 per kilowatt . hOllr for
e lectrica l energy, determin 巳 th 巳 cos t of h巳atin g the wate r.
c.
Reasoning The amollnt Q of heat needed to raise th 巳 wa te r te mperatllre can b巳 fO llnd fro lll
the relation Q = cm ð. T , since the spec ifi c heat capacity , Ill ass , and temperatur巳 c h a n g巳 of
the water are know n. To d e terlllin 巳 th 巳 cost of thi s 巳 n 巳 rgy ， we Ill llltipl y the cost per lI nit of
e nergy ($0.10 per kilowatt . hOllr) by th 巳 a mollnt of e ne rgy lI sed , ex pressed in energy lI ni ts
of kilowatt. hours .
Solution (a) The alllollnt of heat needed to
h巳at th 巳 wa ter
IS
Q = cm ð. T = [41 86 J/(kg ' C )]( 120 kg)(6 1 oC - 15 oc) = I 2.3 X 10 7 J
O
(1 2 .4 )
(b) The kilowatt. hOllr (kW h) is th巳 lI nit ofe n 巳 rgy that lI ti lity cOlllpanies u s巳 i n YOllr electric bi ll.
To ca l c lll at巳 th e cost , w巳 n eed to deterllljne the nlllllb巳 r of j Ollles in one ki lowatt . hOll r. Recall
that 1 kilowatt is 1000 watts (1 Kw = 1000 W) , 1 watt is 1 jOllle per second (l W = 1 J/s; s巳E
Table 6.3), and 1 hour i s 巳gll al to 3600 seconds (1 h = 3600 s). ThllS,
I 1000 W \ I 1 J/s \ I 36心O 叭
1 kWh = (l kWh) 1 一一一一 11 一一一 11 一一一=- 1 = 3.60
飞 1 kW / \ 1 W } \ 1 h
}
The nlllllber of kilowatt . hours of energy lI sed to heat
Figure 12.23 Cats , such as thjs Bengal
tl g巳 1'， often pant to g巳t rid of 巳xce ss
hea t. (PhotoDisc Blue/PhotoDisc , Tnc.l
Gelty Images)
(2.3 X 10' J) 1
k认/ h
飞 3.60 X
At a cost of $0.10
κ
th 巳 water
1 =
10 õ J }
x
，
10 õ J
is
6 .4 kWh
P川Wh， the bi 灿川巳川 1 $0.641 or 64 cen
Example 1 1 Heating a Swimming Pool
Figllre 12.24 shows a swimming pool o n a sunny day. If the water absorbs 2.00 X 109 J of heat from the slln , what is the
cha nge in the volum巳 of the wat巳 r?
。
m
o
m
o
Sp巳cifìc H巳 at
lic
k
lic
C
w
c u -tr a
Solving Eq uatio n 12 .4 for the specifìc h巳 at capacity shows that c = Q/(m ð. T) , so the
unit for specifìc heat capacity is J/(kg ' C O ). Table 12.2 r巳 veal s that the value of the specifìw .cd o
.c
c u -tr a c k
heat capacity d巳p巳nds on the n atUl毡 of the materi a l. Examples 9 , 10, and 11 illustrate the
u s巳 of Eq uation 12 .4.
w
Table 12.2 Specific Heat Capac ities a
Solids and Liquids
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. d o of Some
ck
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HEAT AND TEMPERATURE CHANGE: SPECIFIC HEAT CAPACITY F 371
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Description
H巳at absorb巳d
Value
Q
2.00 X 109 J
D. V
?
vo lum巳 ofwater
m
Figure 12.24 When water
absorbs heat Q from the sun , th巳
water expands. (David Wa]]J
Danita Delimont)
Unknown Variable
Change in
c u -tr a c k
probl巳mar巳·
Symbol
by water
.d o
o
m
w
c u -tr a c k
Knowns and Unknowns The data for this
C
Reasoning
As the water heats up , its volum巳 increases .
.c
According to th巳 relation D. V = 卢 Vo D. T (Equation 12.3) ,
th巳 chang巳D. V in volume dep巳nds on the change D. T in
temp巳rature. The change in temperatur巳， in turn , depeods
00 出巳 amouot of heat Q absorbed by the water and 00 th巳
mass m of water b巳ing heat巳d ， sinc巳 Q = cm D. T
(Equatioo 12 .4). To 巳valuate the mass , we r巳cognize that
it depends on the density p and th巳 initial volume Vo,
SIDC巳 ρ = m/Vo (Equation 11.1). Thes巳 three relations will
be used to deterrnin巳 th巳 chang巳 in vol ume of the wat旺
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12.7
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Modeling the Problem
E圃 阳
Voolum
um
田
囚me川叫巾
The
阳阳
E忧e町
川
lern
e町口
rrm
mal
an am
丑loun
川
lt D. T , th巳 volum巳 ofth巳 water changes by ao amouot D. V, as given by Equation
12.3: D. V = ßVo D. T, wh巳r巳 Vo is the initial volume and ß is th巳 co巳fficient of volume
expansion for water. Both Vo and D. T ar巳 unknown ， how巳V巳r， so w巳 wiUd巳 al with D. T in
Step 2 and Vo in Step 3.
=H E
(1 2.3)
E圃 H
阳阔创……
阳山叫
taa缸
削
m
nd
咀dCαωhan
吨叩
a缸n吨叫呻
E咆灿川
ge
耶盼内
ein
i叮
TI 创
Tempe
era
by an amount D.T. The relation betw巴巴n Q and D. T is giv巳n
by Equation 12 .4 as Q = cm D. T, wh巳re c is th巳 specifìc heat capacity and m is th巳 mass of
th巳 water. Solving this 巳quation for D. T yields
t由
h巳 water 优
t emp巳ratur巳 chang巳s
to the mass density p by Equation 1 1. 1 as p = m/Vo. Solviog this
mass yields
巳quation
for the
E三日
for m can be substituted into Equation 1, as indicated in the right column.
Solution Algebraic aU y combining the
V
二
，
nn
(
r 巳sults
of the
Q川 ，
nn
thr巳巳 steps ，
I
Q
I
D. V 二 ßVo D. T 二 ßVo \ C:l 尸 ßVo l 石可J
we h
ßQ
~;
=
Not巳 that
the initial volum巳 Vo of the water is eliminated algebraically, so it does not
expression for D. V. Taking valu巳s of β= 207 X 10- 6 (C O ) 一 I from
Table 12.1 , c = 4186 J/(kg. C O) from T￥ble 12.2 , and p 二1. 000 X 103 kg/m 3 from
Table 1 1.1, w巳 have that
βQ
[207 X 10- 6 (C O)- I](2.00 X 109 1)
In
， n_O τ l
D. V=.!:一
【
~= I 9.89 x 1O - ~ m J I
cp
[4186 J/(kg.CO)]( l. OOO X 10 3 kg/m3)
I ._-_
appearin 出巳 final
M
"
o
L..
Related Homework: Problems 53, 55
J
th巳 h巳 at
E回 M制 Dens均川…川川叫 of ……elated
expr巳ssion
(12.3)
、l
which can be substituted into Equation 12.3 , as shown at th巳 righ t. We know
Q absorbed by the water, and we will consider the mass m in th巳 n巳xt step.
This
D. V = 卢 Vo D. T
=去|
唱EA
D. T
(1 2.3)
rs-、、
|
ßVo D. T
D. V =
_ _ _- - - - '
( -A)
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As we wiJI see in Section 15.6 , th巳 valu 巳 of th巳 specific heat capacity depends ono c u - t r a c k
whether the pressure or volum巳 is held constant while energy in the form of heat is add巳d
to or removed from a substance. The distinction betwe巳n constant pressure and constant
volum巳 is usually not important for solids and liquids but is significant for gases. As we
wi lI see in Section 15.6 , a greater value for the specific heat capacity is obtained for a gas
at constant pressure than for a gas at constant volume.
HEAT UNITS OTHER THAN THE JOUlE
There are three heat units oth巳r than the joule in common us巳. One kilocalorie
(1 kcal) was defin巳d historicalI y as the amount of heat needed to raise the temperature of
one kilogram of water by one Celsius degree.* With Q = 1. 00 kcal , m = 1. 00 kg , and
6. T = 1.00 C O , the equation Q = cm 6. T shows that such a definition is 巳quivalent to a specific heat capacity for water of c = 1. 00 kcal/(kg . C O ) . Similarly, on巳 calorie (1 cal) was
defined as the amount of h巳 at ne巳ded to raise th巳 temperatur巳 of one gram of water by on巳
O
C巳 Isius degre巳 ， which yields a value of c = 1. 00 cal/(g' C ). (Nutritionists use th巳 word
"Calori巳，" with a capital C , to specify th巳巳 nergy content of foods; this use is unfortunate ,
since 1 Calorie = 1000 calories = 1 kca l.) The British thermal unit (Btu) is the oth巳rcom­
monly used h巳at unit and was defined historically as the amount of heat ne巳ded to rais巳 th巳
temperature of on巳 pound of water by one Fahrenheit degree.
It was not until the time of James Joule (1 818-1889) that the relationship betw巳巳n
energy in the form of work (in units of joules) and 巳n巳rgy in the form of heat (in units of
kilocalories) was firmly 巳stablished . Joule ' s 巳xperim巳nts reveal巳d that the performance
of mechanical work, like rubbing your hands together, can make the temp巳rature of a sub
stance rise , just as the absorption of h巳 at can. His 巳xp巳nm巳 nts and those of later workers
have shown that
1 kcal = 4186 joules
or
1 cal = 4.186 joul巳s
Because of its historical significance , this conversion factor is known as the mechanical
equivalent 01 heat.
CAlORIMETRV
川勺Fj
Calorimeter
cup
Insulating
container
Unknown
material
Figure 12.25 A calorimeter can be used
to measure 由e specific heat capacity of
an unknown material.
1n Section 6.8 we encountered the principle of conservation of energy, which states
that energy can be neither cr巳 ated nor destroyed , but can only be converted from one form
to another. There w巳 dealt with kinetic and pot巳 ntial energies. 1n this chapter we hav巳 ex­
panded Ollr conc巳pt of en巳rgy to incllld巳 h巳 at ， which is en巳rgy that flows from a highertemperature object to a lower-temperature object because of the difference in t巳 mperatur巳­
No matter what its form , whether kinetic en巳 rgy， potential energy, or heat , energy can b巳
neither created nor destroyed. This fact governs th巳 way objects at different temperatures
come to an equilibrium t巳 mperature when they are plac巳d in contac t. 1f ther巳 is no h巳 at loss
to the 巳xternal surroundings , the heat lost by the hotter objects equals the heat gained by
the cooler on巳 s ， a process that is consistent with the conservation of en巳rgy. J llSt this kind
of process occurs within a thermos. A p巳rfect th巳rmos would prevent any heat from leakt h巳 foαrm
ing Ol1 t or in. How 巳V巳町r， 巳 n巳r鸣gy in 由
t由
h巳町rmos tωot由
h巳 巳xt臼巳 川
nlt 由
t ha
创tt由
h巳叮
y have dωif征
fe
臼ren
川
lt 优
t emp巳町ra
创tures; for example , between ice cubes
and warm tea. Th巳 transfer of 巳n巳rgy continues l1 ntil a common temp巳rature is reached at
thermal equilibrillm
The kind ofh巳at transfer that occurs within a thermos of iced tea also occurs within a
calorimeter, which is the experimental apparat l1 s used in a technique known as calorimetry. Figure 12.25 shows that, like a thermos , a calorimeter is essentially an insulat巳d
contain 巳r. 1t can be used to determine th巳 specific heat capacity of a substance , as the next
example illustrates.
"From 14.5
to
15.5 oc
o
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GASES
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CHAPTER 12 TEMPERATURE AND HEAT
O
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er
372
PD
h a n g e Vi
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PD
F-
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h a n g e Vi
ew
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ew
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y
to
bu
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bu
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Th巳 ca
刻l o rime
创te
缸r
cup in Figure 口
1 2.2 5 βi s m ad巳 fro m 0 .15 kg of a luminum and co nta ins 0.20 kg
of wate r. Initi ally, the wate r and the cup have a common te mperature of 1 8. 0 。巳 A 0 .040-kg
mass of unknown mate ri al is heated to a temperature of 97.0 oC and then added to th巳 w ater.
The temperature of the wate r, the cup , and the unknown mate rial is 22.0 oC after the rmal equilibrium is reestablished . Ig noring the sm all amount of heat ga in 巳d by the th e rmo m et町， find th巳
specific heat capacity of the unknown materia l.
.d o
m
o
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o
c u -tr a c k
C
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υ Exa
副曹叫冒
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373
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F-
HEAT AND PHASE CHANGE : LATENT HEAT
N
12.8
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c u -tr a c k
Reasoning Since energy is conserved and there is negligible heat flow between th巳 calorim ete r
and the outside surroundings , the heat gained by the cold water and the aluminum cup as 出 ey
warm up is equ al to the heat lost by the unknown m ateri al as it cools down. Each quantity of heat
can be calculated using the re lation Q = cm ð. T, where , in calorimetry, th巳 ch an ge in te mperature ð. T is always th巳 hi gher te mperature minus the lower temp巳 rature . The equ ation "H eat
gained = Heat lost" co ntai ns a single unknown qu antity, the desired specific heat capac ity.
Problem-solving insight
Solution
m- -3 -au
1l
t
l
-7
at
町
町
EanH
，
+
、
t
cAlmAI ð. T A1
n川
Cunknown
一已。
、
Heat gained by
aluminum and water
H
W
-创
(cm ð. T )water = (cm ð. T)unknown
-
+
一HM
(cm ð. T )AI
Cwalermwnter ð. Twnter
"""",..
un Kn o飞.V Il
ð.
_ T.
... ...l.
U Il Kno、v n
The changes in temperature for the th ree substances are ð. TA1 = ð. Twa,er = 22.0 oC - 18.0 0 C
= 4 .0 C O, and ð. Tunknown = 97 .0 oC - 22.0 oC = 75.0 C O. Table 12.2 contains values fo r the
specific heat capacities of a luminum and wate r. Substitutin g these data in to the equ ati o n above ,
we find that
[9.00 X 10 2 J/( kg ' CO)](0. 15 kg)(4 .0 CO) + [4 186 J/(kg' C O)](0.20 kg)(4 .0 C O)
Cunknown
(0.040 kg)(75.0 C O)
=11 300 除g ' Cη
~ CHECK V 。
ωUR 川
U川
ND
‘唱4ωDERSTA 酌NDII
川
N、唱4
(The answers are given at the end of the book.)
10. Two different objects are supplied with equal amounts of heat. Which one or more of
the following statements explain why their temperature changes would not necessarily be
the same? (a) The objects have the same mass but are made from materials that have
different specific heat capacities. (b) The objects are made from the same material but
have different masses. (C) The objects have the same mass and are made from the same
materia l.
11. Two objects are made from the same material but have different masses. The two are
placed in contact, and neither one loses any heat to the environmen t. Which object ex periences the temperature change wit h the greater magnitude , or does each object experience
a temperature change of the same magnitude?
ó. T( CO)
m(kg)
12. Consider an object of mass m that ex periences a
(a)
2.0
15
change ó. Tin its temperature. Various possibilities for
these variables are listed in the table . Rank these
(b)
1.5
40
possibilities in descending order (l argest first) according
(C)
3.0
25
to how much heat is needed to bring about the change in
(d)
2.5
20
temperature.
一-g lf
HEAT ANB PHASE CHANGE:LATENT
HE盯
Surprisingly, th ere 缸'e situatio ns in which the additio n or rem oval of heat does no t
te mperature change. C o ns ider a well-stined g lass of iced tea that has come to the rmal
equilibrium. E ven thoug h heat e nters the glass 仕om the warmer roo m , the t巳 mp巳rature of the
tea does not rise above 0 oC as lo ng as ice c ubes are pres巳n t. Apparently the heat is be ing used
for some purpose other than raising the te mperature. In fac t , the heat is be ing used to me lt the
ice , and only when all of it is me lted will the temperature of th巳 )j quid b巳gin to rise.
cau s巳 a
In the equation "Heat gained = Heat lost ,"
both sides must have the same algebraic sign.
Therefore , when calculating heat contributions ,
always write any temperature changes as the
higher minus the lower temperature.
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to
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bu
An important point illustrat巳d by the iced tea example is that ther巳 is more than one
w
.c
.d o
matter. For instanc巴 ， som巳 of the water in the glass is in the solid phas巳
c u -tr a c k
(ice) and some is in the liquid phas巳. Th巳 gas or vapor phas巳 is th巳 third familiar phase of
matter. In th巳 gas phase , water is referred to as water vapor or steam. All thre巳 phase s of
water ar巳 present in the scene d巳plct巳d in Figure 12.26 , although th巳 water vapor is not vi sible in the photograph.
Matter can chang巳 from one phase to anoth巳r， and heat plays a role in th巳 change .
Figure 12.27 summarizes the various possibilities. A solid can melt orfuse into a liquid if
heat is added , while the liquid canfreeze into a solid if heat is removed. Similarly, a liquid can evaporate into a gas if heat is supplied , while the gas can condense into a liquid
if heat is tak巳n away. Rapid 巳vaporation ， with the formation of vapor bubbles within the
liquid , is call巳d boiling. Finally, a solid can sometimes change directly into a gas if heat is
provided. We say that th巳 solid sublimes into a gas. Examples of sublimation ar巳(1) solid
carbon dioxid巳 ， CO 2 (dry ic时 ， tuming into gaseous CO 2 and (2) solid naphthalen 巳 (moth
balls) turning into naphthal 巳 ne fumes. Convers巳ly， if heat is remov巳d under the right con
ditions , the gas will condense directly into a solid.
Figure 12.28 displays a graph that indicates what typically happens when heat is
added to a material that changes phase. The graph records temperature v巳rsus heat added
and refers to wat巳r at the normal atmospheric pr巳ssur巳 of 1. 01 x 105 Pa. The water starts
off as ice at th巳 subfreezing temp巳rature of - 30 o As heat is add巳d ， th巳 temp巳rature of
O
th巳 ice increases , in accord with the speci 自 c heat capacity of ice [2000 J/(kg' C ) ] . Not until the temperature r巳ach巳s th巳 normal melting/freezing point of 0 oC does th 巳 water begin
to change phase. Then , when heat is added , th 巳 solid changes into the liquid , the temp巳ra­
ture staying at 0 oC until all the ice has melted. Once all th巳 material is in the liquid phas巳，
additional heat causes the t巳 mperature to incr巳 ase again , now in accord with the specifìc
heat capacity of liquid water [4186 J/(kg' C O )]. When the temperature reaches th巳 normal
boiling/condensing point of 100 oC , the water b巳gins to change from th巳 liquid to the gas
phas巳 and continues to do so as long as h巳 at is added. The temperature remains at 100 oC
until all liquid is gone. When all of the material is in the gas phas巳， additional heat Once
again causes the t
Figure 12.26 Th巳 three phases of water:
solid ice is ftoating in liquid wat巳r， and
water vapor (invisibl巳) is present in the
aír. (K.I ein/Peter Arnold , Inc.)
c.
。些些哩芒些mple 13 Saving Energy
哩'型=­
Figure 12.27 Thre巳 familiar phases of
matter-solid , Ii quid , and gas-and
the phase changes that can occur
between any two of them
Suppose you ar百 cooking spaghetti , and the instmctioos say "boil 由巳 pasta in water for ten minlltes."
To cook spagh巳tti in an open pot using 由巳 least amount of en巳rgy， should you (a) tllrn lIP th巳 bl1 rner
ωits fl111巳st so the water vigoro l1 s1y boils or (b) tl1 rn down th巳 burner so th巳 water barely boils?
Reasoning Th巳 spaghetti n巳巳ds to cook at the temperature of boiling water for t巳 n min l1 tes. ln
an op巳n pot the pressure is atmosph巳ric pre s sur毡， and water boils at 100 oC , regardless of
whether it is vigoro l1 s1y boiling or jllst barely boiling. To convert water into steam req l1 ires en
ergy in th巳 form of h巳 at from the bllrn町， and th巳 greater the amO l1 nt of water converted , th巳
greater the amount of energy needed.
Answer (a) is incorrect. Cal1 sing the water to boil vigorously jllst was tes
All it accomplishes is to conv巳rt more water into steam.
巳nergy l1 nn 巳cess arily.
Answer (b) is correct. Keeping the water j l1 st barely boiling l1 S巳 s the 1巳 ast amount of e n巳rgy
to k巳巳p the spaghetti at 100 oC , becal1 s巳 it minimizes the amO l1 nt of water convert巳d into steam
υ100
Water boils
o
Ql
』
=
Water warms up
H
ro
」
αJ
Figure 12.28 The graph shows th巳
way 由e temperatur巳 of water changes
as heat is added , starting with ice at
30 oc. The pressure is atmospheric
pr巳 s sure .
0.
E
Ql
•
- 30
Heat
Water vapor
warms up
m
o
m
o
typ巳 or phas巳 of
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to
k
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c u -tr
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ack
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CHAPTER 12 TEMPERATURE AND HEAT
F-
er
374
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h a n g e Vi
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m
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When a s ub stanc巳 cha n ges fro m o ne phas巳 to another, the amoun t of heat that must b巳
.c
added
or removed depends o n t h巳 ty p巳 ofmat巳 r i a l and th 巳 natur巳 ofth 巳 pha s e chang巳. The
k
c u -tr a c
heat per Iá logram associated with a phase c h ang巳 is referred to as latent heat:
w
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F-
HEAT AND PH AS E CHANGE: LATEN T HEAT 375
er
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HEAT SUPP Ll ED OR REMOVED IN CHANGING THE PHASE OF A SUBSTANCE
The h巳at Q that must be supplied or removed to change the phase of a mass m of a
substance is
Q =mL
where L is the latent heat of th 巳
(1 2.5)
substance
SI Unit of Latent Heat: J/k g
The latent heat of fusion Lr refers to the change between solid and Ii quid phases , the
latent heat ofvaporization Lv applies to the chang巳 between liquid and gas phases , a nd th巳
latent heat of sublimation Ls refers to the chang巳 b巳tween solid and gas phases .
Table 12.3 gives some typical values of latent heats of fusion and vapori zation . For instance , the latent h巳at of fusion for water is Lr = 3.35 X 105 J/kg. Thus , 3.35 X 105 J of
h巳 at must b巳 s upplied to melt on 巳 kilogram of ice at 0 oc into liquid water at 0 oC; convers巳Iy， this amount of heat must be remov巳d from o n 巳Iá l ogra m of liquid water at 0 oc to
freeze the liquid into ic巳 at 0 oC. In co mpaJ怡o n ， the latent heat of vaporization for water
has the much larger value of Lv = 22.6 X 10 5 J/kg. When water boils at 100 oc , 22.6 X
105 J of heat must be supplied for eac h ki logram of liquid turned into steam. And when
steam condenses at 100 oc , this am o u 川 of h巳at is releas巳d fro m 巳ac h Iá logram of steam
that changes back into liquid . Liquid wat巳r at 100 oc is h ot 巳no u gh by itself to cause a bad
burn , and th巳 additional 巳仔巳ct of the large late nt heat can cau s巳 S巳vere tissue dam age if
condensation occurs on th巳 slán .
By taláng advantage of th 巳 latent heat of fusion, d巳signers can now e ngi neer clothing
that can absorb or release heat to help maintain a comfortable and approx imately constant
temperature close to your body. As the photograph in F i g ur巳 12.29 shows , the fabric in this
typ巳 of clothing is coated with microscopic balls of heat-resistant plastic that contain a
substance known as a "phase-change material" (PCM). When yo u are e njoyi ng your favorite winter sport, for example , it is easy to beco m巳 overh巳 ated. The PCM prevents this
by melting , absorbing excess body heat in the process. When you are taking a break and
cooling down , however, th巳 PCM fr巳巳Z巳 s and releases heat to keep you w aJ.m. The temperature range over wruch the PCM can maintain a comfort zo n 巳 i s re lated to its
melting/freezing t巳mperature ， which is determin 巳d by its ch巳 mical com posi tio n.
Examples 14 and 15 illustrate how to take into account the 巳ff，巳ct of latent h巳 at when
usi ng th巳 cons巳 rvation-of-energy principle.
冒
The physics of
~
steam burns.
The physics of
high-tech clothing.
Table 12.3 Latent Heats. of Fus ion a nd Va porization
Sub s tanc巳
Melting Point
(OC)
Ammonia
一77. 8
Be n z巳n 巳
5.5
1083
Copper
Ethy l alcohol
Gold
Lead
Mercury
Nitrog巳n
Oxyg巳 n
Water
一 114 .4
1063
327.3
-38.9
-210.0
-218.8
0.0
"The vallles pertain to I atm pressure.
Latent Heat
of Fusion, Lr
(J /kg)
Boiling Point
(OC)
Latent Heat of
Vaporization , Lv
(J/kg)
33.2 X 10 4
12.6 X 10 4
20.7 X 10 4
10.8 X 10 4
6.28 X 10 4
2.32 X 10 4
1. 14 X 10 4
2.57 X 10 4
1. 39 X 10 4
33.5 X 104
-33 .4
80.1
2566
78.3
2808
1750
356.6
- 195.8
-1 83.0
100.0
13.7 X 10 5
3.94 X 10 5
47.3 X 10 5
8.55 X 10 5
17.2 X 10 5
8.59 X 10 5
2.96 X 10 5
2.00 X 10 5
2.1 3 X 10 5
22.6 X 105
Figure 12.29 Th.i s highly magn.i fied
image shows a fabric that has been
coated with microscopic balls of
heat-resistant plastic. The balls contain
a substance known as a "phase-change
Ill aterial," the Ill elting and freezing of
which absorbs and releases hea t. Such
fabrics autolllatically adjust in reaction
to your body heat to help Ill aintain a
constant temperature next to your ski n
(Collrtesy Olltlast Technologies ,
Boulder, CO)
.c
TEM PERATU RE AN 0 H EAT
O
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c u -tr
y
c
k
lic
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0 oC is pJaced in a Styrofoam cup containing 0.32 kg of lemonade at 27 oc. Th 巳 specific
heat capacity of lemonade is virtually the same 也s that of wat巳r; that is , c = 4186 J/(kg . C O ).
After the ice and lemonade reach an equilibrium temperatur毡 ， som巳 ice still remains . Th巳 latent
heat of fusion for wat巳r is Lr = 3.35 X 10 5 J/kg . Assum巳 that the mass of the cup is so small
that it absorbs a negligible amount of heat , and ignore any heat lost to the surroundings
Det巳rmine the mass of ice that has melted .
Ic巳 at
m
o
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ack
C
w
o
m
to
bu
y
bu
to
k
lic
C
。 Exam回e 14 Ice-Cold Lemonade
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CHAPTER
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c u -tr a c k
Reasoning According to the principle of 巳 n 巳rgy conservation , the h巳at gained by the melting
ice equals th巳 heat lost by the cooling lemonade. According to Equation 12.5 , the heat gained
by th巳 melting ic巳 is Q = mLr, where m is the mass of the melted ice , and Lr is the latent heat
of fusion for wate r. The h巳 at lost by the lemonade is giv巳n by Q = cm /::"T, where /::"T is the
higher temperature of 27 oC minus the lower equilibrium temperature. The equilibrium t巳mper­
ature is 0 oC , because ther 巳 is some ic巳 remaining ， and ice is in equilibrium with liquid water
when th巳 temp巳rature is 0 oC.
(mLrhe = (cm /::，. T)lemom由
Solution
Th巳 mass
肌β
m
、----v--'
'-----v------'
Heat gained
by ice
H 巳at Iost
by 1 巳 monade
mice of ice that has melted is
(
cm /::，. T)le阳阳
e白l1川
1
E阳
C出
Lr
3.妇
35
x
斗
L一一一二二
10
铲5 川J/川kg
。
0些??町
mp
附
E恒
e 臼
1 5 Gωωω
创创
t忱
t川
A 7.00-kg glass bowl [c = 840 Jν/(仪
kg'C。η)] contains 16.0 kg of punch 瓜
a t 25.0 o 巳
C. Two-and-a-half
kilograms of 比
i c巳 [c = 2.00 X 10 3 Jν/(仪
kg ' C。η)] 缸
ar巳 add巳d to the punch. The ic巳 has an initial temperature of -20.0 oC , having been kept in a very cold 仕'eez巳r. The punch may be treated as if it
O
W巳re water [c = 4186 J/(kg' C ) ], and it may be assumed that th 巳re is no h巳at f1 0w between th巳
punch bowl and the extemal 巳nvironmen t. The latent heat of fusion for water is 3.35 X 10 5 J/k g.
When thermal equilibrium is reached , all the ice has melted , and the final temperature of the
mixture is above 0 oC. Det巳rmine this temperatur巳.
Reasoning Th巳 final temperature can b巳 found by using the cons巳rvation of energy principle:
the h巳 atgam巳d is equal to the heat los t. H巳 at is gained (a) by the ice in warming up to th巳 melt­
ing point, (b) by th巳 lC巳 in changing phase from a solid to a liquid , and (c) by the liquid that results from the ice warming up to the final temperature; h巳at is lost (d) by th巳 punch and (e) by
the bowl in cooling down. The heat gained or lost by each component in changing t巳mp巳 rature
can be d巳termined from the relation Q = cm /::"T, where /::,. T is the high巳r temperature minus the
lower t巳 mp巳rature. The h巳 at gained wh巳n water changes phase from a solid to a liquid at 0 oC
is Q = mLr, where m is the mass of water and Lr is the latent heat of fusion.
Solution The heat gained or lost by each component is
list巳 d
as follows:
(a) Heat gained when
ice warms to 0.0 oC
= [2.00 X 10 3 J/(kg' C O )](2.50 kg)[O.O o C 一 (-20.0 o C)]
(b) Heat gain巳d when
ice melts at 0.0 oC
= (2.50 kg)(3 .3 5 X 10 5 J/kg)
(c) Heat gained when melt巳d
ice (liquid) warms
= [4186 J/( 峙的 ](2.50 kg)(T - 0.0 oc)
to temperature T
(d) Heat lost when
punch cools to
t巳mperature T
(e) Heat lost when
bowl cools to
temperatur 巳 T
= [4186 J/(kg 'C O ) ] (l 6.0 kg)(25 .0 oC - T)
=
[840 J/(kg . C O )](7 .00 kg)(25.0 oC - T)
.c
h a n g e Vi
ew
N
y
bu
to
lic
k
gained equal to the heat lost gives:
c u -tr a c
(a) 十 (b)
+ (c) =
(d)
.d o
m
w
+ (巳)
o
o
m
th巳 heat
c
k.
w
w
.d o
w
w
w
C
Setting
C
lic
k
to
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N
377
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PD
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F-
12 .9 EQUI Ll BRIUM BETWEEN PHASES OF MATTER
O
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h a n g e Vi
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PD
F-
c u -tr a c k
、--、~
Heat gained
Heat lost
This eq
。
The physics 01 a dye-sublimation color printe r. An interesting application of the phase
change b巳 tween a solid and a gas is found in on巳 kind of color print巳r used with computers. A dye-sublimation printer uses a thin plastic film coated with separate panels of cyan
(blue) , yellow, and magenta pigment or dye. A full sp巳ctrum of colors is produced by using combinations of tiny spots of these dyes. As Figure 12.30 shows , th巳 coat巳d film pass巳S
in front of a print h巳 ad that 巳xtends across th巳 width of the paper and contains 2400 h巳 at­
mg 巳 lements. Wh en a heating element is turned on , th巳 dye in front of it absorbs heat and
goes from a solid to a gas-it sublimes-with no liquid phase in between. A coating on
the paper absorbs the gaseous dye on contact, producing a small spot of color. The intensity of the spot is controlled by the heating element, since each elem巳 nt can produce 256
different temperatures; the hotter the element, the greater the amount of dye transf，巳rred to
the paper. The paper makes three separate pass巳s across the print head , once for each of
the dyes. The final result is an image of near-photographic quality. Som巳 printers also employ a fourth pass , in which a clear plastic coating is deposited over th巳 photograph . This
coating makes the print wat巳rproof and also helps to prevent premature fading.
~ CHECK YOUR UNDERSTANDING
川加
m1
(The answers are given at the end of the book.)
13. Fruit blossoms are permanently damaged at temperatures of about - 4 oC (a hard
freeze) . Orchard owners sometimes spray a film of water over the blossoms to protect
them when a ha rd freeze is expected. Why does this technique offer protection?
14. When ice cubes are used to cool a drink , both
Temperature
Mass of ice
their mass and temperature are important in how
of ice cubes
cubes
effective they are. The table lists several possibili一 6.0 C
ties for the mass and temperature of the ice cubes
(a)
used to cool one particular drink. Rank the possi(b)
-12 C
bi lities in descending order (best first) according to (c)
- 3.0 C
their cooling effectiveness. Note that the latent
heat of phase change and the specific heat capac
ity must be considered.
Gaseous dye
Figure 12.30 A dye-sublimation printer.
As the plastic film passes in front of 由巳
print head ， 出e heat from a given heating
element causes one of three pigments or
dyes on the film to sublime from a solid
to a gas. The gaseous dye is absorbed
onto 由e coated pap巳r as a dot of color.
The size of the dots on the paper has
been exaggerated for clarity.
o
0
0
(a)
CJ} II 吧。 UI Ll BRIUM
N
BETWEEN PHASES OF MATTER
Under specific conditions of t巳mperature and pressure , a substanc巳 can eXlst at
equilibrium in mor巳 than one phas巳 at the sam巳 time. Consider Figure 12.31 , which shows
a container k巳pt at a constant temp巳rature by a large reservoir of heated sand. Initially the
contain巳r is evacuated. Part a shows itjust after it has be巳 n partially filled with a liquid and
a few fast-moving mol巳cules are escaping the liquid and forming a vapor phas巳. Thes巳
mol巳cules pick up th巳 required 巳 nergy (the latent heat of vaporization) during collisions
with neighboring molecules in the liquid. However, the reservoir of heated sand replenishes the 巳n巳rgy carried away, thus maintaining the constant temperature. At first , the
movement of molecules is pr巳dominantly from liquid to vapor, although som巳 molecules
in the vapor phas巳 do reent巳r the liquid. As the molecules accumulate in the vapor, the
number reentering the liquid eventually equals the number entering the vapor, at which
point 巳quilibrium is establish巳d ， as in part b. From 出is point on , the concentration of molecules in the vapor phase does not change , and th巳 vapor pressure remains constan t. The
pressure of the vapor that co巳xists in equilibrium with the liquid is called the equilibrium
vapor pressure of the liquid.
(b)
Fig ure 12.31 (α) Some of the molecules
begin entering the vapor phase in 由巳
evacuated space above th巳 liquid.
(b) Equilibrium is reach巳dwh巳n 由巳
number of molecules entering 由e vapor
phase equals the number retuming to
由巳 liquid .
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3 x 10 5
2 x 10 5
1.01 x 10 5
0.53 x 10 5
O
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50
]
自
83 100
150
Temperature , oc
The equilibrium vapor pressure do巳 s not dep巳 nd on the volu 日1巳 of space above the liquid. If more spac巳 were provided , more liquid would vaporize , until equilibrium was
reestabbshed at the same vapor pressure , assuming the sam巳 temperature is maintained. In
fact , th巳 equilibrium vapor pressure d巳pend s only on th巳 temperature of the bquid; a higher
temp巳rature causes a higher pr巳 ssure ， as the graph in Figure 12.32 indicates for the spe
cifìc cas巳 of water. Only when th 巳 temperature and vapor pr巳 ssure corr巳 spond to a point
on the curved bne , which is called th 巳 vapor pressure curve or th巳 vaporization curve, do
liquid and vapor phases co巳xist at 巳 quilibrium.
To illustrate th巳 use of a vaporization curv巳， let's consider what happens when water
boils in a pot that is open to the ai r. Assume that the air pr 巳 ssure acting on the water is
1.01 X 10 5 Pa (on巳 atmospher巳). When boiling occurs , bubbles of wat巳r vapor form
throughout the liquid , ris巳 to the surfac巳， and break. For these bubbles to form and ri se ,
th 巳 pressure of th巳 vapor inside th巳 m must at least equal th 巳 air pressure acting on the surfac巳 of the water. According to Figur巳 12.32 ， a valu巳 of 1.01 x 10 5 Pa corresponds to a
temperature of 100 o Cons巳qu 巳 ntl y， water boils at 100 oC at one atmosph 巳 re of pres
sure. In gene时， a liquid boils at the temperature for which its vapor pressure equals the
external pressure. Water wiU not boil , th 巳 n ， at s巳 a level if the temperature is only 83 oC ,
because at this temperature the vapor pr巳ssur巳 of water is only 0.53 X 10 5 Pa (s巳巳 Figur巳
12.32) , a value less than th巳巳xt巳rnal pr巳ssur巳 of 1. 01 X 10 5 Pa. How巳ver， water does boil
at 83 oC on a mountain at an altitude of just under fìve kilometers , because the atmospheric pressure ther巳 is 0.53 X 10 5 Pa.
The fact that water can boil at a temperature less than 100 oC leads to an inter巳 sting
ph巳nom巳 non that Conceptual Example 16 discusses.
c.
也叫
5 1
4
令
一一-Water
Conceptual
Example 冒 6
How to Boil Water That Is Cooling Down
Figure 12 .33a shows water boiling in an op巳 n flask . Shortly after the flask is removed from the
burner, the boiling stops. A cork is then placed in the neck of the flask to seal it , and wat巳 r is
pour巳 d over the n 巳ck of the flask , as in part b of th巳 drawing. To restart the boiling , should the
wat巳r poured ov巳r the neck be (a) cold or (b) hot-b l1 t not boiling?
Reasoning When the open flask is removed from the burner, the water b巳gins to cool and the
pressure above its surfac巳 is on巳 atmosph巳re (l. 01 X 10) Pa). Boiling quickly stops , b巳cause
water cannot boil when its t巳 mperatur巳 is less than 100 oC and the pressure abov巳 its surface is
one atmosph巳re. To 1 巳 staft the boiling , it is necessary either to reheat the water to 100 oC or reduce the pr 巳 ssure abov巳 th巳 water in the corked flask to something less than one atmosph巳I 巳 s o
that boiling can occur at a temp巳rature less than 100 oC.
Water boiling again
Figure 12.33 (α) Wat巳r is boiling at a
temperatllre of 100 oc and a pressure of
one atmosphere. (b) The water boils at a
temperature that is 1巳ss than 100 oc ,
because the cool water reduces the
pressure above the water in the flask
(b)
Answer (b) is incorrect. Certainly, pouring hot water over th巳 corked flask wiU reheat 由e wate r.
However, since the water being pour 巳d is not boiling , its temperature must be less than 100 o
Thel 巳fore， it cannot reheat the water within the flask to 100 oC and restart 由巳 boiling.
c.
Answer (a) is correct. wh巳 n cold water is pour巳d ov巳 r the corked flask , it causes some of 出e
water vapor inside to condense. Consequently, th巳 pressur 巳 above the l.i quid in the flask drops.
When it drops to the value of the vapor pressure of the water in the flask at its current t巳 mpe r­
ature (which is now less than 100 OC) , th巳 bo ilin g restarts.
m
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4 x 10 5
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bu
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bu
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k
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C
Figure 12.32 A plot of the eqllilibrillm
vapor . cpress lI re v巳rS lI s t巳 mperatur巳 I S
c u -tr a c k
called th巳 vapor pr巳 ssure curve or th巳
vaporization curve, the example shown
b巳ing that for the liquid/vapor
eqllilibri l1 m of wate r.
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TEMPERATUR 巨 AND
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CHAPTER 12
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1 2.9 EQUI Ll BRIU M BETWEEN PHASES OF MATTER
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High-pressure
propellant
vapor
Tube
Uquid
propeliant
plus product
(α
Figure 12.34
(α)
A
Liqu id propellant
plus product
(b)
closed spray can containing liquid and vapor in equilibrium. (b) An open spray can
而
且
a
ω
』2的的ω
』且
Temperature , oc
(0)
而
ι
Normal meltin g!lreezing
point 01 H20
a
ω』2的的ω』
止
The physics of spray cans. The operati on of spray cans is based on the equilibrium between a liquid and its vapor. Fi gure 12 .3 4a shows that a spray can contain s a liquid propellant that is mi x巳 d with the product (s uch as hair spray). In side th 巳 can ， propell ant
vapor forms over the liquid. A propell ant is chosen that has a n 巳quilibrium vapor pressure that is greater than atmospheri c press ure at room temperature. Co n sequ 巳 ntl y， when
th 巳 n ozz le of th 巳 c an is pressed , as in part b of the draw ing , the vapor pressure forces the
liquid p rop巳 ll a nt and product up the tube in th巳 c an and out th 巳 n ozz l 巳 as a spray. When
the nozzle is released , the coiled spring r巳 seal s the can and the propell ant vapor build s
up once again to its equilibrium value.
As is the case fo r liquid/vapor equilibrium , a solid can b巳 in equilibrium with its liquid phase only at specific conditi ons of temperature and pr巳 s s ure. For each te mp巳 ra ture ，
there is a single press ure at which the two phases can co巳 x i s t in equilibrium. A pl ot of
the equilibrium pressure versus equilibrium t巳 mp eratu r巳 i s referred to as the fu sioll
curve, and Figure 1 2 . 3 5 αs h o w s a typi cal c urv巳 fo r a normal substance. A normal substance expands on meltin g (巳 g . ， carbon di ox id 巳 and sulfllr). Since hi gher pr巳 ss ur巳 s make
it more difficult for slI ch materials to 巳 x p a nd ， a higher melting temperature is n e巳 d e d for
a higher press ure, and th 巳 fll s i o n cllrve slopes upward to th巳 ri gh t. Part b of the pictllre
illustrates th 巳 fu sio n curve for water, one of the few substances that contract when they
melt. Higher p ress ur巳 s make it easier fo r such s ub s tanc巳 s to melt. Consequently, a lower
melti ng temperature is associated with a hi gher pressure , and th巳 fll si o n c urv巳 s l o p es
dow nward to the right.
It shollld be noted thatjust because two phases can coex ist in eqllilibrium d o巳 s not
n ec巳 ssaril y mean that they wilL Other fac tors may pr巳ve nt it. Fo r 巳 xa mpl e ， wat巳 r in an
open bowl may n 巳v e r com 巳 into eqllilibrium with water vapor if air currents are present.
Under such conditions the liquid , perhaps at a temperatur巳 of 25 oC, attempts to establi sh the corr巳s pondin g eqllilibrium vapor press Ul 巳 of 3.2 x lO :l Pa. If air currents continllally blow th巳 water vapor away , h ow ev町，巳quilibrium will n 巳V巳 rb巳巳 s t abli s h 巳 d ， and
eventuall y th 巳 water will evaporate co mpletely. Each kilogram of water that goes into the
vapor phase takes along the latent heat of vaporization. Becall se of thi s h 巳 at loss , th巳 re ­
maining liqllid wOllld
Solid
1. 01 x 10" 1 - --- - -- 1
/ U
O
Temperature , oc
(b)
Figure 12.35 (α) The fusion curve for
a normal substance that expands on
melting. (b) Th巳 fu s i o n curve for water,
o n巳 of the few substances that contract
on melting
The physics of
evaporative cooling
of the human body.
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It takes less heat to boil water high on a
mountain , b巳cause the boiling point
becomes less than 100 oC as the air
pressur巳 decreases at higher elevations.
(Gregg Adams/Stone/Getty Images)
The physics of
relative humidity.
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(The answers are given at the end of the book.)
c u -tr a c k
15. A camping stove is used to boil water high on a mountain , where the atmospheric
pressure is lower than it is at sea leve l. Does it necessarily follow that the same stove can
boil water at sea level?
16. 节. Medical instruments are sterilized at a high temperature in an autoclave , which is
曹 essentiallya pressure cooker that heats the instruments in water under a pressure greater than one atmosphere. Why is the water in an autoclave able to reach a very
high temperature , while water in an open pot can only be heated to 100 OC?
'在 A jar is half filled with boiling wate r. The lid is then screwed on the jar. After the jar has
cooled to room temperature , the lid is difficult to remove. Why? Ignore the thermal expan
sion and contraction of the jar and the lid.
18. A bottle of carbonated soda (sealed and under a pressure greater than one atmosphere) is left outside in subfreezing temperatures , although the soda remains liquid .
When the soda is brought inside and opened immediately, it suddenly freezes. Why?
19. When a bowl of water is placed in a closed container and the water vapor is pumped
away rapidly enough , why does the remaining liquid turn into ice?
~ II
一 ω
*HUMIDITY
Air is a mixture of gases , in cI uding nitrogen , oxyg巳n ， and water vapor. The totaI pres
sure of the mixture is the sum of th巳 partial pressures of the component gases. The partial
pressure of a gas is the pressure :it would 巳xert if it alone occupied the entire volume at th巳
sam巳 temperaωre as th巳 mixture. The parti a1 pressure of wat巳r vapor in air depends on
weather conditions. It can be as low as zero or as high as th巳巳quilibrium vapor pressure of
water at th巳 giv巳 n temperature.
To provide an indication of how much water vapor is in the air, weath 巳 r forecast巳 rs
usually give the relative humidity. If the relative humidity is too low, the air contains such
a small amount of water vapor that skin and mucous membranes tend to dry ou t. If the r巳 1ative humidity is too high , especial1 y on a hot day , we become v巳ry uncomfortable and our
skin feels "sticky." Under such conditions , the air holds so much wat巳 r vapor that the water exuded by sw巳 at glands cannot evaporate effici巳 ntly. Th巳 relative humjdity is defined
as th巳 ratio (巳xpressed as a percentage) of the partial pressure of water vapor in the air to
th巳 equilibrium vapor pressure at a given temp巳rature.
Percent
relativ巳
humidity
Parti a1 pressure
of water vapor
Eq uilibrium vapor pressure of
wat巳r at 出e
x 100
(I 2.6)
eXJstmg temperature
The term in the denominator on the right of Equation 12.6 is given by th巳 vaporization
curve of water and is the pressure of the water vapor in equilibrium with th巳 liquid. At a
glV巳 n temperature , th巳 partial pressure of the water vapor in the air cannot exc巳ed this
value. If it did , the vapor would not b巳 111 巳quilibrium with the liquid and would condense
as dew or rain to reestablish 巳quilibrium.
When the partial pr巳ssure of th巳 water vapor 巳quals the equilibrium vapor pressur巳 of
water at a giv巳n temperature , the relative humidity is 100%. In such a situation , th 巳 vapor
is said to be saturated because it is present in the maximum amount , as it would b巳 abov巳
a pool of liquid at equilibrium in a cI os巳d container. If th巳 relative humidity is less than
100% , th巳 water vapor is said to be unsaturate d. Example 17 demonstrates how to find the
relative humidity.
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e
On巳 day， the partial pressure of wa与r vapor in th巳 air is 2.0 X 10 3 Pa. Using the vaporization
curve for wa阳 in Figure 12.36 , determine the I毛lative humidity if the temp巳rature is (a) 32 oC
and (b) 21 oc.
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CHAPTER 12 TEMPERATURE AND HEAT
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Relative humidity at 32 oC
2.0 X 10 3 Pa
4.8 X 10 3 Pa
=飞
4 .8 x 103
e
x
r-:-:-::-:-l
100 = 142% 1
'-一一」
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Reasoning and Solution (a) According to Figure 12.36 , the equilibriurn vapor pressure of
water
at 32 oC is 4.8 X 10 3 Pa. Equation 12.6 reveals that th巳 relative humidity is
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12.11 CONCEPTS & CALCULATIONS
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(b) A similar caIculation shows that
Relative hurnidity at 21 oC =
-
2.5 x
10。
2.0 X 10 3 Paπ-:-:-:-l
、
x 100 =问0% 1
2.5 X lO J Pa
'-一一」
O
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When air containing a given amount of water vapor is cooled , a ternperature is reached
in which 出ep缸tial pressure of the vapor equals the equilibriurn vapor pressure. This temperature is known as the dew point. For instance , Figure 12.37 shows that if th巳 parti al
pressure of water vapor is 3.2 X 10 3 Pa, the dew point is 25 oC. This partial pressure would
correspond to a relative humidity of 100% , if the ambient t巳 rnperature were equal to th巳
dew-point temp巳rature . The physics of fog formation. Hence , the dew point is the temperature below which water vapor in the air cond巳nses in the forrn of liquid drops (d巳w or fog)
The closer the actual temp巳rature is to the dew point , the closer the relative humidity
is to 100%. Thus , for fog to form , the air temperature must drop below the d巳w poin t.
Similarly, water condenses on the outside of a cold glass wh巳 n the temperature of th巳 air
next to the glass falls below the d巳w poin t. The physics of a home dehumidifie r. The cold
coils in a home dehumidifi巳r (see Figure 12.38) function very much in the same way that
th巳 cold glass does. The coils are kept cold by a circulating refrigeran t. When the air blown
across them by the fan cools below the dew point, water vapor condenses in the form of
droplets , which collect in a receptacle.
,(' CHECK
32
Temperature , oc
Figure 12.36 The vaporization curve of
water.
4 x 10 3
m
c..、
ai
3.2 x
10~
2x
1O ~
2
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ct'"
(l)
O
40
O
YOUR UNDERSτANDING
(The answers are given at the end of the book.)
20. A bowl of water is covered tightly and allowed to sit at a constant temperature of
23 oC for a long time. What is the relative humidity in the space between the surface of the
water and the cover?
21. Is it possible for dew to form on Tuesday night and not on Monday night , even
though Monday night is the cooler night?
22. Two rooms in a mansion have the same temperature. One of these rooms contains an
indoor swimming poo l. On a cold day the windows of one of the two rooms are 飞teamed
up." Which room is it likely to be?
Figure 12.37 On the vaporization
curve of water, the dew point is the
temp巳rature that co盯esponds to the
actual partial pressure of water vapor
in the air.
Dehumidified air
This section contains examples that discuss one or more conceptual questions , followed by a related quantitative problem. Example 18 provides insight on the variables involved when the length and volume of an object change due to a temp巳rature change.
Example 19 discusses how different factors affect th巳 te mperature change of an object to
which heat is being added.
0
…
Concepts ~垦 Calculations Example
Lin
nd Volur川lermal Expansion
18
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Concept Questions and Answers D。自由e change in the vertical height of a block depend
only on its height, or do巳 s it also depend on its width and depth ? Without doing any calculations , rank the blocks according to their chang巳 in height , largest firs t.
mmmw-m
•
(-Muecmh
WmAFMmrEll:
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二
•
则
Figure 12.39 sho问时 three rectangular blocks made from the same material. The initial dimensions of each are expressed as multip 1es of D , where D = 2.00 cm. The blocks are heated and
their temperatures incr巳 ase by 35.0 C O. The coefficients of lin巳ar and volume expansion are
α= 1. 50 X 10
(CO) - I and ß = 4.50 X 10 (CO)-I , respectively. Determin巳 the chang巳 in
the (a) vertical heights and (b) volumes of 由巳 blocks
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Figure 12.39 The temp巳ratures of the
three blocks are raised by the same
amoun t. Which one(s) 巳xperi巳 nce the
greatest chang巳 in height and which the
greatest change in volume?
o
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Does the change in the volume of a block dep巳 nd only on its height , or does it also d巳p巳nd on
its width and d巳 pth ? Without doing any calculations , rank th巳 blocks according to their greatest change in volume , largest fìrst
Answer According to Eq l1 ation 12.3 , ð. V = ßVo ð. T , the change ð. V in the volum 巳 of an
object dep巳 nds on its original volum巳 Vo ， the change in temperature ð. T , and the coefficient
of volum巳 expansion ， ß. Thus , the change in volume d巳p巳 nds on height , width , and depth ,
because the original vol l1 me is the product of th巳 se three dim巳 nsions. Th巳 initial volumes
of A , B , and C , are , respectiv巳Iy， D x 2D x 2D = 4D 3 , 2D X D X D = 2D 3 , and
2D X 2D X D = 4D 3 . Blocks A and C have equal volum邸， which are greater than that
ofB. Thus , w巳 expect A and C to 巳对libit th巳 great巳 st 1I1 creas巳 in volume , while B exhibits
the small巳 st increase.
Solution (a) The change in the height of each block is
giv巳 nby
Equation 12.2 as
ð. L A = αD ð. T = [1.5 0 X 盯5 (C勺一 1](2.00 cm)(35 .0 CO) = I 1. 05 X 10- 3 cm I
ð. L ß = α (2D) ð. T = [1. 50
X
10→ (C O ) 一 1](2
ð. Lc = α(2D) ð.T = [1.5 0
X
10- 5 (C O )- I](2 X 2.00 叫 (35.0 C O ) = I2.10 X 10- 3
th 巳 heights
As expected ,
(b) The change in the
ð. VA = ß(D
X
of B and C
vol l1 m 巳 of
X
2.00 cm)(35.0 C O ) = I2 .10 X 10- 3 cm I
increas巳 more
叫
than the height of A.
each block is given by Equation 12.3 as
2D X 2D) ð. T
= [4.50 X 10- 5 (CO) - I][4(2.00 叫3](35.0 C可 =15.04 X 盯2 时|
ð. Vß = ß(2D
X
D X D) ð. T
= [4.50 X 10- 5 (CO) - I][2(2.00 cm)3](35 .0 CO) = I2 .52 X 10- 2 cm3 1
ð. Vc
= 卢 (2D X
2D X D) ð. T
= [4.50 X 10- 5 (C O ) 一 1][4(2. 00 cm)3](35 .0 CO) = I5.04 X 10- 2 cm 3 1
As
discuss巳d 巳 arlier， th巳 great巳 st chang巳 in
vol l1 me occurs with A and C followed by B.
。
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Concepts & Calculations Example 1 9
Heat and Temperature Ch 叫 es
Ob吁巳cts A and B in Figure 12 .40 are made from ∞pp巳r， bl1 t the mass of B is three times the
mass of A. Object C is made from glass and has the same mass as B . The same amount of heat
Q is suppliedω 巳ach one: Q = 14 J. Determin 巳 th巳 rise in 臼 mperature fOl巳ach.
Concept Questions and Answers Which object, A or
B ， 巳xperience s
the
gl 巳 at巳
ri s e
in
temperatul 巳?
Figure 12.4 0 Wh ich block has the
greatest chang巳 in temperature when
the same heat is supplied to each?
Answer Consid巳r an extrem巳巳xample. Suppose a cup and a swimming pool 创毛白lI ed with
water. For the same heat input , w0111d you intuitively 巳xpect the temperature of th 巳 C l1 p to
nse mor 巳 than the t巳 mperature ofthe pool? Yes , because th巳 cup has less mass. Another way
to 创 rive at this conclusion is to solve Equation 12 .4 for the change in temperature:
ð. T = Q/(cm). Since the heat Q and the specifìc heat capacity c are the same for A and B ,
ð. T is inversely proportional to the mass m. So th 巳 0均巳ct with the small巳 r mass exp巳[1enc巳 s the larger temperature chang巳. Therefore , A has a greater t丑 mp巳rature change than B.
Which object , B or C , exp巳riences the greater rise in
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Answer According to Equation 12.2 , ð. L = αLo ð. T , the change ð. L in th 巳 height of an
o均 ect depends on its original h巳ight Lo , the chang巳 in temperature ð. T , and the coeffìci 巳 nt
w
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ack
of lin巳创·巳xpansion ， α It do巳 s not d巳pend on the width or depth of the objec t. Sinc巳 block s u - t r
B and C have twice the height of A , their h 巳ights will incr巳 as巳 by twic巳 as much as that of
A. The heights of B and C , howev町， wiU incr巳 as巳 by th巳 same amount , even though C is
twice as wide.
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CHAPTER 12 TEMPERATURE AND HEAT
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temp巳 rature?
Answer Objects B and C are made from different materials. To see how the ri se in
temperature depends on th巳 type of material , let's again l1 se Equation 12 .4: ð. T = Q/(cm) .
Since the heat Q and the mass m are the same for B and C , ð. T is inversely proportional to
th巳 specifìc heat capacity c. So the object with th 巳 small 巳r specifìc heat capacity 巳xper卜
enc巳 s the larger temp巳ratur巳 chang巳. Table 12.2 indicates that th巳 sp巳cifìc heat capacities
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Solution According to Equation 12 .4, the
ð.T^
门
ο14
= 一二L一=
ð. T =
D
temp巳rature
CA Jn A
J
lic
k
of copp巳r and glass ar巳 387 and 840 J/(kg . C勺， respectively. Since B is made from copp巳r，
which has the smaller specific heat capacity, it has the great巳 r t巳 mperature chang巳.
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= 11 8 C 叮
L一一」
[387 J/(kg' CO)](2.0 X 10
οl4J
----=:..一=
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__ 1
同
= 1 6.0C 叮
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J
[387 J/(kg' C )](6.0 X lO- kg)
LI -- - - '
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J
I _ _ __ I
ð. T，户立一口 .8 C 叮
Ccmc
[840 J/(kg' C O)](6.0 X lO- J kg)
L一一一」
As anticipated , ð. TA is greater than ð. T ß , and ð. T B is greater than ð. Tc.
CONCEPT SUMMARV
If you need more help with a concept , use the L巳arning Aids noted next to the discussion or equation. Examples (Ex.) are in the text
of this chapter. 00 to www.wiley.com/college/cutnell for the following Learning Aids:
Interactive learningWare
(I l\的 一 Additional
Concept 5imulations (C5) Interactive 5olutions (15) -
examples solved in a five-step interactive forma t.
An imated text figures or animations of important conc巳pts.
Models for certain types of problems in the chapter homework. The calculations are carried out interactively.
Topic
Celsius lemperalure scale
Fahrenheillemperalure scale
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Discussion
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冒 2.1 COMMON TEMPERATURE SCALES On the Celsills temp巳rature scale , there are 100 equal Ex. 1 , 2
divisions between the ice point (0 o c) and the st巳am point (1 00 o c). On th巳 Fahrenheit temperature
scale , there are 180 eqllal divisions between the ice point (32 OF) and the steam point (212 OF)
12.2 THE KELVIN TEMPERATURE SCALE For scientific work , the Kelvin temperature scale is the
scale of choice. One kelvin (K) is equal in size to one Celsius d巳gree . However, the temperature T
on the Kelvin scale differs from the temperature Tc on the Celsius scale by an additive constant of
273.15:
T= 汇+
(12.1)
273 .15
The lower Ii mit of temperature is called absolute zero and is designated as 0 K on the Kelvin scale.
Absolule zero
Thermomelric properly
12.3 THERMOMETERS The operation of any thermometer is based on the change in some physical property with temperature; this physical property is called a thermometric property. Examples
of thermometric properties are th巳 length of a colllmn of mercury, electrical voltage , and electrical
reslstance.
12.4 Ll NEAR THERMAL EXPANSION
Most substances expand when heated. For linear expansion ,
temperaωre changes by !1 T:
an object of length Lo experiences a change !1 L in length when the
句a
『
l --nnua t knHasr mal nEwAnurnea nunH
!1 L
whereαis
the
co巳ffìcient
(12.2) Ex. 3 , 18
= α Lo !1 T
of Ii near expansion.
Thermal slress
For an object held rigidly in place , a therm al stress can occur when the obj 巳ct attempts to expand Ex. 4
or contrac t. Th巳 stress can be large , even for small temperature changes.
How a hole in a plale expands
or conlracls
When the temperatllre changes , a hole in a plat巳 of solid material expands or contracts as if the hole Ex. 5 , 6 , 7
were fì lled with the surrollnding materia l.
12.5 VOLUME THERMAL EXPANSION For
object of volume Vo is given by
!1 V
Volume Ihermal expansion
whereβis
How a cavity expands or
conlracls
vo lum巳 expansion ， th巳 change
the
co巳fficient
= β Vo !1 T
!1 V in the
volllm巳 ofan
(1 2.3)
Ex.8 , 18
IlW 12.1
of volume expansion.
When the temperature changes , a cavity in a piece of solid material expands or contracts as if the 15 12.29
cavity were filled with the surrounding materia l.
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币 2.6
HEAT AND INTERNAL ENERGY The internal energy of a substance is th巳 sum of the kinetic ,
potential, and other kinds of energy that the molecules of 出 e substance have. H巳at is energy that
ftows from a higher-temperature obj 巳ct to a lower-temperature object because of the difference in
temperatures. The SI unit for heat is the joule (J).
c u -tr
Inlernal
energy
Heal
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12.7 HEAT AND TEMPERATURE CHANGE: SPECIFIC HEAT CAPACITY The heat Q that must be supplied or remov巳d to chang巳 th巳 temperature of a substance of mass m by an amount I1 T is
Heal needed 10 change Ihe
lemperalure
Q = cm l1 T
(12 .4)
Ex. 9. 10. 11.
12:
1 占，
'"J
where c is a constant known as the specific heat capacity.
Energy conservalion and heal
When materials are placed in thermal contact within a perfectly insulated container, the principle 15 12.47, 12.51
of energy conservation requires that heat lost by warmer materials equals heat gained by cooler
materials.
Heat is sometimes measured with a unit ca l1 ed th巳 kilocalorie (kcal). The conversion factor between kilocalories and joul 巳s is known as the mechanical equivalen t of hea t:
1 kcal = 4186 joules
Mechanical equivalenl 01 heal
ILW 12.2
冒 2.8 HEAT AND PHASE CHANGE: LATENT HEAT Heat must be supplied or removed to make a
material change 台om one phase to another. The heat Q that must be supplied or removed to change
th巳 phas巳 of a mass m of a substance is
Heal needed 10 change Ihe
phase
(1 2.5) Ex. 13, 14, 15
Q = mL
where L is the latent heat of the substance and has SI units of J/kg. Th巳 latent h巳ats of fusion , va- 1512.95
porization , and sublimation refer, respectively, to the solid/liquid , the Iiqui d! vapor, and the
solid/vapor phase changes
Vapor pressure curve
Fusion curve
12.9 EQUI Ll BRIUM BETWEEN PHASES OF MAπER The equilibrium vapor pressure of a sub- Ex. 16
stance is the pressur巳 of the vapor phase that is in equilibrium with the liquid phase. For a given
substance, vapor pressure depends only on temperature. For a liquid, a plot of the equilibrium vapor pressure versus temp巳rature is called the vapor pressure curve or vaporization curv巳.
The fusion curve gives the combinations of
solid and liquid phases.
t巳 mperature
and pressure for equilibrium between
12.10 HUMIDITY The relative humidity is defined as follows:
Partial pressure
of water vapor
Percent
Relalive humidily
r巳lative
humidity
Dew poinl
Equi1i brium vapor pressure of
water at th巳 existing temperature
x 100
(1 2.6) Ex. 17
The dew point is the temperature below which the water vapor in the air condenses. On the vaporization curve of water, the dew point 自由e temperature that coπesponds to th巳 actual pressure of
water vapor in the air
FOCUS ON CONCEPTS
avω lable
Note to lnstructors: The numbering of the questions shown here reflects the fact thαt they are only a representatiνe subset of the total number that are
online. Howeve r， αII of the questions are available for αssignment via an online homework manαgement program such as WileyPLUS or \<\帖'Assign
Section 12.2
The Kelvin Temperature Scale
Section 12.4
Li near Thermal Expansion
1. Which one of the following statements correctly describes the
Celsius and the Kelvin temperature scales? (a) Th e size of the degree
2. The drawing shows two thin rods , one made 企om alurr山lUm
[α= 23 X 10- 6 (CO)- I] and the 0由er 仕omst臼I[α= 12 X 10- 6 (CO)- I].
on the Celsius scale is larger than 出at on the Kelvin scale by a factor
of 9/5. (b) Both scales assign the sarne temperature to 出 e lce p Ol nt ,
but they assign different temperatures to the steam point
(c) Both scales assign the same temperature t。由e stearn point, but they
assign di仔erent temperatures to the ice poin t. (d) Th巳 Celsius scale
assigns the same values to 由e ic巳 and the ste创n points 由 at 由巳 Kelvin
scale assigns. (e) The size of the degree on each scale is the sarne
Each rod has 由e sarne length and
attached at one end to an immov
able wall , as shown. Th e tempera
tures of 由e rods are increased ,
both by 由巳 sarne arnount , until 由e
gap betw巳en 由e rods is CI osed
Where do th巳 rods m巳巳t when 出巳
由e
same initial
temperaωre
Midpoint
and is
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CHAPTER 12 TEMPERATURE AND HEAT
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(d)α。 αA ， αB
Section 12.5
(e)αA ， αB ， αc
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4. A ball is slightly too large to fit
A
through a hole in a f1 at p l at巳 . The
τ丁~\
C
drawing shows two arrangements B
of this situation. In Arrangement 1
飞飞~之咀伊/
出巳 ball is made from metal A and
Arrangement 11
Arrangement 1
the plate from metal B. When
both the ball and the plate are cooled by the same number of Celsius
degrees , the ball passes throu gh the hole. 1n Arrangement II the ball
is also made from metal A , but the plate is made from metal C. Here ,
the ball passes thro l1 gh the hole when both the ball and th巳 plate are
heated by the same number of Celsius degrees. Rank the coefficients
of lin巳ar thermal expansion of metals A , B , and C in d巳scendin g order (l argest first): (a)α日 ， αA ， αc (b)αB ， αC'αA (c)αC' αB ， αA
巳xperiences
k
the smallest drop in temperature , and which one experiw
ences the largest drop?
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Sample A. 4.0 kg of water [c = 41 86 J/(kg ' C )]
Sample B. 2.0 kg of oil [c = 2700 J/(kg ' C O)]
Sample C. 9.0 kg of dirt [c = 1050 J/(kg' C O)]
(a) C smallest and A
smallest and B larg巳 st
and A largest
Section 12.8
larg巳st
(b) B small巳st and C largest (c) A
(d) C smallest and B largest (e) B smallest
Heat and Phase Change: Latent Heat
13. The latent heat of fusion for water is 33.5 X 10 4 J/kg , while the
latent heat of vaporization is 22.6 X 10 5 J/kg. What mass m of water
ml1 st be froz巳n in ord巳 r to release the amount of heat that l. 00 kg of
steam releases when it condenses?
Section 12.9
Equilibrium Between Phases of MaUer
15. Which one or more of the following
freeze water?
Volume Thermal Expansion
•
can be used to
A. Cooling the water below its normal freezing point of 0 oC at
the normal atmospheric press ure of l. 01 X 10 5 Pa
6. A solid spher巳 and a solid cube ar巳 made from the sam巳 material.
The sphere wou ld just fit within the cl1 be, if it could. Both begin at the
same temperature , and both are heated to th巳 sa me temp巳ratl1 re.
Which object, if either, has the greater chang巳 in vo l l1 m巳? (a) The
sphere. (b) The cube. (c) Both have the same change in vol l1 me.
(d) 1ns l1 fficient information is given for an answe r.
7. A container can be made from steel [卢= 36 X 10- 6 (CO)-I] or lead
[卢= 87 X 10- 6 (C O )一 1]. A liq l1 id is poured into 由巳 contamer， 自 llin g it
to the brim. The liquid i s 巳ith er water [β= 207 x 10 (CO)- I] or ethyl
alcohol [卢= 1120 X 10- 6 (CO)- I]. When the fl1 11 container is heat时，
some liquid spills ou t. To keep the overflow to a minim l1 m, from what
material sho l1 ld the container be made and what sho l1 ld the liquid
be? (a) Lead , wat巳r (b) Steel , water (c) L巳ad ， ethyl alcohol
(d) Ste巳 1 ， ethyl alcohol
t巳chniql1e s
B. Cooling the water below its freezing point of -1 oC at a pressure great巳r than l. 01 X 10 5 Pa
C. Rapidly pl1 mping away the water vapor above the liq l1 id in an
ins111ated container (The insulation prevents heat f1 0wing
from th巳 s l1 rro l1 ndings into the remaining liquid.)
(a) Only A
(e) Only C
Section 12.10
(b) Only B
(c) Only A and B
(d) A , B , and C
Humidity
17. Which of the fo llowing three statements concerning relativ巳 hu­
midity val l1 es of 30% and 40% are true? Note that when the relative
humidity is 30% , the air temperatl1 re may be di旺巳rent than it is when
the relative hl1 midity is 40%.
A. It is possible that at a relative hl1 midity of 30% there is a
smaller partial press l1 re of water vapor in the air than there is
at a relativ巳 humidity of 40%
Section 12.7 Heat and Temperature Change:
Specific Heat Capacity
9. Which of the following cases (if any) req l1 ires the greatest amount
of heat? 1n each case 由 e material is the s缸ne. (a) l. 5 kg of th巳 ma­
terial is to be heat巳d by 7.0 CO. (b) 3.0 kg of th巳 m aterial is to be
heated by 3.5 C"' (c) 0.50 kg ofthe material is to be heated by 21 CO.
(d) 0.75 kg of the material is to be heated by 14 C O. (e) The amount
of heat req l1 ired is the same in each of the four previo l1 s cases.
10. The following three hot samples have the same temperatl1 re. The
same amO l1nt of heat is removed from each sample. Which one
B. 1t is possible that there is th巳 same partial pressure of water
vapor in the air at 30% and at 40% relative humidity
C. It is possible that at a relative hl1 midity of 30% th巳re IS a
greater partial pressure of water vapor in th巳 air than ther巳 i s
at a relative humidity of 40%.
(a) A , B , and C (b) Only A and B
and C (e) Only A
(c) Only A and C
(d) Only B
在DB
Note
ω Instructors:
Most of the homework problems in this
chαpter a时 ava ilable
for assignment
viα an
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Note: For problems in this set, use
the νalues
of a
and 卢 in
Table 12.1 , and the
ssm Solution is in the Student Solutions Manual.
www Solution is available online at www.wiley.comlcollege/cutnell
Section 12.1
Section 12.2
Section 12.3
Common Temperature Scales ,
The Kelvin Temperature Scale,
Thermometers
1. ssm A t巳mperatl1re of absolute zero occurs at -273.15 oc. What
is this temperature on th巳 Fahrenheit scale?
νalues
of c, L[, and Lv in Tables 12.2 and 12.3, unless stated otherwise
γThis icon repre… biomedi叫plication
2. S l1ppO S巳 yo u are hiking down the Grand Canyon. At the top , the
temperature early in the morning is a cool 3 oC. By late afternoon , the
temperature at the bottom of the canyon has warmed to a sweltering
34 oc. What is the difference betwe巳n the higher and lower temperatures in (a) Fahrenheit degrees and (b) kelvins?
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(b) The
left of
由e
C
m
gap is c1 osed? (a) The rods meet exactly at the midpoin t.
(c) 白1巳 rods meet to
rods
. c m巴巴t to th巳 right of th巳 midpoin t.
c u -tr a c k
the midpoin t.
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PROBLEMS
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CHAPTER 12 TEMPERATURE AND HEAT
(a) When the temperature of th巳 plate is increased , wi lJ the radius of
the h o l 巳 be larger or smaller than the racl ius at J I OC? Why?
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(b) When the plate is heated to 110 oc , by what fractio n !:::"rlro will u - t r
the rad ius of the hole change?
k
+ _
14. A thi ck, vertical iron pipe has an inner diameter of 0.065 m. A
thin alllminulll di sk , heated to a temperature of 85 oc , has a d iameter
that is 3.9 x LO- 5 m gr巳ater than the pipe's inner di ameter. Th巳 disk
is laid on top of th巳 ope n upper end of th巳 pipe ， perfectly centered on
it, and a ll owed to coo l. What is the temperature of the aluminum disk
when the disk falls into th巳 pipe ? Ignore the te mp巳 rature change of
the pipe.
15. ssm When the telllperatllre of a coin is raised by 75 C O, the
+1300 。B
coin 's diameter increases by 2.3 X 10- 5 m. If the ori ginal diameter
of the coin is 1.8 x 10- 2 m, find th巳 coefficient of linear expansion .
0
16. One January morning in 1943 , a warm chinook wind rapidly
raised the te mperature in Spe缸 fish ， South Dakota , from below 仕.ee z­
ing to + 12.0 o As the chinook died away, the temperature fe ll to
-20.0 oc in 27.0 minutes. Suppose that a 19-m a lum.i num 日 agpo l e
were subj ected to this telllperature change. Find the averag巳 speed at
which its h巳 i g ht wou ld d巳crease， assumin g th巳f1 agpole responded insta ntaneously to the chan ging temperature.
c.
5. 嗖p'
ssm Derm atologists often remove small precancerous ski n
lesions by freezing them quickly with liquid nitrog巳 n ， which
has a temperature of 77 K.认Ih at is thi s temperature on the
(a) Celsius and (b) Fahre由巳it scales?
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and 巳x p巳创11巳n ce the same c h a n g巳 111 忧
t emp巳ratωur，
陀
e. The
length of each rod is the same. If the ini tial length of the
lead rod is O.LO m, what is the initial length of the quartz rod?
c h an g巳 in
6 . 唱r
What's your normal body temperature? It may not be 98.6 oF,
the oft-quoted average 出 at was determ.i ned in the n.ineteenth
century. A more recent study has reported an average te mp巳rature of
98.2 OF. What is the difference between these averages , expressed in
Celsi us degrees?
* 18.
Concrete s id巳wa lk s are always laid in sections , with gaps besection. For exa mple , the drawing shows three identi cal
2.4-m sections, th巳 outer two of which ar巳 against immovab le walls.
The two identical gaps between the secti ons are providecl so that thermal ex pansion will not create the therma l stress that could lead to
cracks. What is the Illinimum gap width necessary to account for an
increase in temperature of 32 C O?
twee n 巳 ac h
!' 7. ~ .A.. ...........
co n s. . tant-volumeσas
thermomet巳 r (see Figures 12.3 and
"..... ..
12.4) has a pressure of 5.00 x 10 3 Pa when the gas te mperature is
0.00 o What is the temperature (i n oc) when the pressure is
2.00 X 10 3 Pa?
‘~
.....， .~.....J ....，
/;)
飞
c.
*8.
If a nonhuman civilization were to develop on Saturn's largest
moon, Titan , its scientists might well devise a te mperature scale
based on the properties of methane , which is much more abundant on
the slllface than water is. M巳thane fre巳zes at - 182.6 oc on Titan , and
boils at -155.2 oC. Taking the boiling point of methane as 100.0 oM
(degrees Methane) and its freezing point as 0 oM , what te mp巳rature
on the Methane scale con.巳 sponds to the absolute zero point of the
Kelvin scale?
ssm On the Rankine temp巳rature scale , which is sometimes used
in engineering applications , the ice point is at 49 1.67 OR and the
steam point is at 67 1.67 oR. Determine a re lationship (analogous to
Equation 12. 1) between the Rankine and Fahrenheit temperaω re
scales.
~ 9.
Section 12.4
Linear Thermal Expansion
10. An aluminum basebaU bat has a length of 0.86 m at a temp巳ra­
ture of 17 o When the temperature of the bat is raised , the bat
length巳n s by 0.000 16 m . Determine the fin al telllperature of the ba t.
c.
11. ssm www Find the approximate length of the Golden Gate
Bridge if it is known that the steel in the roadbed 巳xpands by 0.53 m
when the telllperature changes frolll +2 to +32 o
c.
→{卡­
Gap
司 19 .
The brass bar and the aluBrass
minum b缸 in the drawing are
each attached to an immovable
2.0 m 一一一叫
wall. At 28 oc the air gap between the rods is 1.3 X 10- 3 m
At what temperature will the gap be closed?
' 20. Multiple-Concept Example 4 illustrates the concepts that are pertI nent to 由 is problem. A cylindrical brass rod (cross-sectional area =
1.3 X 10- 5 m2) hangs vertically straight down fro m a cei ling.
When an 860-N block is hung from th巳 lower end of the rod , the rod
Stl 巳 tc h es. The rod is then cooled such that it contracts to its origi nal
length. By how Ill any degrees must the temperature be lowered?
提 2 1.
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13. Conceptual Example 5 provides backgrouncl for this problem. A
hole is clri l/ ecl throllgh a copper plate whose telllperature is 11 o
ssm www A simple pendulum consists of a ball connected to
one end of a thin brass wire. The period of th巳 pend u l um is 2.0000 s
The temperature rises by 140 C O, and the length of th巳 wire increases.
Determine the period of the heated pendulum.
t LHr
e dstLU
nomhh
MP
o k hM
-Steel
AEE
huhv
nu-hu
E
Ot
Aluminum
2h
12. A steel aircraft carrier is 370 m long wh巳 n moving through the
icy North Atlanti c at a temp巳rature of2.0 O By how much d o巳 s the
carrier l e n g th巳n when it is travelin g in the warm Medit巳rranean Sea
at a telηperature of 21 OC?
忏一-2 .4 m 一-叫
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T he drawing shows
4. (
Boiling
two thermometers , A and
B , whose te mperatures are +60.0 OA_ ~ _ E巴巴 - li measured in 0 A and oB .
I I
rI
The ice and boiling points
of water are also indicated.
(a) Using the data in the
draw ing , determine the
Ice
number of B degrees on th巳
point
B scale that co町es po nd to 一30.0 A1 A 0 on th巳 A scale.
(b) If the temperature of a
substance reads +40.0 0 A
A
B
on the A sca l巳， what would
that temperature read on
the B sca l 巳?
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3. On the moon the swface temperature ranges from 375 K during
the day
to 1.00 x 102 K at nigh t. What are these temperatures on the
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23. Consult Conceptllal Exalllple 5 for background pertinent to this
problelll. A lead sphere has a diameter that is 0.050% larger than the
inner diameter of a steel ring when 巳 ach has a temperature of 70.0 o c.
ThllS, the ring will not slip over the s pher巳 . At what common telllperature will the ring just slip over the sphere?
A ball ar们仙1 plate al 巳Illade from differeot mat巳rials and
h av巳 t he same initial telllperature. The ball does not fit throllgh a hole
in the plate , becallse the diameter of the ball is slightly larger than the
diameter of the hole. However, the ball will pass throllgh the hole
when th巳 ball and th巳 plate are both h eat巳d to a common higher temperatme. In each of th巳 arra ng巳ments in th巳 drawing the diallleter of
the ball is 1.0 X 10- 5 m larσer than the diameter of the hol巳 in the
thin plate , which has a diallleter of 0.10 m. The initial temp巳ratllre of
巳ach arrangement is 25.0 oC. At what temperature will the ball j注 II
through the hol巳 in each arrangement?
"24.
(
。
Gold
Lead / '
Aluminum
飞
Steel
SilveιD
飞\
Arrangement A
Arrangement B
Arrangement C
** 25. ssm A steel rlller is calibrated to read true at 20.0
o
c. A drafts
man lI ses the rll 1er at 40.0 oc to draw a lin巳 on a 40.0 oc copper plate.
As indicated on the warm ruler, the length of the line is 0.50 m. To
what telllp巳rature shollld the plat巳 be cooled , sllch that th巳 length of
the line truly becomes 0.50 Ill ?
i'* 26. A steel bicycle wheel (withollt the rllbber tire) is rotating freely
with an anglllar speed of 18.00 rad/s. The temperature of the wheel
changes from -100.0 to +300.0 oC. No net external torq l1 e acts on
the wh巳巳 1 ， and the mass of the spokes is negligible. (a) Does the ang l1 lar speed increase or decrease as the wheel heats IIp? Why?
(b) 明而 at is the angular sp臼d at th巳 hi g h er t巳 mp巳rature ?
不牛 27.
Consult M111tip 1e -Conc巳 pt
Aluminum wire
Example 4 for insight into solving
this problem. An al l1 minllm wir 巳
of radills 3.0 X 10- 4 III is stretched
between the ends of a concret巳
block, as the drawing illllstrates. When the system (wire and concret巳) is at 35 oc , the tension in the wire is 50.0 N. What is the tension in the wire when the systelll is h巳at巳d to 185 OC?
Section 12.5
Volume Thermal Expansion
28. A ftask is filled with 1.500 L (L = liter) of a liq l1 id at 97 .1 o c.
When th巳 liq l1 id is cooled to 15 .0 oc , its volllm巳 is only 1.3 83 L ,
howeve r. Neglect the contraction of th巳 tlask and lI S巳 Tabl e 12.1 to
identify the liquid.
29. Interactive Solution 12.29 at www.wiley.comlcollege/cutnell
presents a Ill odel for solving problems of this type. A thin sph巳 rical
s h巳 11 of silver has an inner rad ills of 2.0 x 10
m when the telllperature is 18 o The shell is heated to 147 o Find th巳 change in th巳
interior vo111me of the shel l.
c.
c.
•
30. A test tube contains 2.54 x 10- 4 1113 of liq l1 id carbon tetrachloride at a temperature of 75 .0 o c. The test tllbe and the carbon t巳tra­
chloride ar巳 cooled to a temperature of - 13.0 oc , which is above the
freezing point of carbon tetrachloride. Find the volllllle of carbon
tetrachloride in the test tub巳 at -13.0 oC.
31. ssm A lead object and a qu 创 tz object 巳ach have th巳 sa lll e initial
vol l1 me. The vo111me of each incr巳ases by th巳 sam巳 alllO l1 nt， beca l1 se
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the telllperaωre increases. If the temperaωre of the lead object increases by 4.0 C O , by how Ill uch does th巳 temperature of thew .quartz
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same temperature. On巳 is steel, and the other is aluminum. The steel
strip
is 0.10% longer than the alulllinulll strip. By how Ill llch shollld
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the telllperature of the strips b巳 i ncreased , so that the strips have th巳
same length?
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32. At a temperature of 0 oc , th巳 mass and volume of a ftllid
are 825 kg and 1. 17 m 3 . Th巳 coeffici巳nt of vo1ume expansion is
1.26 X 10- 3 (CO) - l. (a) What is the density of the ftllid at this temperature? (b) What is the density of the ftuid when the telllperatlll 巳
has risen to 20.0 OC?
33. Consu lt Interactive LearningWare 12.1 at www.wiley.com/
college/cutnell for help in solving this problem. During an all-night
cram session , a student heats up a one-half liter (0.50 X 10- 3 m3)
glass (Pyrex) beaker of co1d coffee. Initia l1 y, the temperatm巳 IS
18 oc , and the beaker is fi ll 巳d to the brilll. A short time later when the
student returns , the temperature has risen to 92 o The coeffici巳 ot of
volume expansion of coffee is the same as that of water. How Ill uch
coffee (in c l1 bic meters) has spilled Ollt of the beaker?
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t h巳 sa!盯
III巳
拟mηlO un t. Th巳 can'、s initial volul
a
川
I口
mη1巳 at 5 oc 归
i s 3.5 X 10- 4 m3 . The co
efficient of vo111me expansion for alumin l1 m is 69 X 10- 6 (CO)- l
When the can and th巳 liquid are heated to 78 oc , 3.6 X 10- 6 11] 3 of
liquid spills over. What is the coefficient of volume expansion of th巳
liquid?
35. ssm Suppos巳 that th巳 steel gas tank in your car is completely
filled when the temperatul 巳 is 17 o c. How many gallons will spill out
ofth巳 twenty-gallon tank wh巳 n the temperature rises to 35 OC?
36. Interactive LearningWare 12.1 at www.wiley.comlcollege/cutnell
provides some L1 seful backgrollnd for this problem . Many hot-water
heating systems have a reservoir tank connected directly to the
pip巳line ， to al1 0w for expansion wh巳 n the water becomes ho t. The
heating system of a hO Ll se has 76 m of copper pip巳 whose inside radius is 9.5 X 10- 3 m. Wh 巳 n th巳 wat er and pipe are heated frolll 24 to
78 oc , what must be the minilllum vo1ume of the reservoir tank to
hold the overftow of water?
ι' 37.
A solid alllminum sphere has a rad ius of 0.50 III and a telllperature of 75 o Th巳 sphere is then completely immersed in a pool of
water whose temperature is 25 o c. The sph巳r巳 cools ， while the watel
telllperature 1 巳ma ll1 sn巳arly at 25 oc , because the pool is very large.
The sph巳re is weighed in th巳 water immediately after being sllb
merged (before it begins to coo l) and then again after cooling to
2
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of mercury in a barometer (see Figure 1 1. 11) has a
height of 0.760 m when the pressure is one atmosph巳re and the temperature is 0.0 oc. Ignoring any change in the glass containing the
mercury, what will be the height of the mercury column for the sam巳
one atmosphere of pr巳 ssure wh巳n the temperature rises to 38.0 oC
on a hot day? (Hint: The pressure in the barometer is given by
Pressure = pgh, and the density p 01 the mercury changes when the
temperature changes.)
Section 12.6 Heat and Internal Energy,
Section 12.7 Heat and Temperature Change:
Specific Heat Capacity
43. ssm Ideally, when a thermometer is used to measure the temperature of an object, the temperature of the object itself should not
change. However, if a significant amount of heat fiows from the object to the th巳rmometer， the temperature will change. A thermometer
has a mass of 31.0 g , a specific heat capacity of c = 815 J/(kg' C勺，
and a temperature of 12.0 oc. It is immersed in 119 g of water, and
the final temperature of the water and thermometer is 41.5 oc. Wh at
was the temp巳rature of the water before the ins巳rtion of the thermometer?
44. If the price of electrical energy is $0.10 per kilowatt. hour, what
is the cost of using electrical 巳nergy to heat the water in a swimming
pool (12.0 m X 9.00 m X 1. 5 m) from 15 to 27 OC?
45. An ic巳 chest at a beach party contains 12 cans of soda at 5.0 oc.
Each can of soda has a mass of 0.35 kg and a sp巳cific heat capacity
of 3800 J/(kg . C O). Someone adds a 6.5-kg watermelon at 27 oC to
the ches t. The specific heat capacity of watermelon is nearly the sam巳
as that of water. Ignore the specific heat capacity of the chest and
determine the final temperature T of the soda and watermelon.
46 . 电r
When you drink cold wat町， your body must exp巳nd meta
bolic energy in order to maintain normal body temperature
(37 oc) by warming up the water in your stomach. Could drinking ice
wat町， then , substitute for exercise as a way to "burn calories?"
Suppose you 巳xpend 430 kilocalories during a brisk hour-long walk.
How many lit巳rs of ice wat巳r (0 oc) would you have to drink in order
to use up 430 kiloc
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;' 5 1. Interactive Solution 12.51 at www.wiley.comlcollege/cutnell
deals with one approach to solving probl 巳ms such as this. A 0.35-kg
coffee mllg is made from a material that has a specific heat capacity
of 920 J/(kg . C O) and contains 0.25 kg of wate r. The cup and water
are at 15 oc. To make a cup of coffeee , a small electric heater is immersed in th巳 water and brings it to a boil in three minlltes. Assllme
that the cup and water always have the same temperature and determine the minimum power rating of this heate r.
* 52. ③ Three 阳tions of the same liquid are mixed in a container
that prevents the exchange of heat with the environmen t. Portion A
has a mass m and a temperature of 94.0 oC , portion B also has a mass
m but a temperature of 78.0 oC, and portion C has a mass mc and a
temperature of 34.0 oc. What must be the mass of portion C so that
tt而且 nal temperature T f of the three-portion mixture is T r = 50.0 OC?
Express your answer in terms of m; for exampl巳， mc = 2.20 m.
封 53.
M lI ltiple-Concept Example 11 deals with a situation that is sim
ilar, but not identical, to that here. When 4200 J of heat are added to
a 0.15-m-long silver bar, its length increases by 4.3 X 10- 3 m. What
is the mass of the bar?
在 54.
The heating element of a water heater in an apartm巳nt building
has a maximum power output of 28 kW. Four residents of the building take showers at the same time , and each receives heated water at
a volume flow rate of 14 X 10 m3/s. If the water going into the
heater has a temperature of 11 oC , what is the maximum possible
temperature of th巳 hot water that each showering resident receives?
•
" 55. ssm Multiple-Concept Example 11 uses th巳 sam巳 physics principles as those 巳mployed in this problem. A block of material has a mass
of 130 kg and a volume of 4.6 X 10- 2 m3 . The material has a specific
heat capacity and coefficient of volume expansion , respectively, of
750 J/(kg . C O) and 6 .4 X 10 (C O) - 1. How much heat must be added
to the block in order to increase its volume by 1. 2 X 10- 5 m3 ?
•
i叫 56.
An insulated container is partly filled with oi l. The lid of the con
is removed , 0.125 kg of wat
t创 ner
Section 12.8
Heat and Phase Change: Latent Heat
57. ssm How much heat must be added to 0 .45 kg of aluminum to
chang巳 it from a solid at 130 oC to a liquid at 660 OC (i ts melting
point)? The latent heat of fusion for aluminum is 4.0 X 10 5 J/kg.
58. To help prevent frost damage , fruit growers sometimes protect
their crop by spraying it with water when overnight temperatures are
expected to go below freezing. When th巳 water turns to ice during the
night , heat is released into the plants , thereby giving a measure of
protection against the cold. Suppose a grower sprays 7.2 kg of wat巳r
at 0 oC onto a fruit tree. (a) How much heat is released by the wat巳r
when it freez巳s? (b) How much would the temperature of a 180-kg
盯ee rise if it absorbed the heat releas巳d in part (a)? Assume that the
specific heat capacity of th巳 tre巳 is 2.5 X 103 J/(kg . C O) and that no
phase change occurs within the tre巳 itsel f.
59. Assume that the pressure is one atmosphere and determine the
heat required to produce 2.00 kg of water vapor at 100.0 oC , starting
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50. A piece of glass has a temperature of 83.0 oc. Liquid that has a
temperature of 43.0 oC is poured over the glass , completely covering wit. d, o
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and the temp巳rature at 巳quilibrium is 53.0 oc. The mass of the glass and
the liquid is the same. Ignoring 由巳 container that holds 出巳 glass and Liquid and assuming 由at 由巳 heat lost to or gained from the surroundings
is negligible, determine 由e specific heat capacity of the liquid.
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41. ssm www Two identical thermometers made of Pyrex glass con, identical volumes of mercury and methyl aIcoho l.
tain , crespectively
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greater is the distanc巳 b巳创tw巳巳n 由
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CHAPTER 12 TEMPERATURE AND HEAT
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60. (
(a) 0均ects A and B have the same mass of 3.0 kg. They
melt wh巳n 3.0 X 10 4 J of heat is added to A and when 9.0 X 10 4 J is
add巳d to B. Determine the latent heat of fusion for the substance from
which each object is made. (b) Find the heat required to melt object
A when its mass is 6.0 kg.
61. ssm Find the mass of water that vaporizes when 2.10 kg of mercury at 205 oC is added to 0.110 kg of water at 80.0 oc.
62. A mass m = 0.054 kg of b巳nzene vapor at its boiling point of
80.1 oC is to be condensed by mixing the vapor with water at 41 oC
What is the minimum mass of water required to condense all of the
benzene vapor? Assume that th巳 mixing and condensation take place
111 a p巳rfectly insulating container
63. 吨，.
A person eats a container of strawbe町 yogurt . The
i Nutritional Facts label states that it contains 240 Calories
(1 Calorie = 4186 J). What mass of p巳rspiration would one have to
lose to get rid of this energy? At body t巳 mp巳rature ， the latent heat of
vaporization of water is 2.42 X 10 6 J/kg
64. A woman finds the front windshield of her car covered with ic巳
at -12.0 oc. The ice has a thickness of 4.50 X 10- 4 m , and the windshield has an area of 1. 25 m2 The density of ice is 917 kg/m 3. How
much heat is required to melt the ice?
65. A thermos contains 150 cm 3 of coffe巳 at 85 oc. To cool the coffee , you drop two ll-g ice cubes into the thermos . Th巳 ice cubes ar 巳
initially at 0 oC and melt completely. What is the final temp巳rature of
the coffee? Treat the coffe巳 as if it wer 巳 wate r.
;, 66.
A snow maker at a resort pumps 130 kg of lake water per minute
and sprays it into the air above a ski run . The water droplets freeze in
出巳 air and fall to the ground , forming a layer of snow. If all the water pumped into the air turns to snow, and the snow cools to the ambient air temperature of -7.0 oC , how much heat does the
snow-making process release each minute? Assume that the temperature of the lake water is 12.0 oC , and use 2.00 X 10 3 J/(kg . C O) for
the specific heat capacity of snow.
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ssm It is claimed that if a lead bullet goes fast 巳nough ， it can melt
completely when it comes to a halt suddenly, and all its kin巳tlC energy
is converted into heat via friction. Find the minimum speed of a lead
bullet (initial temperature = 30.0 oc) for such an event to happen.
ar巳 at 0 oC , and another two grams are
at 100 oc. Heat is removed from the water at 0 oC , completely fre缸"
ing it at 0 oc. This heat is then used to vaporize some of th巳 water at
100 oc. What is the mass (in grams) of the liquid water that remains?
;, 72. Two grams of liquid water
;'* 73. A locomotive wheel is 1. 00 m in diameter. A 25 .0-kg steel band
has a t巳mperature of 20.0 oC and a diameter that is 6.00 X 10- 4 m
less than that of the whee l. What is the smallest mass of water vapor
at 100 oC that can b巳 condensed on the st巳巳 1 band to heat it, so that it
will fit onto the whe巳 I? Do not ignore the water that results from 由巳
condensation.
Section 12.9 Equilibrium Between Phases of Matter,
Section 12.10 Humidity
74. Use the vapor pressure curve
that accompanies 由is problem to
determine the t巳 mperaωre at which
liquid carbon dioxide exists in 巳qUl­
librium with its vapor phase when
the vapor pressure is 3.5 X 10 6 Pa.
8 x 10 6
'"
x 10 0
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ω
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4 x 10 0
l)l
也2
CL2
×
106
75. ssm At a temperature of 10 oC
O
th巳 percent relative humidity is
-50
0
50
R IO , and at 40 oC it is R 40' At each
Temperature , oc
of these temperatures 出e partial
Problem 74
pressure of water vapor in the 缸r is
the same. Using the vapor pressure curve for water that accompanies
this problem , determine the ratio R IO /R 40 of th巳 two humidity values .
8000
7000
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2町的E 且 L
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1063 OC (its melting point). The water boils away, forming st巳am at
100.0 oC and leaving solid gold at 1063 oc. What is the minimum
mass of water that must be used?
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;' 68. ③ Wa阳 at 23.0 oC is sprayed onto 0.180 kg of mo阳 gold at
民U
ssm Ice at -10.0 oC and steam at 130 oC are brought together at
atmospheric pr巳s s ur巳 in a p巳d巳ctly insulated container. After thermal
equilibrium is reach巳d ， the liquid phas巳 at 50.0 oC is presen t.
Ignoring the container and the equilibrium vapor pressure of the liquid at 50.0 oC , find the ratio of the mass of steam to th巳 mass of ice.
The specific heat capacity of steam is 2020 J/(kg . C O)
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字 69.
*70. Occasionally, huge icebergs are found ftoating on the oc巳a n ' s
currents . Suppose one such iceberg is 120 km long , 35 km wide, and
230 m thick. (a) How much heat would be required to melt this iceberg (assumed to be at 0 oc) into liquid water at 0 OC? The density of
ice is 917 kg/m 3 . (b) Th巳 annual energy consumption by th巳 United
States is about 1. 1 X 10 20 J. If this en巳rgy were delivered to the
would it take before the ice
melted?
在 67.
ssm www An unknown material has a normal melting/freezing
point of -25 .0 oC, a nd 由 e liquid phas巳 h as a specific heat capacity of
160 J/(kg' C O). One-tenth of a kilogram of the solid at -25.0 oC is put
into a 0.150-kg aluminum calorimeter cup that contains 0.100 kg of
glycerin. The temp巳rature of the cup and the glycerin is initially
27 .0 oc. All the unknown material melts , and the final temperature at
equilibrium is 20.0 oc. The calorimeter neither loses energy to nor
gains energy from the external environmen t. What is the l at巳nt heat of
fusion of the unknown material?
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76. On a certain 巳vening ， the dew point is 14 oC and the relative hu
midity is 50.0%. How many Celsius degrees must the temperature
fall in order for the relative humidity to increase to 69%? Use the vapor pressure curv巳 for water that accompanies Problem 75 as needed.
As s um巳 that the dew point does not change as the temperature falls
77. 哼，...
Suppose that air in the human lungs has a temperature of
37 oC , and the partial pressure of water vapor has a value of
5.5 X 10 3 Pa. What is the relative humidity in the lungs? Consult the
vapor pressure curv巳 for water that accompanies Problem 75
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A container is fìtted with a movabl巳 pi ston of
and radills r = 0.061 m. Inside the
container is Li qllid water in eqllilibrillm with its vapor， 豁 出巳 drawin g shows. Th巳 piston remains stationary with a 120-kg block on top of i t. Th巳 air
pressure acting on the top of 由巳 piston is one atmos pher巳 . By lI sing th巳 vaporization cllrv巳 for water in
Figllr巳 12.32 ，自 nd the t巳mp巳ratllre of the water.
n巳g ligibl巳 mass
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ssm At a picnic , a glass contains 0.300 kg of t巳 a at 30.0 oC ,
which is the air temp巳 ratllre. To make iced tea , so meo n巳 add s
0.0670 kg of ice at 0.0 oC and stirs th巳 mixture. When all the ice
me lts and th e final t巳 mpera tllre is reach巳d ， the glass begins to fog
IIp , becall s巳 wat巳 r vapor condenses on the ollter glass sllrface.
Using the vapor press lI r巳 cllrve for wat巳r that accompanies Problem
75 , ignoring the specific heat capacity of the glass , and tr巳a tin g the
tea as if it were w ate r，巳 s timate the relativ巳 hllmidity.
中' 84.
A tall colllmn of water is open to the atmospher 巳. At a depth of
10.3 m below the sllrface , th巳 w a ter is boiling. V，而 at is the temperatllre at this depth? Use the vaporization curv巳 for wat巳r in FigllI 巳
12.32 , as needed.
AIDDI Tl ONAL PROBLEMS
85. A steel section of the Alaskan pipeline had a length of 65 m 创ld
a temperatllre of 18 oC when it was installed. What is its change io
length when the t巳 mpera ture drops to a frigid -45 OC?
The latent heat of vaporization of H20 at body temp巳ratllre
ij (37.0 oc) is 2 .42 X 10 6 J/kg. To cool th巳 body of a 75-kg jogger [average specific heat capacity = 3500 J/(kg' CO)] by 1.5 C气 how
many kiJograms of water in th巳 form of s w巳at have to b巳 evaporated ?
86.
"'"苦，..
87. ssm Liqllid nitrogen boils at a chilly -195.8 OC wh巳 n the pres
sllre is on巳 atmosphere . A silver coin of mass 1. 5 X 10- 2 kg and temperature 25 oC is dropped into th巳 boiling Ii qllid. What mass of
nitrogen boils off as the coin cools to -195.8 OC?
88. A 0.200-kg piece of a1l1 minllm that has a temperature of - 155 oC
is added to 1. 5 kg of water that has a temperatllre of 3.0 o At eqllilibrillm th巳 temp巳 rature is 0.0 o Ignoring the container and assllming that the heat exchanged with the sllrrollndings is negli gibl巴，
determine the mass of water that has been frozen into ice.
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89. A commonly lI sed method of fastening one part to another part
is called "shrink fìttin g." A steel rod has a diameter of 2.0026 cm , and
a flat plate contains a hole whose di ameter is 2.0000 cm. The rod is
cooled so that it jllst fìts into the hole. Wh en the rod warms IIp ， 出巳
enormolls th巳 rm a l stress exerted by the plate holds the rod sec ur巳 I y
to the plate. By how many Celsills degrees shollld th巳 rod be cooled?
90. A thin rod consists of two parts joined together. One-third of it is
silver and two-thirds is gold. The temp巳rature decreases by 26 C 。
I1 L
D e termin巳 th巳 frac tional decr巳as巳
in th巳 rod ' s
L o. Sil ver + L o. Gold
length , where L o. Sil vcr and Lo , Gold are the initial lengths of th巳 silve I
and gold rods
91. ssm A copper kettle contains water at 24 o c.认rhen th巳 wa ter IS
heated to its boiling point, the volllme of the kettle expands by
1. 2 X 10- 5 m3 Determine the volllme of th巳 kettle at 24 oC.
92. When yOll t冰e a bath , how many kilograms of hot water (49.0 OC)
mllst yOll mi x with cold water (1 3.0 oc) so that the temperature of the
bath is 36.0 OC? The total mass of water (hot pllls cold) is 19 1 kg
Ignore any heat flow betw巳en the water and its 口t巳rnal sllITollndings
93. ssm What is the relati ve hll I1lidity 00 a day when th巳 t巳mpera­
tllre is 30 oC and the dew point is 10 OC? Use the vapor pressllre
curve that acco mpanies Problem 75 .
功 94.
A copp巳 r-co n s tanta n thermocollple generates a voltage of
4.75 X 10- 3 volts wh巳 n th巳 t巳 mperatllre of the hot jllnction is
110.0 oC and the refere nce jllnction is kept at a t巳 mperature of 0.0 o
lf the voltage is proportional to the difference in temperature betw e巳n
the jllnctions , what is the temp巳rature of th巳 hot jllnction when the
vo ltage is 1. 90 X 10- 3 volts?
c.
地 95.
Interactive Solution 12.95 at www.wiley.comlcollege/cutnell
provides a modeI for solving problems sllch as this. A 42-kg block of
ice at 0 oC is sliding on a hori zontal s lI rface. Th巳 i IlÍ tial s pe巳d of the
I C巳 i s 7. 3 mJs and the fìnal speed is 3.5 m/s. Assllm巳 that the part of
the block th at melts has a very small mass and that all the heat gen巳 rated by kinetic fri cti on goes into the block of ice. D巳 termin e th 巳
mass of i c巳 t hat melts into water at 0 o
c.
*96.
Eqllal masses of two different liqllids have the same temperaωre
of 25.0 o Liqllid A has a fre巳zing point of -68.0 oC and a specific
heat capac ity of 1850 J/(kg' C O). Liqllid B has a freezing point of
96.0 oC and a specific heat capacity of 2670 J/(kg . C 勺 . The same
amollnt of h巳 at mllst be removed from each liqllid in order to freeze
it into a solid at its respectiv巳 fr巳巳zing poin t. De termin 巳 th e differ巳 nc巳 Lr. A 一 Lr， ß b etwe巳 n the lat巳 nt h巳ats of fllsion for the s巳 liqllid s
1' 97.
c.
电曹~
Refer to Interactive LearnìngWare 12.2 at www.wiley.com/
concepts that play roles
in this problem. The box of a well-known breakfast c巳 real states that
one ollnce of the cereaJ contains 110 Calori 巳 s (1 food Calori 巳 =
4186 J). If 2.0% of this e o巳rgy cOllld be converted by a weight lifter's
body into work don 巳 in lifting a barbell , what is th巳 h eav i est barbell
that cOllld b巳 lifted a distanc巳 of 2.1 m?
H college/cutnell for a r巳view of th巳
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' 82. The te mp巳 rature of th巳 a ir in a room is 36 o A perso n tllrn s
on a dehllmidi fier and notic巳 s th at wh巳 n th巳 coolin g coil s 1 巳ac h
30 oC , water begins to co ndense on th 巳 m. What is the relative hllmidity in th巳 room? Use the vapor pre ss ur巳 curve that accompanies
Problem 75
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79. The temp巳ratllre of 2.0 kg of water is 100.0 oC , bllt th巳 wat巳 r is
not boiling , becallse th巳巳xternaJ pressure acting on the water sllrface
is 3.0 X 10 5 Pa. Using the vapor press ure curve for water given in
Figllre 12.32 , determi 日巳 th巳 amollnt of heat that mllst be added to th巳
water to bring it to the point where it jllst b巳gins to boil
A woman has been olltdoors where the temperature is 10 o She
walks into a 25 oC h O ll s巳， a nd h巳r glasses "steam IIp.'' Using the va
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por pressllre curv巳 for water that accompanies Problem 75 , fìnd the c u - t r a
smallest possible va llle for the relative hllmidity of th巳 roo m.
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percentage
of atmosph巳 llC preSSllI 巳 iβs 由
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a t 10 oC is dll巳 to water vapor when th巳 relati ve hllmidity is
100 9盯
(c) The vapor pressllre of water at 35 oC is 5500 Pa. 认rhat
is the rel ative hllmidity at this temperatllre if 出巳 partial pr巳ssure of
water in the air has not chanσ巳
d from what it was at 10 oC when th巳
D
relative hllmidity was 100%?
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TEMPERATURE AND HEAT
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忡 102.
An 85.0-N backpack is hung from the middle of an aluminum
wire , as th巳 drawi n g shows. The temperature of th巳 wire then drops
by 20.0 C"' Find the tension in the wire at the lower temperature.
Assllme that the distance between the sllpports does not chang巳， and
19nor巳 any th巳 rmal stress.
。
kμ 100.
A wir巳 is mad巳 by attaching two segments together, end to 巳 nd.
One s巳gment is made of allllninllm and the other is steel. The e旺ec ­
tive coefficient of linear expansion of the two-segment wire is
19 X 10- 6 (CO)- I. What fraction of the length is aluminum?
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A steel rod (ρ= 7860 kg/m 3 ) has a length of 2.0 m. It is bolted
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at both ends betwe巳 n imrnobile sllpports . Initially ther巳 is no tension
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in th巳 rod ， becallse the rod just fits between the supports. Find the ten
sion that develops wh巳 n the rod loses 3300 J of h巳at
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' 99. ssm A rock of mass 0.20 kσfalls from rest from a hei 2: ht of
15 m into a pail containing 0.35 kg of water. The rock and water have
the sarne initial temperature. The speci 自 c heat capacity of th巳 rock is
1840 J/(kg' C O). Ignor巳 the heat absorbed by the pail itself, and detenni ne the rise in the tem p巳rature of the rock and water
年 半 10 1.
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Multiple-Concept Ex创llple 4 reviews the concepts that are inin this problem. A ruler is accurate when the temperatllre is
volved
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25 oc. When the temperature drops to -14 oC , the ruler shrinks and
no longer m巳asur巳s distances accurately. However, the rlller can be
mad巳 to read correctly if a force of magnitllde 1. 2 X 10 3 N is applied
to each end so as to stretch it back to its originallength . The rlller has
a cross-sectional area of 1. 6 X 10- 5 m2, and it is made from a mat巳­
rial whose co巳fficient of linear expansion is 2.5 X 10- 5 (C O ) 一 1 . What
is Young's modullls for the material from which the rlller is made?
C
ψ 98.
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ADD ITION AL PROBLEMS
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THE TRANSFER OF HEAT
These workers are pouring a (very)
hot liquid into forms . Their helmets ,
aprons , and clothing are designed to
protect them from excess heat. This
chapter considers the processes by
which heat is transferred from one
place to another. (@ Oale O'Oell/
SUPERSTOCK)
F.MWF
CONVECTION
Wh en heat is transferred to 0 1' from a substance , the internal energy of the substance can change , as we saw in Chapter 12. This change in internal 巳 nergy is accompanied by a change in temperature 0 1' a change in phase. The transfer of heat affects us in
many ways. For instanc巴， within our homes furnac巳 s distribut巳 heat on cold days , and air
conditioners remove it on hot days . Our bodies constantly transfer heat in on巳 direction 0 1'
another, to prevent the advers巳巳ffects of hypo- and hyperthermia. And virtually all our energy originates in th巳 sun and is transferred to us ov巳l' a distanc巳 of 150 million kilometers
through the void of spac巳. Today's sunlight provides the energy to drive photosynth巳 sis in
the plants 由 at provide our food and , henc巳， metabolic energy. Anci巳 nt sunlight nurtured
the organic matter that becam巳 the fossil fuels of oil , natural gas , and coal. This chapter
examines the 仕rre巳 process巳 s by which heat is transferred: convection , conduction , and
radiation.
When part of a fluid is warmed , such as th巳 air above a fìre , the volume of that part of
the fluid expands , and th巳 d巳 nsity decreases. According to Archimedes ' principle (see
Section 11.6) , th巳 surrounding cool巳r and dens巳l' fluid 巳xerts a buoyant forc巳 on the
warmer fluid and pushes it upwru.吐 As warmer fluid rises , the surrounding cooler fluid replaces it. This cooler fluid , in turn , is warmed and push巳d upward. Thus , a continuous flow
is established , which carries along heat. When巳ver heat is transferred by the bulk movement of a gas 0 1' a liquid , th巳 heat is said to be transferred by convection. The fluid flow
itself is call巳d a convection current.
CONVECTION
Convection is the process in which heat is carried from place to place by the bulk movement of a fluid.
392
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Conceptual Example
and Refrigerators
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The smoke rising from the industrial fìre in Figur巳 13.1 is one visible result of convection.
Figure 13.2 shows th巳 less visible example of convection curr巳nts in a pot of water
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being heated on a gas burner. The currents distribute the heat from 由巳 burning gas to a11 p缸ts
of the water. Conceptu a1 Example 1 de a1s with some of the important roles 由 at convection
plays in the home
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13.1 CONVECTION
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Water 8aseboard Heating
Hot water baseboard heating units are 仕equently us巳d in homes , and a cooling coil is a m句 or
component of a refrigerator. The locations of these heating and cooling d巳vices are diff，巳rent
becaus巳巳ach is d巳sign巳d to maxirnize 出 e production of convection currents. Where should the
heating unit and the cooling coil be located? (a) Heating unit near the ftoor of the room and
cooling coil ne缸出e top of the refrigerator (b) Heating unit near th巳 ceiling of 出巳 room and
cooling coil near the bottom of the refrigerator
Reasoning An important goal for the heating system is to distribut巳 heat throughout a room.
The analogous goal for the cooling coil is to remove heat from all of the space within a r巳frig­
erator. 1n each case , the heating or cooling device must be position巳d so that convection m 地巳S
the goal achievable.
Answer (b) is incorrect. If the heating unit were placed near the ceiling of the room , warm air
from the unit would remain 出ere ， becaus巳 warm air does not fall (i t rises). Thus , th巳re would
be very little natural movement (or convection) of air to distribute the heat throughout th巳 room .
1f the cooling coil were located n巳 ar the bottom of the refrigerator, the cool air would remain
由ere ， because cool air does not rise (it sinks). There would be very little convection to c缸ry
the heat from other parts of th巳 refrigerator to the coil for remova l.
Figure 13.1 The plumes of thick , gray
smok巳 fro m this industrial fire rise
hundreds of meters into the air because
of convection. (Scott Shawrrhe Plain
Dealer/@APlWide World Photos)
The physics of
heating and cooling by convection.
Answer (a) is correct. Th巳 air above th巳 bas巳board unit is h巳ated， like the air above a fire.
Buoyant forces from the surrounding cooler air push 由巳 warm air upward. Cool巳r air near the
c巳iling is displaced downward and then warmed by 由 e baseboard heating unit , causing the convection current illustrated in Figure 13 .3a. Within the refrigerator, air in contact with the topmounted coil is cooled , its volume decreases , and its density increases. The surrounding
warmer and less d巳nse air cannot provide sufficient buoyant force to support th巳 cooler 组 r，
which sinks downward. 1n the proc巳ss ， warmer air near the bottom of the re仕igerator is displaced upward and is then cooled by th巳 coil ，巳stablishing 由e convection current shown in
Figur巳 13. 3b.
Figure 13.2 Convection currents are set
up when a pot of water is h巳ated .
heating unit
(α)
(b)
Figure 13.3 (α) Air warmed by the
baseboard heating unit is pushed to the
top of 由巳 room by the cooler and
denser air. (b) Air cooled by the cooling
coil sinks to the bottom of 出巳
refrigerator. In both (α) and (b) a
convection current is 巳stablished.
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Another example of conv巳ction occurs wh巳n the ground , heated by 出巳 sun 's rays ,
warms the neighboring air. Surrounding cooler and d 巳 nser air pushes th巳 heat巳 d air up-w . d o
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ward. Th巳 I 巳 sulting updraft or " thermal" can b巳 quite strong , depending on the amount of
heat that the ground can supply. As Figure 13 .4 illustrates , th巳 se thermals can b巳 us巳d by
glider pilots to gain consid巳rable altitud巳 . Birds such as eagles utilize thermals in a similar
fashion
It is usual for air temp巳rature to decrease with incr巳 asing altitude , and the resulting
upward conv巳ction currents are important for disp巳rsing pollutants from industrial sources
and automobile exhaust systems. Sometimes , however, m巳teorological conditions cause a
layer to form in th巳 atmosphere wh巳 re the temp巳rature increases with increasing altitud巳
Such a lay巳r is called an illversioll layer because its temperature profile is inverted com
pared to the usual situation. An inv巳rsion layer aITests the normal upward convection cur
rents , causing a stagnant-air condition in which the concentration of pollutants incr巳 ases
substantially. This condition leads to a smog layer that can oft巳n be s巳en hov巳ring over
large cities.
We hav巳 be巳n discussing Il atural cOllvectioll, in which a t巳 mperature differ巳nce
caus巳s the d巳nsity at one place in a f1 uid to be different from the density at another.
Sometim巳s ， natural conv巳ction is inadequate to transfer sufficient amounts of hea t. 1n such
casesforced cOllvectioll is often us巳 d， and an external devic巳 such as a pump or a fan mixes
th巳 warmer and cool巳r portions of the f1 uid. Figure 13.5 shows an application of forced
convection that is revolutionizing th巳 way in which the effects of overheating are being
tr巳 ated. Athletes , for 巳xampl巳， are esp巳cially prone to overheating , and th巳 devic巳 illus­
trated in Figure 13.5 is appearing more and more frequently at athletic events. The t巳 ch­
nique is known as rapid thermal 巳xchange and takes advantag巳 of specialized blood vessels
called arteriov巳 nous anastomoses (AVAs) that are found in the palms of th巳 hands (and
soles of the feet). These blood v巳ss巳ls ar巳 used to help dissipate unwanted h巳 at from th巳
body. Th巳 device in th巳 drawing consists of a small chamb巳r containing a curv巳d metal
plate , through which cool water is circulated from a refrigerated supply. Th巳 overh 巳 ated
athlete inserts his hand into th巳 chamber and places his palm on the plat巳. The chamber
seals around the wris
The physics of
"thermals."
The physics of
an inversion laye r.
Figure 13.4 Updrafts , or thermals , are
caused by th巳 convective movement of
air that the ground has warmed.
The physics of
rapid thermal exchange.
Figure 13.5 An overheated athlete uses
a rapid-thermal-exchange device to
cool down. He places the palm of his
hand on a curved metal plate within a
slightly evacuated chamber. Forced
convection circulates cool water
through the plate , which cools the
blood flowing through the hand. The
cooled blood returns through veins to
the heart, which circulates it throughout
the body
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CHAPTER 13 THE TRANSFER OF HEAT
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Figure 13.6 shows the application of forced convection in an automobile engine. As
in
the
previous application , forc巳d convection occurs in two ways. First, a pump circulates
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radiator fluid (water and antifre巳ze) through th巳巳ngine to remove excess heat from
the combustion process. Second , a radiator fan draws air through th巳 radiator. Heat is
transferred from the hotter radiator fluid to th巳 cooler air, ther巳by cooling the f1 uid.
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13.2 CONDUCTIONF
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ωUR 川
U ND
唱..DERSTANDI川酌
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(The answer is given at the end of the book.)
1. The transfer of heat by convection is smallest in
(a) solids , (b) liquids ,
(c) gases.
d~ CONDUCTION
Figure 13.6 The forced convection
generated by a pump circulates radiator
fluid through an automobile engine to
remove excess hea t.
-': 11
Anyone who has fried a hamburger in an a lI -metal ski Il et knows that the metal
handle becomes ho t. Somehow, heat is transferred from th巳 burn 巳r to th 巳 handle. Clearly,
heat is not being transferred by the bulk mov巳m 巳 nt of th 巳 m巳tal or the surrounding air, so
convection can b巳 ruled ou t. Instead , heat is transferred directly through the metal by a
process called conduction.
CONDUCTION
Conduction is the process whereby heat is t:ransferr巳d directly through a material , with
any bulk motion of the material playing no role in the transfer.
One m巳chanism for conduction occurs when the atoms or molecuJes in a hotter part
of the material vibrat巳 or move with greater energy than thos巳 in a cooJer par t. By means
of coJJisions , the mor巳巳n 巳rgetic moJ 巳cul巳 s pass on some of the町 energy to their less en 巳 r­
g巳 tic neighbors. For 巳xample ， imagin巳 a gas filI ing the space between two walls that face
each oth巳r and ar巳 maintained at different temperat盯巳 s. MoJecules strike the hott巳 r waIl,
absorb energy from it , and rebound with a greater kinetic 巳 n巳rgy than when they arrived.
As th巳 se more energetic mol巳cuJes coIlid巳 with their less en巳rgetic neighbors , they transfer some of their energy to them. EventuaIl y, this energy is passed on untiJ it reaches the
molecules next to the cooler wal l. These molecules , in turn , collide with the wall , giving
up som巳 of their energy to it in the proc巳 ss. Through such molecular coIIisions , heat is
conducted from the hotter to th巳 cooler wal1.
A similar mechanism for 由 e conduction of heat occurs in metals. Metals are di 旺erent
from most substanc巳 s in having a pool of electrons that are more or less fr臼 to wander
throughout the metal. These free electrons can t:ransport energy and allow metals to transfer heat v巳ry w巳1 1. The free electrons are also r巳 sponsible for th巳巳xceIl巳 nt electrical conductivity that metals hav巳
Those materials that conduct heat weJJ are called thermal conductors, and those that
conduct heat poorly 缸巳 known as thermal insulators. Most metals are exceIIent thermal
conductors; wood , glass , and most plastics are common thermal insulators. Thermal insulators have many important applications. Virtually all new housing construction incorporates thermal insulation in attics and walls to reduce heating and cooling costs. And the
wooden or plastic handles on many pots and pans reduce th巳f1 0w of heat to th巳 cook's hand.
To iJJustrate th巳 factors that influenc巳 th巳 conduction of heat , Figure 13.7 displays a
rectangular bar. Th巳 ends of th巳 bar are in thermal contact with two bodi町， on巳 of which
Figure 13.7 Heat is conducted through
the bar when the ends of 由 eb盯盯E
maintained at different temperatures
The heat flows from the warmer to the
cooler end
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Heat flow
Heat flow
is kept at a constant higher temperature , while the other is kept at a constant low巳r temperature. Although not shown for the sake of clarity, the sides of the bar are insulated , so the
heat lost through them is negligible. The amount of heat Q conducted through the bar from
the warmer end to the cooler end depends on a number of factors:
1. Q is proportional to the time t during which conduction takes place
heat ftows in longer tim巳 periods
(Qα t).
More
2. Q is proportional to the temp巳rature difference !:::.T between the ends of the bar
(Q c汇!:::.T) . A larger difference causes more heat to ftow. No heat ftows when both
ends have th巳 same temperature and !:::.T = 0 C O.
Table 13.1 Thermal Conductivitiesa
of Selected Materials
Substance
Metals
Aluminum
Brass
Copp巳r
Iron
Lead
Silver
Ste巳1 (stainless)
AMATERIAι
The heat Q conducted during a time t through a bar of length L and cross-sectional
areaA is
飞
'vh-
/-
，
一L
Q
T 一
"Except as noted , the values pertain to
temperatures near 20 o c.
CONDUCTION OF HEAT THR。υGH
A
0.090
0.20
1.1
2450
0.80
0.025
2.2
0.010
0.60
0.15
0.040
ductivi，砂.
'飞-
0.0256
0.180
0.0258
0.0265
These proportionaliti巳s can b巳 stated togeth巳r as Q α (A !:::.T) t/L. Equation 13.1 expresses
this result with the aid of a proportionality constant k , which is call巳d the thermal con-
J'
Other Materials
Asbestos
Body fat
Concrete
Diamond
Glass
Goose down
Ice (0 OC)
Styrofoam
Water
Wood (oak)
Wool
240
110
390
79
35
420
14
4. Q is inversely proportional to the length L of the bar (Qα lIL). Greater lengths of
material conduct less heat. To exp巳rience this e仔巳ct ， put two insulated mittens (th巳
pot holders that cooks ke巳p near the stove) on the same hand. Then , touch a hot pot
and notice that it feels cooler than when you wear only one mitten , signifying that
less heat pass巳 s through the greater thickness ("length") of material.
A 一
Gases
AiI
Hydrogen (H 2)
Nitrogen (N 2)
Oxygen (0 2)
Thermal
Conductivity, k
[1/(s. m. C O)]
3. Q is proportional to th巳 cross-sectional area A of the bar (Qα A). Figur巳 13.8 helps
to 巳xplain this fact by showing two identical bars (insulated sides not shown) placed
betwe巳n the warmer and cooler bodies. Clearly, twice as much heat ftows through
two bars as through one , b巳cause the cross-sectional area has been doubled.
'e
'-
(13.1)
where !:::.T is the temperature differenc巳 between the ends of th巳 bar (the higher temperature rninus the lower temperature) and k is the thermal conductivity of the mat巳rial .
SI Unit of Thermal
Conductiv句:
1/(s . m . C O )
Since k = QLI(tA !:::.T) , the SI unit for thermal conductivity is 1. m/(s . m 2 • C O) or
1/(s' m . C O). The SI unit of power is the joule per second (1/s) , or watt (W) , so the thermal
conductivity is also given in units of W /(m . CO).
Different materials hav巳 different thermal conductivities , and Table 13.1 gives some
representative values. Because metals ar巳 such good thermal conductors , they have larg巳
thermal conductivities. In comp缸ison ， liquids and gas巳s generally have small th巳rmal conductivities. In fact , in most ftuids the heat transferred by conduction is n吨 ligible compared
to that transferred by convection when there ar巳 strong convection currents. Air, for instance , with its small thermal conductivity, is an excellent thermal insulator when confined
to sm a1 1 spaces where no appreciabl 巳 convection currents can be established. Goose down ,
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Figure 13.8 Twice as much heat flows
through
two identical bars as through
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1 3 THE TRANS FE R O F HEAT
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Styrofoam , and wool d巳rive th巳ir fin巳 insulating prop巳rties in p征t from the small dead-air
spaces
within them , as Figure 13.9 illustrates. The physics of dressing warmly. We also take
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advantag巳 of dead-air spaces when we dress "in layers" during very cold weather and put
on several layers of relatively thin clothing rather than one thick layer. The air trapp巳d
between th巳 layers acts as an excellent insulator.
Example 2 deals with the role that conduction through body fat plays in regulating
body temp巳rature.
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Heat Transfer in the Human Body
th巳rmal
conductivity of body fat is given as
Q = (kA !1T) t
一
L
Q
[020
1/(s-m-c。)](17II12)(37O 。c
L - -- - -- ,- --- 0.030 m
/ .J \ . - -
-34.0 。C)(1800
)
-", - --- 5
-/
6.1|X4
10 4
J
For comparison , a jogger can generat巳 ov巳r ten times this amount of h巳at in a half hour. Thus ,
conduction through body fat is not a particularly effective way of removing excess hea t. Heat
transfer via blood ftow to th巳 skin is more 巳ffl巳ctive and has the added advantage that the body
can vary the blood ftow as n巳eded (see Problem 7)
户、
Reasoning and Solution In Table 13.1 the
k = 0.20 J/(s. m. CO ). According to Equation 13.1 ,
户、
within the body, it must be transferred to the skin and dispersed if the temperature
at the body interior is to be maintained at th巳 normal valu巳 of 37.0 。巳 On巳 POSSl­
ble mechanism for transfer is conduction through body fa t. Suppose that heat travels through
0.030 m of fat in reaching the skin , which has a total surface area of 1. 7 m2 and a temperature
of 34.0 oc. Find the amount of heat that reaches the skin in half an hour (1 800 s).
~
Figure 13.9 Styrofoam is an excellent
thermal insulator because it contains
many small , dead-air spaces. These
small spaces inhibit heat transfer by
conv巳ction currents , and air itself has a
very low thermal conductivity.
。
Example 3 uses Equation 13.1 to deterrnin巳 what the
the warmer and cooler ends of the bar in Figure 13.7.
temp巳rature
is at a point between
…………………………………………………………………………………………………………………………………… .
A NALVZING MULTIPLE-CONCEPT PROBLEM
Example 3
The Temperature at a Point Between the Ends of a Bar
In Figure 13.7 the temperatures at the ends of the bar are 85.0 oc at the warmer end and 27 .0 oc at the cooler end. The bar
has a length of 0.680 m. Wh at is the temperature at a point that is 0.220 m from the cooler end of the bar?
Reasoning The point in question is closer to th巳 cooler end than to the warmer end of the bar. It rnight be expected , therefore ,
that th巳 temperature at this point is less than halfway between 27.0 oc and 85 .0 oC. We will demonstrate that this is , in fact , the
case , by applying Equation 13. 1. This expression applies because no heat escapes through the insulated sides of the bar, and we
will us巳 it twice to deterrnine th巳 desired t巳mperature.
Knowns and Unknowns The available data are as follows:
Description
Temperature at warmer end
Temperature at cooler end
Length of bar
Distance from cooler end
Unknown Variable
Temperature at distance D
from cooler end
Symbol
Value
Tw
Tc
L
D
85.0 0 C
27.0 0 C
0.680 m
0.220 m
T
q
Continued
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The Conduction of H创 Theh创 Q conducted in a time t 阳t the point in
question (which is a distance D from the cooler end of 由e bar) is given by Equation 13.1 as
Q=~A(T 一 Tc)t
-
D
''
A
.
、‘ J
QD
T= 耳+~一一
，，‘、
where k is th巳 thermal conductivity of th巳 material from which the bar is made , A is th 巳
bar's cross-sectional ar巳a ， and T and Tc are , resp巳ctively， 由 e temperature at the point io
question and at the ∞oler end of the bar. Solving for T gives Equation 1 at the righ t. The
variables Q , k , A , and t are unknown , so we proce巳d to Step 2 to deal with them.
kAt
E圃 川n灿
d也1u归削
E伽n叫ofH
川阳阳
ωat山巾阳川川川
e阅
阳阳灿叫
Ree肝
附v材由闹
M
州
i沁i剑i
si刷
t优
ted 川臼创叫叫叫
山咖
tω
仙
Qω
由
that
灿川川川
a创川山川
t川川川……
盯
iiscs oωc
∞O
-A
、 1 ，J
rs-、
po
创io
川t io qu 巳s创t
叫ion tωo the ∞
c 00ωle
衍r end of the bar originates at the war
肌
rmτn陀阳巳创町r end of the bar
Thus , since no heat is lost through the sides , we may apply Equation 13.1 a secood time
to obtain an expression for Q:
I ~ kA(Tw - Tdt I
l
.
..
一
I
!
L
where Tw and Tc ar巳， respectively， 出e temperatures at 由巳 warmer and cooler ends of the bar,
which has a length L. This expression for Q can b巳 substituted into Eq uation 1, as indicated
at 由e righ t. The terms k , A , and t remainωbe dealt with. Fortunately, however, values for
出em are unnecessary, because 由ey can be eliminated alg巳braic a1ly from 出e 缸lal calculatioo.
Solution Combining the results of each step algebraic a1l y, we fìod that
...,..…,....
,
,
T='= Tr
OD
+ 二二一 =
kAt
Tr
I
+-=
kA(Tw - Tc )~ID
L
」
Simplifying this result gives
k A (Tw T = Tr
= 27.0 oC
As
exp巳cted ，
L
+
kA t
+
Tc)t ~
= Tr +
(Tw - Tc) D
L
o
(85.0
C - 27.0 oC) (0.220 m)
口7ττ刀
,- - - - - - - ,,- -- - -- , = 145.8 OCI
0.680 m
'-一一一」
this temperature is less than halfway
betw巳en
c.
27.0 oC and 85.0 o
Related Homework: Problem 16
Virtually all homes contain in sul ation in the walls to reduce heat loss. Example 4
how to determine this loss with and without insulation.
illustrat巳 s
Inside
(25.0 0 C)
Figure 13.10 H巳at f1 0ws through the
insulation and plywood from the
warmer inside to the cool巳r outside.
Example 4
Layered Insulation
The physics of layered insulation. One wall of a house consists of 0.019-m-thick plywood
backed by 0.076-m-thick insul ation , as Figure 13. 10 shows. Th巳 temperature at the inside
surface is 25.0 oC , while the temp巳 rature at the outside surface is 4.0 oC , both being constan t.
The thermal conductivities of the insulation and th巳 plywood ar毡 ， r毡 spectively， 0.030 and
0.080 1/(s . m . CO ) , and the area of th 巳 waIl is 35 m2 . Find the heat conduct巳 d through th巳 wall
in on巳 hour (a) with the insulation and (b) without the insulation.
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Modeling
the Problem
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CHAPTER 13 THE TRANSFER OF HEAT
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Th巳 temp巳rature T at th巳 insulation-plywood interface (se巳 Figure 13.10) must b巳
before th巳 h巳at conducted through th巳 wall can be obtained. 1n calculating this temc u -tr
perature , we use th巳 fact that no heat is accumulating in th巳 wall because the inner and outer
t巳mp巳ratures are constan t. Therefore , th巳 h巳 at conducted through the insulation must 巳qual the
Qplywood' Each
heat conduct巳d through th巳 plywood during th巳 same time; that is , Qin s山山 011
ofth巳 Q values can be expressed as Q = (kA è!. T)t/L , according to Equation 13 .1, leading to an
expression that can be solved for the int巳rfac巳 t巳mperature. Once a value for T is available ,
Equation 13.1 can be used to obtain th巳 heat conducted tlu-ough the wall.
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Problerr卜 solving insigh t: When heat is conducted through a multi-Iayered material and the high and low temperatures are constant , the heat conducted through each layer is the same
Solution (a) Using Equation 13 .1 and the fact that Q川slllation
Qplywood, we find that
[旦旦1l
1 slI latiol'n 1 [旦旦1v
L
Le.
L
L"
oI .., ., =
[0.030 J/(s'm'C O )]A(25.0 oc - T) t
[0 .080J/(s'm. C" )]A(T - 4.0 oC) t
0.076 m
0.019 m
Note that on each side of th巳 equals sign we have written è!. T as the higher temperature minus
the lower temp巳rature . Eliminating th巳 area A and time t algebraically and solving this equation
for Treveals that the temperature at the insulation-plywood interface is T = 5.8 oc .
Th巳 heat conducted through th巳 wall is either Qi川 latiol1 or Qplywood, since the' two quantities
ar毡巳qual. Choosing Qinsulatio l1 and using T = 5.8 oc in Equation 13.1 , w巳自 nd that
Q Însulation0.030 J/(s ' m . C )](35 的(25.0 C - 5.8 oC) (3600 s
0.076 m
O
0
L=I 9.5 X
10 5 J
I
(b) It is straightforward to use Equation 13.1 to calculat巳 the amount of heat that would f1 0w
through the plywood in on巳 hour if the insulation w巳r巳 a bsen t:
Qplywood
川OJ巾 mm5ZZY 。c-40 。C)(36005L=I 110 x 的|
Figure 13.11 During the night of
25 , 2005 , temperatures in some
areas of F1 0rida dipped below freezing.
Farmers sprayed strawbe盯y plants with
water to put a coat of ice on them and
illsulate 出巳m against the subfreezing
t巳 mpera阳 res. (Chris O ' Meara!@AP.刑ide
World Photos)
Janu缸 Y
Tubular space contalning
refrigerant fluid
Without insulation , the heat loss is increased by a factor of about 12.
The physics of protecting fruit plants from freezing. Fruit growers som巳 times protect
their crops by spraying th巳m with water wh巳n overnight temperatur巳 s ar巳 exp巳cted to drop
below fr巳ezing. Som巳 fruit crops , like th巳 strawberries in Figure 13.11 , can withstand temp巳ratures down to fre巳zing (0 oc), but not below freezing . When water is sprayed on the
plants , it can fre巳Z巳 and release h巳 at (see Section 12. 时 ， som巳 of which go巳s into warming
th巳 plan t. 1n addition , both water and ic巳 hav巳 r巳lativ巳 Iy small thermal conductivities , as
Table 13 .1 indicates. Thus , they also protect th巳 crop by acting as thermal insulators that
reduce heat loss from the plants .
Although a layer of ice may be b巳 neficial to strawberri 白 ， it is not so desirable inside
ar巳frigerator， as Conceptual Example 5 discusses.
Conceptual Example 5
An Iced-up Refrigerator
1n a refrigerator, heat is removed by a cold refrig巳rant f1 uid that circulates within a tubular space
embedded inside a metal plate , as Figure 13.12 illustrates. A good refrig巳rator cools food as
quickly as possible. Which arrangement works b巳 s t: (a) an aluminum plate coated with ic巳，
(b) an aluminum plate without ice , (c) a stainless steel plat巳 coat巳 d with ice , or (d) a stainless
steel plate without ic巳?
Reasoning Figure 13.12 (see the blow-ups) shows the metal coobng plate with and without a
layer of ice. Without ice , heat passes by conduction through the metal plate to the refrigerant
f1 uid within. For a given temperature difference across the thickness of the metal , the rat巳 of
heat transfer depends on the thermal conductivity of th巳 meta l. When th巳 plate becom巳 s coated
with ice , any heat that is remov巳 d by 出巳 refrig巳rant fluid must first b巳 transferred by conduction through th巳 ice before it encounters the metal plate.
Refrigerant
fluid
Heat
Metal
Ice
Figure 13.12 In a 1 巳frigerator， cooling
is accomplished by a cold refrigerant
ftuid that circulates through a tubular
spac巳巳 mbedded within a metal plate
Sometimes the plate becomes coated
with a layer of ice.
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and (d) are incorrect. For answers (a) and (c) , the re lation Q
= 一一τ一一
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(Equation 13. 1) indicates that the heat conducted per unit time (Q/t) is inversely proportional
to the thickness L of the ice. As ice builds up , th巳 heat removed per unit tim巳 b y th巳 cooli n g
plate decreases. Thus, when covered with ice, the cooling pJate- regardless of whether it's
mad巳 from alurninum or stai nless steel-d oes not work as well as a plate that is ice-fTee .
An sw巳r (d)-the stai nless steel pl at巳 w ith o ut ice- is incorrect, because heat is transferred more
readily through a plate that has a greater thermal conductivity, and s tainl 巳ss steeJ has a smaller
th巳rm al conductivity than does aluminum (s巳巳 Tabl e 13. 1).
Th 巳 re l ati o n
(kA t1T) t
13.1) shows that th巳 h eat
L
conducted per unit time (Q/t) is directly proportional to 出巳 therrnal conductivity k of the metal pJate.
Sinc巳 the therrnal conductivity of alurninum is more than 17 也nes greater th an 由巳 thermal conductivity of stainless steel (see Table 13. 1), alurninum is the preferred pJate. The alurninum plate
arrangement works best without an ice buildup. Wh en ic巳 builds up , th巳 heat removed p巳r unit time
decreases b巳c au s巳 of the increased thickness of material through which 出e heat must pass.
Answer (b) is correct.
Q
= 一一一一一一 (Equ atio n
。
~ CHECK VOUR UNDERSτANDING
(The answers are given at the end ofthe book.)
2. A poker used in a fireplace is held at one end , while the other end is in the fire. In terms
of being cooler to the touch , should a poker be made from (a) a high-thermal-conductivity
material , (b) a low-thermal-conductivity material , or (c) can either type be used?
3. Several days after a snowstorm , the outdoor temperature remains below freezing . The
roof on one house is uniformly covered with snow. On a neighboring house , however, the
snow on the roof has completely melted. Which house is be忧er insulated?
4. Concrete walls often contain steel re inforcement bars. Does the steel (a) enhance ,
(b) degrade , or (c) have no effect on the insulating value of the concrete wall? (Consult
Table 13.1.)
5. To keep your hands as warm as possible during skiing , should you wear mittens or
gloves? (Mittens , except for the thumb , do not have individual finger compartments.)
Assume that the mittens and gloves are the same size and are made of the same materia l.
You should wear: (a) gloves , because the individual finger compartments mean that th e
gloves have a smaller therma l conductivity; (b) gloves , because the individual finger
compartments mean that the gloves have a larger thermal conductivity; (c) mitte ns ,
because they have less surface area exposed to the cold air.
6. A water pipe is buried slightly beneath the ground. The ground is covered with a thick layer
of snow, which contains a lot of small dead-air spaces within it. The air temperature suddenly
drops to well below freezing. The accumulation of snow (a) has no effect on whethe r the water in the pipe freezes , (b) ca uses the water in the pipe to freeze more quickly than if the
snow were not there , (c) helps prevent the water in the pipe from freezing .
在 Some an imals have ha ir strands that a re hollow, air-filled tubes. Othe rs have hair
strands that are solid. Which kind , if e ithe r, wou ld be more likely to give an animal an
advantage for surviving in very cold climates?
8. Two ba rs are placed betwee n plates whose temperatures are Thot and 兀 old (see the drawing). The
thermal conductivity of bar 1 is six times that of bar
2 (k, = 6k2 ) , but bar 1 has only one-third the crosssectional area (A , = ~ A2 )' Ignore any heat loss
through the sides of the bars. What can you
conclude about the amounts of heat 0 , and O2 ,
respectively, that bar 1 and ba r 2 conduct in a given
amou 川 oftir阿
me
(怡
c)
0 , = 20 2
(d) 0 , = 40 2 (e) 0 , = O2
9. A piece of Styrofoam and a p
m
(剑， (C) ,
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RADIATION
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Energy from the sun is brought to earth by large amounts of visible light waves , as
well as by substantial amounts of infrared and ultraviolet waves. These wav巳 s are known
as electromagnetic waves , a class that also includes th巳 microwaves used for cooking and
the radio wav巳 s used for AM and FM broadcasts. Th巳 sunbather in Figure 13.13 f，巴巴ls hot
b巳cause her body absorbs 巳 nergy from the sun's electromagnetic waves. Anyone who has
stood by a roaring fire or put a hand near an incand巳 scent light bulb has 巳 xperienced a
similar eff，巳c t. Thus , fir巳s and light bulbs also 巳mit el巳ctromagn巳 tic waves , and when the
energy of such waves is absorb巳 d ， it can hav巳由e same eff，巳ct as heat
Th e process of transferring 巳nergy viael巳ctromagnetic waves is called radiation, and , unlike conv巳ction or conduction , it do巳s not requ让e a material medium. Electromagnetic waves
from the sun , for exampl巳， travel through 由e void of space during their joumey to 阳th
RADIATION
Radiation is
th巳 process
in which energy is transferred by means of electromagnetic
wav巳 s.
All bodies continuously radiate energy in the form of electromagnetic waves. Even an
ice cube radiates energy, although so little of it is in th巳 form of visible light that an ice
cube cannot be seen in th巳 dark. Likewise , th巳 human body emits insufficient visible light
to be seen in the dark. However, as Figures 12.6 and 12.7 illu stra饵， the infrar巳d waves radiating from th巳 body can be det巳cted in the dark by electronic cameras. Generally, an obj 巳ct do巳 s not emit much visible light until th巳 temperatur巳 of th巳 obj 巳ct 巳xceeds about
1000K. Then a characteristic red glow appears , like that of a heating coil on an electric
stove. When its temp巳rature reaches about 1700K, an obj 巳 ct begins to glow whit巳 -hot， like
the tungsten filam巳 nt in an incandescent light bulb.
1n th巳 transfer of 巳nergy by radiation , the absorption of electromagnetic waves is just
as important as 出巳ir emission. The surface of an obj 巳ct pl ays a significant role in determining how much radiant energy the object will absorb or emi t. The two blocks in sunlight in Figure 13.14, for example , are identical , except that on巳 has a rough surface coated
with lampblack (a fine black soot) , while the other has a highly polished silver surface. As
th巳 thermometers indicat巳， the temperature of the black block rises at a much faster rat巳
than that of th巳 silvery block. This is because lampblack absorbs about 97% of the incident radiant energy, while the si lv巳ry surface absorbs only about 10%. Th巳 remaining part
ofth巳 incident energy is reflected in 巳 ach case. We s巳e the lampblack as black in color because it reflects so little of th巳 light falling on it, while th巳 silv巳ry surface looks like a mirrorb巳caus巳 it reflects so much ligh t. Since the color black is associat巳d with nearly
complet巳 absorption of visible light , the term pe矿ect blackbody or, simply, blackbody is
us巳d when referring to an 。同巳ct that absorbs all the electromagnetic waves falling on i t.
Allobjects 巳mit and absorb elec位omagnetic waves simultan巳ously. VI厅len a body has the
same constant t巳mp巳rature as its sUlToundings , the amount of radiant energy being absorbed
must balance the amount being emitted in a given int巳rval of time. The block coated with
lampblack absorbs and emits th巳 same amount of radiant ener町， and the silvery block does
too. In either case , if absorption were greater than emission , the block would experience a
net gain in energy. As a result, the temperature of the block would ris巳 and not be constan t.
Simila
Figure 13.13 Suntans ar巳 produced by
ultraviolet rays. (Ron Chapple厅hinkstockl
Alamy lmages)
sZ; •
111
Lampblack-coated
Silver-coated
block
block
Figure 13.14 The temperature of the
block coated with lampblack rises
faster than the temperature of the block
coated with silver becaus巳 the black
surface absorbs radiant en巳rgy from the
sun at the greater rate.
The physics of
summer clothing.
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Q=σT 4At
The relationship above holds only for a pe出ct emitter. Most obj ects are not p巳rfect
however. Suppose that an 0均巳ct radiates only about 80% of the visible light energy that a perf，巳ct emitter would radiate , so Q (for th巳 object) = (0.80)σT 4At. The factor
such as the 0.80 in thi s eq uation is called the emissivity e and is a dim 巳 n s ionJess numb巳r
betwe巳 n zero and on 巳. The ernissivity is the ratio of the en 巳 rgy an obj 巳ct actually radiates
to the 巳
valu巳 of e foαr 由
t h巳 humηlan body , for instance , varies between about 0.65 and 0.80 , the
small巳r values pertaining to ]j ghter skin colors. For infrared radiation , e is n 巳 ar忖 on巳 fo r
all skin colors. For a perfect bl ackbody emitter, e = 1. lncluding th巳 factor e on th 巳 right
si d巳 of the expression Q =σT切 leads to the Stefan- Boltzmann law of radiation.
ernitte邸 ，
(b)
Figure 13.15 (α) Most lemurs , like this
one , are noctu ll1 al and have dark fur.
(Wolfgang Kaehler/Corbis-lmages)
(b) The species of lemur called th巳
wh ite sifaka, h ow巳V缸 ， is activ巳 during
the day and has white fur. (Nigel Dennis/
Wildlife Pictures/Peter Arnold , Inc.)
THE STEFAN-BOLTZMANN LAW OF RADIATION
The radi ant 巳nergy Q, ernitted in a time t by an object that has a Kelvin
a surface area A , and an emissivity e, is given by
temp eraωre
Q=eσT 4At
T,
(13.2)
where σ= 5.67 X 10- 8 1/(s' m2 . K4) is the Stefan-Boltzmann constant.
In Equation 13.2 , the St巳fan-Bo ltzmann constant σis a uni versal constant in the sense that
its value is the same for alI bodies , regardless of the nature of their surfaces . The ernissivity e , how巳ver， d ep巳 nd s on the condition of the surface.
Example 6 shows how th巳 Stefa n-Boltzmann law can be used to deterrnine th巳 siz巳 of
a star.
Example 6
A Supergiant Star
The supergiant star Bete l g巳 u s e has a surface temperature of about 2900 K (about one-half
that of our sun) and emits a radia nt power (in joules per second , or watts) of approximately
4 X 10 30 W (about 10 000 times as great as that of our sun). Assuming that Betelgeus巳 IS a
perfect emÜter (emissivity e = 1) and sp h巳ri cal ， fìnd üs radius.
Reasoning According to 由巳 Stefan - Boltzmann law， 由巳 power emitted is Q/t = eσT 4A . A star
with a relativ巳ly smaU temperatllre T can hav巳 a relatively large radiant power Q/t only if the area
A is large. As we will 优
s E巳吼， B巳创t优巳
e时l g巳 lI S巳 has a v巳I叩
y large 凯
川
S ur臼c巳 a
缸r巳a ， so its radius is enormolls.
m
lic
C
m
k
lic
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w
(lI)
The u s巳 of light colors for comfort also occurs in n at ur巳. Most I巳murs ， for instance ,
w
.c
.d o
are nocturnal and have dark fur like th 巳 lemur shown in Figure 13.15a. Since they are acc u -tr a c k
tiv巳 at night , th巳 dark fur poses 0 0 disadvantag巳 in absorbing excessive sunlight. Figure
13 .15b shows a species of lemur call巳d the white sifaka , which lives in serniarid regions
where th巳re is Ii ttle shade. Th巳 whi
讪it臼巳 coloωr of the fur ma
叮
yh巳
elp in 由
t h巳创rm
n巳cting sunlight , but during the cool mornings , reflection of su nIi ght would hinder warming up. However, th 巳 se I 巳 murs have black skin and only s p 创 S 巳 fur on their be Il ies , and to
warm up in the morning , they turn their dark bellies toward th 巳 sun . The dark color
enhances the absorption of su nlight.
Th巳 amount of radiant energy Q ernitt巳d by a perfect blackbody is proportional to the
radiation time interval t (Q α t) . The longer the time , th巳 gr巳 ater is the amount of 巳 n巳rgy
radiated . Exp巳 rim巳 nt shows that Q is also proportional to the surface area A (Qα A). An
object with a larg巳 s urfac巳创'ea radiates more 巳nergy than one with a small surface area ,
other things being equa l. FinalI y, experiment reveals that Q is proportional to the fourth
power ofthe Kelvi n. tern.perature T (Q cx T句 ， so the emitted energy increases markedly with
increasing temperature. If, for example , the Kelvin temperature of an o bj 巳c t doubles , th巳
object ernits 2 4 or 16 times more 巳n巳rgy. Combining these factors into a single prop Oltiona Ii ty, w巳 see that Q cx T 4 At. This proportiona]j ty is convert巳d into an eq uation by inserting
a proportionality constantσ" known as the Stefa n. -Boltzrn. a n.n. constant. It has been fo und
experimenta lI y thatσ= 5.67 X 10- 8 1/(s . m2 . K4):
w
The physics of
c
k . sifaka lemur warming up.
ac uwhite
-tr a c
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THE TRANSFER OF HEAT
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CHAPTER 唱 3
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th巳 ar巳 a ， w巳自 nd
c
A
= - Q/t
EσT 4
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Solution SoIving the Stefan-Bo ltzmann Iaw for
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13.3 RADIATION F 403
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But the s Ulfac巳缸巳a of a sphere is A = 4 7Tr 2, so r = VA瓦石. Therefore, we have
4 飞川。飞
A
A
4 7T (1)[5 .67 X lO- ð J/(s. m L • K4)](2900 K)'
For comparison , Mars orbits the sun at a distanc巳 of 2.28 X 10
"supergian t."
11
m. Betelgeuse is certain ly a
o
The next 巳xampl巳 explain s how to apply the St巳fan-Boltzmann law when an
such as a wood stove , simultaneously emits and absorbs radiant 巳 nergy.
0
Problem-solving insight
== | 3 X 10 11 m 1
First solve an equation for the unknown
in terms of the known variables. Then
substitute numbers for the known
variables, as this example shows
0时 ect ，
阳
E
xam
叩
p盯 A
阳n Un川edW
帕阳
∞
O
叫
O
d
stove has a constant 忧
t emp巳 ra
曰tωur陀它 of 18 oc (ρ291 K) , whi比
ch iβs aIso 由
t h巳
of 由
t h巳 room in whiωch 由
th巳 s剑tov巳 stands. Th巳 stove has an emissivity of 0.900 and a
sllrface area of 3.50 m 2 . What is the net radiant power generated by the stove?
An
unu s巳d woodι-b
趴
bu川
rning
t忧阳巳创m
引1p巳ratωur陀!它巳
Reasoning Power is th巳 change in 巳n 巳rgy p巳 r unit time (Equation 6.10b) , or Q/t, which , according to the Stefan-Boltzmann law, is Q/t = eσT4A (Equation 13.2). In this problem , however, we ne巳d to find the net power produced by th巳 stov巳. Th 巳 net power is the power the stove
emits minll s 由巳 power th巳 stove absorbs. The pow巳r th巳 stove absorbs comes from the waI1 s ,
ceiling , and floor of the room , all of which emit radiation.
Solution Rem 巳 mb巳ring that temperature must
Stefan-Boltzmann law, we 白 nd that
b巳 express巳d
in kelvi ns
wh 巳 n
using the
Power emitt巳d
by u巾巳ated
丰 = euT 4 A
stove at 18 ocι
=
(1 3.2)
(0.900)[5.67 X 10- 8 J/(s . m2 • K 4)](29 1 K)4(3.50 m2)
=
Problem-solving insight
In the Stefan-Boltzmann law of radiation ,
the temperature Tmust be expressed in
kelvins , not in degrees Celsius or degrees
Fahrenhei t.
1280 W
The fact that the unheated stove emits 1280 W of power and yet maintains a constant t巳mp巳r­
atllre means that 由巳 stov巳 also absorbs 1280 W of radiant power from its surrollndings. Thus ,
the nel power generated by the lI nheated s tov巳 is zero
Net power
by
stove at 18 oc
g巳nerated
1280W
Example 8
A
1280W
、~
Power emitted
by stove at
18 C
Power 巳 mitted by
room at 18 oC and
absorb巳d by stove
0
A [~Ð.'..~.
-
、~
M
LT I P L E
Heated Wood-Burning Stove
=
10wI
嗣 [11:tR.. :I:a
_:JI illl :.1 :1 \'J]
一­
~
The physics of
a wood - burning stove.
The wood-buming stove in Example 7 (emissivity = 0.900 and surfac巳缸巳a = 3.50 m 2) is
being us巳d to heat a room. The fìre keeps th巳 stove surface at a constant 198 oC (471 K)
and 由巳 room at a constant 29 oC (302 K). Determine the net radiant power g巳 nerated by th巳 stov巳 .
Reasoning Power is th巳 change in en巳rgy per unit tim巴， and according to th巳 Stefan-Boltzmann law, it is Q/t = εσ-y 4A
(Equation 13.2). As in Example 7 , however, w巳 seek the net powe几 which is the power the stove emits minus the power the stov巳
absorbs from its 巳nvironmen t. Since 由e stove has a higher temperature than its environment, the stove ernits more power than it
absorbs. Thus , we wiU fìnd 由 at the net radiant pow巳r effil 忧.ed by th巳 stove is no longer zero , as it is in Example 7. ln fact , it is
th巳 net power 由 at the stove radiat巳 s that has warmed th巳 room to its temperature of 29 oC and sustains that temperature.
Continued
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Emissivity of stove
Surface area of stove
Temp巳rature of stove surfac巳
Temperature of room
Value
e
A
T
0.900
3.50 m2
198 oC (471 K)
29 oC (302 K)
To
Unknown Variable
Net power generated by stove
m
.d o
Comment
Symbol
T巳mp巳ratur巳 in
kelvins must be us巳d .
Temp巳rature in kelvins must be used.
?
P net
Modeling the Problem
J
.
P._;".., - P .ho"巾，、A
、、
Pn.'
嘈 BA
d
that the stove ernits minus the power Pabsorbed that 由e stov巳 absorbs from its environment,
as expressed in Equation 1 at the righ t. In St巳ps 2 and 3 we eva1 uat巳 the emitted and
absorbed powers.
r ·、
…
回罩I Net Power The net 阳时net generated by the stove is the power P,
占 4
E固I Emi削 P
阳
0仰
we盯r Acc
∞C叫n吨gμ
刊
叩叭
川
tω
O ωt阳一』扣缸拍
阮础
B
Bolt
刷1让tzm
删
盯m
n
由
t ha
刽t iβ
s
''A
、
stove (temperature T) ernits a powe町r
f
t白
h巳 heated
J
，
，、、
几旦旦l竖立二 EσT 4 A
毡， 川
This expression can be substituted into Equation 1, as indicated at the righ t. In Step 3,
we discuss the absorbed power.
E固!l 蝴
A bs阳
0呐
r巾bed
巾 Pow
阳
e创町r The 时圳
radian
川
削
m
a创
川
nlt 阳川1川叫ove 抽
a bsor
翩 巾bs fr刀
.om
川
n
i沁s
iden tical tωO 由巳 power 白
t ha
剖t 由
t h 巳 stove would 巳盯1山
i扰t 创
at 由
th巳 constant room 怡
tβmpe
缸ra
创tu
旧
1江n毡~ of
29 oC (302 K). The reasoning here is exactly like that in Example 7. With To representing
th巳 temperature (in Kelvins) of 出e room , the Stefan-Boltzmann law indicates that
p汕sorbed
?
result into Equation 1, as shown in the right column.
Solution The results of each step can be
akus
AM
-
A
T A
、υ
P
A甸
A
ρL
V
aLOS
algebraically to show that
σ
DZ
combin巳d
AU哼
--t ed,
吧
substitute 由 is
~"
4
吧
We can also
飞~ab町、 rbed
---一 =eσT0 A
Thus , the net radiant power the stove produces from the fuel it burns is
P net = eσ A(T 4
-
= 0.900[5.67
T04 )
X
10- 8 J/(s . m2 . K4)](3.50 m2 )[(471 K)4 一 (302 K)4]
= I7.30
X
10 3
wl
Related Homework: Problems 25, 27, 30, 38
Exampl巳 8 illustrates that when an object has a higher temperature than its surroundings , the object 巳mits a net radiant power P net = (Q/t) net. The net power is the power the
object ernits minus the power it absorbs. Applying the Stefan-Boltzmann law as in
Ex缸nple 8 leads to th巳 following expression for P net when the temperature of the object is
Tand th巳 temperature of the environment is To:
P ne, = e σA(T 4
-
T 04 )
(1 3.3)
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Description
C
Knowns
and Unknowns The following data are available:
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CH A PTER 13 THE T RANSFER O F HEAT
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404
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U 酌NDERSTA
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T
One way that heat is transferred from place to place inside the human body is by
the flow of blood . Which one of the following heat transfer processes-forced
convection , conduction , or radiation-best describes this action of the blood?
11. Two strips of material , A and B, are identical, except that they have emissivities of 0 .4
and 0.7, respectively. The strips are heated to the same temperature and have a red glow.
A brighter glow signifies that more energy per second is being radiated. Which strip has
the brighter glow?
12. One day during t he winter the sun has been shining all day. Toward sunset a light
snow begins to fal l. It collects without melting on a cement playground , but melts instantly
on contact with a black asphalt road adjacent to the playground. Why the difference?
(a) Being black , asphalt has a higher emissivity than cement, so the asphalt absorbs more
radiant energy from the sun during the day and , consequently, warms above the freezing
point. (b) Being black , asphalt has a lower emissivity than cement , so it absorbs more
radiant energy from the sun during the day and , consequently, warms above the freezing
pomt.
10.
~
13.
If you were stranded in the mountains in cold weather, it would help to minimize
losses from your body if you curled up into the tightest ball possible.
Which factor in the relation 0 = e σ严At (Equation 13.2) are you using to the best advantage by curling into a ball? (a) e (b)σ(c) T (d) A (e) t
~'"
￥
energy
14. Two identical cubes have the same temperature. One of them , however, is cut in two and the
pieces are separated (see the drawing). The radiant energy emitted by the cube cut into two
pieces is Otwo pieces and that emitted by the uncut
cube is Ocube ' What is true about the radiant energy
emitted in a given time? (a) Otwo pieces = 20cube
(b) Otwo 阳es = ~ Ocube (c) Otwo 阳臼= Ocube
(d) Otwo pieces = ~ Ocube (e) Otwo pieces = ~ Ocube
Cube cut into
two pieces
Uncut cube
15. Two objects have the same size and shape. Object A has an emissivity of 0.3, and
object B has an emissivity of 0.6. Each radiates the same power. How is the Kelvin temperature TA of A related to the Kelvin temperature 马 of B? (a) 瓦 = 兀 (b) 瓦 = 2TB
(c) TA = ~ TB (d) 瓦 = fi TB (e) TA = 恒 TB
APP Ll CATIONS
To ke巳p heating and air conditioning bills to a minimum , it pays to use good thermal insulation in your ho me . l nsulation inhibits convection between inner and outer walls
and minimizes heat transfer by conduction. With respect to conduction , the logic behind
hom巳 in s ulati on ratings comes directly from Equation 13. l. According to this equation , the
heat p巳r unit time Q/t ftowing through a thickness of material is Q/t = kA !:::..T/L. Keeping
the value for Q/t to a minimum means using m aterials that have small thermal conductivities k and large thicknesses L. Construction e n g in巳巳 rs ， how巳ver， prefer to use Equation
13. 1 in the slightly di ff，巳re nt form shown below:
Q
A!:::..T
L/ k
The term Llk in th巳 denomin ator is called the R value. An R value expresses in a single number the combined effects of thermal conductivity and thickness. Larger R values
r巳duce the heat per unit time ftowing through the material and , therefore , mean better insulation. It is also convenient to use R values to describe l ayer巳d slabs formed by sandwiching together a number of m aterials with diffl巳rent thermal conductivities and different
thicknesses. The R values fo r th巳 individual layers can be added to give a single R value
for the entir巳 slab. It should be noted , how巳ver， that R values are expressed using units of
feet, hours, F", and BTU for thickness, time, temperature, and heat, respectively.
When it is in the 巳arth ' s shadow, an orbiting satellite is shielded from the intense electromagnetic waves emitted by the sun. But when it moves out of the earth's shadow, the
satellite experienc巳S 由巳 full e旺巳ct of these wav巳s . As a result , the temperature within a
The physics of
rating thermal insulation
by R values.
The physics of
regulating the temperature
of an orbiting satellite.
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(The answers are given at the end of the book.)
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APP Ll CATIONS 405
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Figure 13.16 Highly reflective metal
foil covering this satellite (the Hubble
Space Tel巳sco pe) minimizes
temperature changes. (Courtesy NASA)
Stopper
Gasket
Silvered surfaces
时I
GONGEPTS & GAlGUlATIONS
-('1)
四
Heat transfer by conduction is gov巳rn 巳d by Equation 13.1 , as w巳 have seen. The next
example il\ ustrates a familiar application of this relation in the kitchen. It also gives us the
opportunity to review the idea of latent heat of vaporization , which Section 12.8 discuss巳s.
Hot or
cold
liquid
Gasket
Glass
Figure 13.17 A thermos bottle
minimizes energy transfer due to
convection , conduction , and radiation.
Concepts
&
Calculations Example 9
Boi li ng W ater
Two pots are identical ，巳xc巳pt that in one case th 巳 flat bottom is aluminum and in th 巳 other it is
coppe r. Each pot contains th巳 sam巳 amount of boiling wat巳r and sits on a h巳atin g el 巳m巳nt that
has a t巳mp巳rature of 155 oC. 1n th巳 aluminum-bottom pot, th巳 water boils away compl巳tely in
360 s. How long does it take th巳 wat巳r in the copper-bottom pot to boil away completely?
Concept Questions and Answers Is th巳 h 巳 at n巳巳 ded to boil away the water in the aluminumbottom pot less than , greater than , or the same as the h巳at needed in th巳 copper-bottom pot?
Answer Th巳 heat Q n 巳ed ed is th巳 same in 巳 ach cas巳 When water boils , it changes from
the liquid phase to the vapor phase. Th巳 heat that is required to mak巳 th e water boil away
is Q = mLv , according to Equation 12.5 , where m is the mass of th 巳 water and Lv is the latent h巳 at of vaporization for wate r. Since the amount of water in each pot is the same , th 巳
mass of water is the sam巳 in each case. Moreov巳r， th巳 latent h巳 at is a characteristic of
water and , th 巳refor巳， is also the same in 巳 ach cas巳.
One of the factors in Equation 13 .1 that influences the amount of heat conducted through the
bottom of each pot is the temp巳 ra ture differenc巳 ðTb巳 tween the upper and lower s urfac巳 s of
the pot's bottom. Is this temperature di仔巳renc巳 for the aluminum-bottom pot less than , greatel
than , or the sam巳 as that for the copper-bottom pot?
Answer For each pot th巳 t巳mp巳rature di 仔巳rence is the same. At th巳 upper surface of each pot
bottom 由巳 t巳 mperature is 100.0 oC , becau s巳 water boils at 100.0 oC under normal conditions
of atmosph巳ric pressure. The t巳 mp巳ratlll 巳 remain s at 100.0 oC until all 出巳 water is gone. For
each pot th巳 temp巳rature at th巳 lower s urfac巳 of the pot bottom is 155 oC , the temp巳rature of
the h巳ating elemen t. Therefore , ðT = 155 oc - 100.0 oC = 55 C O for each pot
Figu re 13.18 In a halogen cooktop ,
emit a large
amount of 巳l巳ctromagnetic energy that
is absorb巳d dir巳ctly by a pot or pan.
qu创 tz-iodin巳 lar口ps
Is the time required to boil away the water completely in th巳 copper-bottom pot
greater than , or the same as that required for the alu l1l1num-botto l1l pot?
1 巳 ss
than ,
Answer The time is less for the copper-botto l1l po t. The factors that influence th巳 amount of
heat conducted in a given tim巳创 e th巳 therl1l a l conductivity, the ar巳a of the bottom ，由巳 tem
perature differ巳nc巳 across the bottom , and the thickness of the botto l1l . All of these factors ar巳
m
o
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c u -tr a c k
satellite would d巳cr 巳as巳 a nd increase sharply during an orbital period and s巳 nSl lLve 巳 lec­
w
.c
.
tronic circuitry would s u旺町， unl 巳 ss precautions are taken. To I1l jnimjze t巳 mperature ftuc-d o c u - t r a c k
tuations , satellites are often covered with a highly reft 巳cting and , hence , poorly absorbing
metal foil , as Figure 13.16 shows. By reflecting much of th 巳 sunlight ， th 巳 foil minimi zes
te l1l peratur巳 rises. Being a poor absorber, th 巳 foil is also a poor e l1l itter and reduces radiant 巳 nergy losses. Reducing these losses keeps th巳 t巳mp巳ratur巳 from falling 巳xc巳 ssively
The physics of a the rm os bottle. A ther l1l os bottle , sometimes referred to as a Dew 盯
ftask , reduc巳 s the rat巳 at which hot liquids cool down or cold liquids warm up. A th 巳 rmos
usually consists of a doubl巳 - wall巳d glass v巳 ssel with silvered inner walls (s巳巳 Figure
13.17) and minimizes heat transfer via conv巳ction ， conduction , and radiation . The s pac巳
betw巳en th巳 walls is 巳vacuat巳 d to reduc巳巳 nergy loss 巳 s due to conduction and conv巳ction
The silv巳red surfac巳 s r巳ftect most of the radiant energy that would oth 巳rWlse 巳 nt巳r or leav巳
the liquid in the thermos. Finally , ljttle heat is lost through the glass or th巳 rubberlik巳 gas ­
kets and stopper, becau s巳 the s巳 materials have relatively small thermal conductiviti 巳 s.
The physics of a hal ogen cooktop stove . Halogen cooktops use radiant energy to h 巳 at
pots and pans. A halog巳n cooktop us巳s sev巳ral quartz-iodine lamps , like the ones in ultrabright automobile headlights. Th巳s巳 lamps 缸·巳 el 巳 ctrically power巳d and mounted below a
ceramic top. (See Figure 13 .1 8.) Th 巳巳 lectromagnetic en 巳rgy they radiate pass巳 s through
the c巳ramic top and is absorbed directly by the bottom of the po t. Cons巳qu 巳 ntly， th巳 pot
heats up very quic k1 y, rivaling the time of a pot on an open gas burner
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THE TRANSFER OF HEAT
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Solution Applying Eqllation 13.1 to th巳 Cοnduction of heat into both pots and lI sing Equation
12.5 to express 由巳 heat need巳d to boil away the water, we hav巳
Q AI
AI
=
QCC lIlI =
J队 IA D. T)队l
mL ,
JkcuA D.T) tcu
mL
In 由巳S巳 two 巳qllations the area A ， 由e temperatllre di旺。 enc巳 D. T， the thickness L of th巳 pot bottom ,
the mass m of the water, and the latent heat of vaporization of watel ιhave the sam巳 vaJ lIes .
Ther巳fore， w巳 can set the two heats eq l1 al and obtain
(kA1A D. T)t A1
L
(k c lI A D.T) t c
\"-\..lI"
-~ ' ''\..u
L
or
kr"t户H
k^ ,t^ ,
Solving for tC lI and taking vallles for the thermal cond l1 ctivities from Table 13.1 , we fìnd
I
As expected , th巳
一主血I
ωk ClI
boil-away
日五百
s1
巳二三二」
O
[240 J/(s ' m. C )](360 s)
390 J/(s' m . CO )
=口 20
time is Jess for the copper-bottom po t.
Heat transfer by conduction is only one way in which heat gets from place to place
Heat transfer by radiation is anoth 巳r way, and it is gov巳 rned by th巳 Stefan-Boltzmann law
of radiation , as Section 13.3 discuss巳 s. Exampl巳 10 deals with a case in which heat loss by
radiation leads to freezing of wat巳r. It stresses th巳 importance of the 创'ea from which th巳
radiation occllrs and also provides a [巳vi巳w of the id巳 a of lat巳 nt heat of fllsion , which
Section 12.8 discusses .
Concepts & Calculations Example 10
Freezing Water
One half of a kilogranl of liquid water at 273 K (0 OC) is placed outside on a day when 由巳 temper­
is 261 K (一 12 oc). Ass l1 me that heat is lost 仕om the water only by means of radiation and
that the emissivity of the radiating sLllface is 0.60. How long does it take for the water to fr巳 eze II1 tO
ice at 0 oC when the surface area from which the radiation occurs is (a) 0.035 m2 (as it could be in
a cup) and (b) 1. 5 m2 (as it could be if 由e water were sp山ed out ωform a thin sheet)?
aωre
Concept Questions and Answers In case (a) is th巳 heat that must be removed to
water less than , greater than , or the same as in cas巳 (b)?
freez巳 th巳
Answer The heat that must b巳 removed is the same in both cases. Wh巳 n water freezes , it
changes from the liquid phase to the solid phase. The heat that mllst be removed to m 创e
the water freeze is Q = mL f , according to Eq l1 ation 12.5 , wh巳 re 111 is the mass of the water and Lf is the latent heat of fusion for water. The mass is the same in both cases and so
is L f , since it is a characteristic of the water
The loss of heat by radiation depends on the temperatur巳
perature of the water change as the water freezes?
of
the radiating
obj 巳c t. Do巳 s
the tem-
Answer No. The temperature of the water does not change as the fl古巴zing process t对<es
place. The heat removed serves only to change th巳 water from the liqllid to the solid phase ,
as S 巳ction 12.8 discusses. OnJy after all the water has frozen does the temperatllre of the
lC巳 begin to fall beJow 0 o
c.
The water both Joses and gains
to freezing of the water?
h巳 at
by radiation. How, then , can
h巳 at
transfer by radiation lead
Answer The water freezes because it loses more heat by radiation than it gains. The gain occurs becallse the environment radiates heat and 出巳 water absorbs it. However, the temperature
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the same for each pot except for the thenn aI conductivity, which is greater for copper (Cu)
than for aluminllm (Al) (see Table 13.1). The greater the 出巳rmal condllctivity, the greater the
heat that is conduct巳d in a given tim巴， oth巳r things being equal. Therefore, less tim巳 is 1 巳qllired
to boil away th巳 water lI sing 由巳 copper-bottom po t.
to
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13.5 CONCEPTS & CALCULATIONS
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WiU it 时也 longer for the water to
the area is larger?
freez巳 in
of 由巳 water.
case (a) when the
aJ巳a
is
As a result, the env ironwater.w . d o
radiation 丘om 由 e
small 巳 r
or in case (b) when
It wiU take longer when the area is sma ll 巳r. This is becaus巳 th巳 amount of energy radiated in a given time is proportional to th巳 area from which the radiation occurs
A smaller 缸 ea means that less en 巳rgy is radiated per second , so more time will be required
to freez巳 the water by removing heat via radiation.
Answer
Solution We use Eq uation 13.3 to 时也 into account that th巳 water both gains and loses heat
via radiation . This expression gives 由 e net power lost, the net power being the net heat divided
by the time. Thus , we have
主 = eoA(俨 - 1ò4 )
V
or
t =
/
e σA(T 4 -
1♂)
Using Eq uation 12.5 to 巳xpress the heat Q as Q = mLf and taking the latent heat of fusion for
water from Table 12.3 (Lf = 33.5 x 104 J/kg) , w巳自 nd
(a) Smaller area
mL ,
t
=eσA(T 4 二 T0 4 )
4
(0.50 kg)(33.5 X 10 J/kgI ，~ . .，~<5 u ~ ，， 1
=1 1.5 X 10 s (42h)1
2
4
8
0.60[5.67 X 10- J/(S' m . K )](0.035 m 2)[(273 K)4 - (261 K ) 4 ] |
(b)
t
La贝ger
area
m L,
=eσA(T 4 二 T04 )
4
(0.50 kg)(33.5 X 10 J/kg)
I " r " ", , _" " u I
= 13.6 X 10 3 s (1 .0 h)1
0 .60[5.67 X 10- 8 J/(S' m 2 . K 4)]( 1. 5 m 2)[(273 K)4 一 (26 1 K)4] |
As expected ,
th巳 fre巳zing
time is longer when the area is smaller.
CONCEPT SUMMARV
If you need more help with a concept, use the Learning Ai ds n ot巳dn巳xt to the discussion or equation. Examples (Ex.) are in the tex t
of this chapter. Go to www.wiley.com/college/cutnell for 由e following Learning Aids:
Interactive LearningWare (ILW) -
Interactive Solutions (l S) -
Additional examples solved in a fìve-step interactive forma t.
An imated text figures or animations of important concepts
Concept Simulations (CS) -
Models for certain types of problems in the chapter homework. The calcul ations are carried out interactively.
。 iscussion
Topic
Learning Aids
13.1
CONVECTJON Convection is the process in which
the bulk movement of a fiuid
h巳at
is carried from place to place by
Nalural conveclion
During natural convection , the warmer, less dense part of a fiuid is pushed upward by the buoyant Ex. 1
force provided by the surrounding cooler and d巳 nser part.
Forced conveclion
Forced convection occurs when an external device , such as a fan or a pllmp, causes
由巳 fi llid
to move
13.2 CONDUCTJON Condllction is the process whereby heat is transferred directly throllgh a
material , with any bulk motion of the material playing no role in the transfer
Thermal conduclors and
Ihermal insulalors
Materials 由 at
由 at
conduct heat well, such as most metals, are known as thermal condllctors. Materials
conduct heat poorly, such as wood , glass , and most plastics , are referred to as thermal insulators.
Th巳 heat
Conduclion 01 heal
lhrough a malerial
Q conducted during a time t throllgh a bar of length L and cross-sectional area A is
Q=J坠主旦L
L
where !l T is the t巳mperature difference between
tivity of the material.
th巳 end s
of th巳
bar
m
T
of heat due to
Ex.2 , 3 , 4 , 5 , 9
(13.1) CS 13.1
ILW 13.1
and k is the thermal conduc- IS 13.9, 13.17
o
th巳 loss
k
th巳 t巳 mp巳rature
radiation cannot offset completely
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CHAPTER 13 THE TRANSFER OF HEAT
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13.3 RADIATION
mag netlc waves
Rad iation is the process in which
en巳rgy
is transferred by means of electro-
Absorbers and emitters
AlI objects , regardless of their temp巳 rature ， simultaneously absorb and emit electromagnetic
waves. Objects that ar巳 good absorbers of radiant energy are also good emitters , and objects that
are poor absorbers are also poor emitt巳 rs.
Ape叫ecl blackbody
An object th at absorbs all the radiation incident lI pon it is ca ll巳dap巳rfect blackbody. A perfect
blackbody, being a perfect absorber, is also a perfect e mitt巳r.
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Di scussion
Topic
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FOCUS ON CONCEPTS 409
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The radiant energy Q 巳mitted during a time t by an object whose surface area is A and whose
Kel vin te mperature is T is given by the Stefan-Boltzmann law of radiation :
Slelan-Bollzmann law
01 radialion
Emssv
Q =eσT 4 At
"'
(1 3.2) EX. 6
. m2 .
w hereσ = 5.67
J/(s
K4) is the Stefan-Boltzmann constant and e is the emissivity, a
dimensionless nllmber characterizing the surface of the 。同巳ct. The emissivity lies between 0 and
1, being zero for a nonemitting surface and one fo l' a perfect blackbody.
X 10- 8
The n巳 t radi ant power is the power an object emits minus the power it absorbs. The net radiant
Ex. 7, 8, 10
power P Il C1 emitted by an object of temperature T located in an environment of temp巳rature To is
Nel radianl power
PIl C1
eσA(T 4
-
1♂(1 3 . 3) ILW 13.2
FOCUS ON CONCEPTS
Note ω Instructors: The numbering of Ih e queslions shown here reflecls Ih e βlCI Ihal Ihey a陀 only a rep resenlα tive subsel of Ihe tolalnumber Ihm are
available online. Howeve r, all of Ihe questions are available for assignmenl viα an online homework management program sllch as Wil巳yPLUS or "告bA ssign
Section 13.2
6. The long single bar on the left in the drawing has a th巳rma1 conduc-
Conduction
1. Th e heat condllcted through a bar depends on which of the foUowing ?
A. The
co巳fficient
c.
of linear expansion
B. The thermal conductivity
C. The specitìc heat capacity
D. The length of the bar
E. The cross-sectional area of the bar
(a) A, B , and D (b) A , C , and D
and E (e) C , D , and E
tivity of 240 J/(s' m ' C O ) . The ends of the bar are at temperatures of
400 and 200 oc , and the temperaωre of its midpoint is halfway between these two temperatures , or 300 o The two bars on the right
are half as long as the bar on the left , and the thermal conductivities
of these bars are different (see the drawing). All of 由巳 bars have the
same cross-sectional area. What can be said about the temperature
at the point where the two bars on the right are joined together?
(a) The temperature at 由e point wh巳re the two bars are joined together is 300 o C. (b) The temperature at the point where th巳 two
bars are joined together is greater than 300 oc. (c) The temperature
at the point where the two bars are joined together is less than 300 o
(c) B , C , D , and E
(d) B , D ,
2. Two bars are condllcting heat from a region of higher temperature
to a region of lower temperature. The bars have identical lengths and
cross-sectional are筒， but are made from different materials. 1n the
drawing they are placed "in parallel" between the two temperature reg lO ns 111 arrang巳ment A , whereas they are placed end to end in
arrangement B. 1n which arrangement is the heat that is conducted the
greatest? (a) The heat conducted is the same in both arrangements.
(b) Arr angement A (c) Arr angement B (d) It is not possible to
determine which arrangement conducts more hea t.
o OC
100 0 C
c.
400 C
200 C
0
0
400 C
200 C
0
0
kj = 240 J/(s . m . CO)
k2 = 120 J/(s . m . CO)
Section 13.3
Radiation
8. Three cube s 缸e made from the same mat巳rial. As th巳 drawing indicates , they have diff，巳rent sizes and temperatures. Rank the cubes
according to the radiant 巳nergy they emit per second , l argest 自 rs t.
(a) A , B , C (b) A , C , B (c) B , A , C (d) B , C , A (e) C , B , A
A
4. The drawing shows a composite slab
consisting of three mat巳rial s through
which heat is conducted from left
to righ t. The materials have identical
thicknesses and cross-sectional areas
Rank the materials according to their
thermal conductivities , largest first.
(a) k\ ， 也， k3 (b) k\ ， 也， k 2 (c) 屿， k\ , k3
(d) 仇，句 ， k\ (e) 屿， 仇 ， k\
B
B
C
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11. Th巳巳 missivity e of object B is 古 that of 0均 ect A , although both
objects 副 e identical in size and s hap巳. If the objects radiate the same
energy per second , what is the ratio Tß/TA of their Kelvin temp巳ra­
tures? (a) 古 (b)
(c) ~ (d) 2 (e) 4
i
PROBlEMS
Note
10
Illstructors: Most of the hOl11 ework proble l11 s in this chapter are
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See
in
this
set， 时 e
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l11 anagement
program sllch
the valuesfor therl11 al CO l1 d u. ctivities given in Table 13.1 unless stated otherwise
ssm Solution is in the Student Solutions Manua l.
www Solution is available online at www.wiley.comlcollege/cutnell
1.
assignment via an online homework
S
川
P re，
φf'a ce fo
ωradd
仇1IμIOnα 1 det阳
ails
Note : For problel旧
Section 13.2
α vailable for
(iþ
唱贺'
~ This
~
icon rep l'esents a biomedical application.
Conduction
The amollnt of heat per second condllcted from the blood
LO
U capillaries ben巳ath the skin to the surface is 240 J/s. The enm through a body whose
ergy is transferred a distance of 2.0 x
surface area is 1.6 m2 Assllming that the thermal condllctivity is that
of body fat , det巳nrun巳 the temperature difference between the capillaries and the sllrface of the skin.
10- 3
3Lo
c.
2. The temp巳rature in an electric oven is 160 o The t巳mperature at the
ollter sllrfac巳 in the kitchen is 50 o The oven (s ll1face area = 1.6 m 2 )
is inslllated with mat巳rial that has a thickness of 0.020 m and a thermal conductivity of 0.045 J/(s' m . CO ). (a) How much energy is
used to operate the oven for six hOllrs? (b) At a price of $0.10 per
kilowatt . hour for electrical energy, what is th巳 cost of op巳rating the
c.
A
B
ov巳 n?
3. ssm Concept Simulation 13.1 at www.λ再riJ ey.comlcollege/cutnell
illustrates the concepts pertinent to this problem. A person's body is
covered with 1.6 m2 of wool cI othing. The thickness of the wool is
2.0 X 10- 3 m. The temperature at the olltsid巳 surface of the wool is 11 oc ,
and the skin t巳mperature is 36 o How much heat per second does
the person lose dlle to condllction?
c.
T'woo州均巳圳
cαωtωs are
陀毛 ma
阳
11a
4. (iþ 刊
and one cold. Two id巳 ntical bars can be attached end to end , as in part
a of the drawing , or one on top of the other, as in part b. When either
of these anang巳m巳 nts is placed between the hot and the cold objects
for the same amollnt of time, heat Q f1 0ws from left to right. Find th巳
ratio Q/Qb
(a)
(b)
5. ssm Dll巳 to a temperature difference t1 T , heat is condllcted
through an alllminllm plate that is 0.035 m thick. The plate is then
replaced by a stainless steel plate that has th巳 same temperature difference and cross-sectional area. How thick shollld the steel plate b巳
so that the same amollnt of heat per second is conducted through it?
川
aw
呐111吨g 阳 dirr
盯阳
T
6. ③ The block in 阳 dωra
wh 巳re
巳 Lo = 0.30 mη. The block has a thermal conductivity of
250 J/(s ' m' CO ). ln drawings A, B , and C , heat is condllcted throllgh
the block in three different directions; in each case the temperatllre of
the warmer sllrface is 35 oc and that of the cooler sllrface is 19 oc
Determine the heat that f1 0ws in 5.0 s for 巳 ach case
Problem 6
C
www In the condllction eqllation Q = (kA t1 T)此， the
combination of factors kA /L is called the conductance. The
hllman body has the ability to vary the condllctance of the tisslle beneath the skin by means of vasoconstriction and vasodilation , in
which the f1 0w of blood to the veins and capillaries llnd巳rlying the
skin is decreased and incr巳ased ， respectively. The condllctanc巳 can be
adjllsted over a range sllch that the tissu巳 beneath the skin is 巳qlliva­
lent to a thickness of 0.080 mm of Styrofoam 0 1' 3.5 mm of air. By
what factor can the body adjllst the conductance?
7.
唱铲 ssm
Ií
8. A copper pipe with an ollter radills of 0.013 m runs from an out
door wall faucet into the int巳rior of a hOllse. The temperature of the
faucet is 4.0 oc , and the temp巳rature of the pipe , at 3.0 m from the
fallcet , is 25 o ln fifteen minutes , the pip巳 condllcts a total of 270 J
of heat to the olltdoor fallcet from the hOll S巳 interior. Find the inner
radills of the pip巳. Ignor巳 any wat巳r insid巳 the pip巳
c.
9. Conslllt Interactive Solution 13.9 at www. wiJ ey.comlcollege/cutneU
to explore a model for solving this problem. One end of a brass bar is
maintained at 306 oc , while the other 巳nd is kept at a constant, bllt lower,
temperature. The cross-sectional area of the b缸 is 2.6 X 10- 4 m2
Becallse of inslllation , th巳re is negligible heat loss through the sides
of the bar. Heat f1 0ws throllgh the bar, however, at the rate of 3.6 J/s .
Whatis th巳 temperature of the bar at a point 0.15 m from the hot end?
10. ( A wall in a hOllse contains a single window. The window
consists of a single pane of glass whose area is 0 .l 6 m2 and whose
thickness is 2.0 rrun. Treat the wall as a slab of the inslllating material
o
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k
same rate. (d) It is not possible to determine which object cools
w
down at the faster rate
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10. An astronaut in the space shuttle has two objects that 缸巳 identi ­
cal cin
. c all respects , except that one is painted black and the 0出 er IS
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painted silver. Initially, they are at the same temperature. When taken
from inside the space shuttle and placed in outer space , which object ,
if either, cools down at a faster rate? (a) The obj 巳ct P创 nted black
(b) The object painted silver (c) Both objects cool down at the
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CHAPTER 13 THE TRANSFER O F HEAT
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Iron
to
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不 ι 18 .
T he draw ing shows a solid cy lindrical rod made from a center
cylinder of lead and an o ut巳 r co ncentric j acket of coppe r. Except for
its ends , th巳 ro d is insul ated (not show n) , so th at the loss of heat
from the curved surface is negligible.
Cop per
Lead
When a temperatu re d i 仔"ere n ce is
m a inta in巳d between its e nds , thi s rod
conducts one-half the amount of heat h
that it would condll ct if it were solid
r2
coppe r. Determine the rati o of the
radii rl/r2'
半 1' 19.
1'12. A cubi cal piece of heat-shie1d til e from the space shuttle measures
0.10 m on a side and has a thermal conductivity of 0.065 J/(s . m . CO).
Th巳 o uter surface of th巳 til e is heated to a t巳 mp巳rature of 1150 oc ,
while the inner surface is maintained at a temperature of 20.0 oc.
(a) How much heat f1 0ws from the outer to the inn巳 r s u rfac巳 of the tile
m 自 ve minutes?
(b) If this amount of heat were transferred to two
liters (2.0 kg) of liquid water, by how many Celsius degrees w0 1l1d the
te mp巳ratllre of the wat巳r rise?
O
也 13. ssm Three buiJding materi als, plasterboard [k = 0.30 J/(s . m . C )] ,
O
brick [k = 0.60 J/(s . m . C )], and
Plaster- 8rick
wood [k = 0.10 J/(s ' m' C O)] , are
sandwi ched together as the drawing illustrates. The temperatu res at
由巳 in s id e and outside surfaces are
27 oc and 0 oc , respectively. Each Inside
Outside
material h as 由 e same thickness (27 OC)
(O OC)
and cross-secti onal area. Find the
te mpera tur巳
(a) at the plasterboard-bri ck interface and (b) at
th巳 bri ck-wood interface
出盹 ③ Acop阳 rod has a 1巳ngth of 1.5 m and a cross-section山ea
of 4.0 X 10- 4 m2. On巳 end of the rod is in contact with boiling water
and the other with a mi xtllre of ice and wate r. What is the mass of ice
per second that melts? Ass ume that no heat is lost through the side
surface of the rod.
咀
*叫
1臼5. ~ Ap阳
O创t 们盹
阳r is boiling 1I川川
时
n1叫d巳
毗町r on盯t盯m 叫
o h阳巳町
ere
忧
陀巳 O向
f印p附 ure
A岱
s s饥
创
u川
m 巳 t由
I口
h创
a t heat enters 由
th巳 po
创t on1 y 由
t hro u g h its bottom , which is copper and rests on a h巳 ating ele ment. In two minlltes , the mass of water
boiled away is m = 0.45 kg. Th巳 radiu s of the pot bottom is
R = 6.5 cm, and the thickness is L = 2.0 mm . W hat is 由e temperature TE of the heating element in contact with the pot?
*16. Multiple-Concept Example 3 discusses an approach to problems
such as thi s. The ends of a thin bar are maintained at diffe rent t巳 m ­
peratures. T he temperature of the coo l 巳r end is 11 oc , while the temperature at a point 0.13 m from the cooler end is 23 oc and the
temp巳 ratu re of the warmer end is 48 oc. Assuming that heat f1 0ws
only aJ ong the length of the b创(由 e sides are in s ul ated ) ，自 n d 出巳
length of the bar
地 17.
Refer to Interactive Solution 13.17 at www.wiley.comJcollege/
cutnell fo r help in solving this problem. In an aluminum pot, 0.15 kg
of water at 100 oc boils away in fo ur minlltes. T he bottom of the pot is
3 .1 X 10 m thick and has a surface area of 0.01 5 m2 • To prevent 由e
water fro m boiling too rapidly, a stai nless steel pl at巳 h as been placed
b巳tween the pot and the heating element. The plate is 1.4 X 10
m
•
•
ssm www Two cy1indrical rods have the same mass. One is
made of sil ver (density = 10 500 kg/m巧， and one is made of iron
(density = 7860 kg/m 3) . Both rod s condu ct th巳 same amount of heat
per seco nd when the same temperature difference is maintained
across their ends. What is th巳 ra ti o (silver-to-iron) of (a) the lengths
and (b) the radii of these rod s?
Section 13.3
Radiation
矶 ~ L厄
胁阳
h1t bl巾 1 叩
0 p阳 at巳s 飞川h川
a川白 l阳
a创l阳
I
wh巳reas light bulb 2 has a 白创l a m巳
创n刊tt怡巳 mηperatur巳 of 2100 K. Both filaments h av巳 the sa m巳 e mi ss i v ity， and both bulbs radiate the same
power. Find th巳 rati o A /A 2 of the fil ament areas of the bulbs.
21. ssm www How many days does it take fo r a perfect blackbody
cube (0.0100 m on a side , 30.0 oc) to radi ate the same amount of
e n巳rgy th at a one-hundred-watt light blllb lI ses in one hour?
22. In an old house , the heating system uses radiators , w hich 缸e hollow meta1 devices through whi ch hot water or steam circul ates. In one
room the radi ator has a dark color (emissivity = 0.75). It has a temperature of 62 oc. The new owner of the house paints the radi ator a
lighter color (巳 mi ss i v ity
0.50). Assuming that it emits the s am巳
radi ant power as it did before being painted , what is the temperature
(in degrees Celsius) of the newly painted radiator?
23. A person is standing outdoors in the shade where the temperature
is 28 oc. (a) What i s 由e radiant energy absorbed per second by his
head when it is covered with hair? The surface area of the hair (assllmed to be fl at) is 160 cm2 and its emi ssivity is 0.85. (b) What
would be the radi ant energy absorbed per second by the same person
if he w e l 巳 b a l d and the emissivity of his head were 0.65?
24. A baking dish is removed from a hot oven and placed on a cooling rack. As the dish cools down to 35 oc from 175 oc , its net radiant
power decreases to 12.0 W. What was the net radiant power of the
baking dish when it was fi rst re moved fro m the oven? Assume that th巳
temperature in th巳 kitc h e n remains at 22 oc as the dish cools down
唱铲 Multiple - Co n ce pt
Example 8 reviews the approach that is
used in proble ms sllch as this. A person eats a dess巳口 that
contains 260 Calories. (This "C aJ orie" uni t, with a capital C , is the
o n巳 u s巳d by nutritionists; 1 Calori e = 4 186 J. See Section 12.7.) The
ski n tem peraωre of this individu aJ is 36 oc and that of her environment
is 21 oc. The emissivity of her skin is 0.75 and its surfac后 area is 1.3 m2
How mllch time would it t冰e for her to emit a net radiant energy 仕'Om
her body that is equal to the ener
25.
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has a length of 0.50 m. The cross secti on of this
composite rod is shown in the drawing and
consists of a s qu ar巳 w ithin a circ1 e. The square
cross section of the steel is 1.0 cm on a side
The temperature at one end of the rod is 78 oC ,
while it is 18 oC at the other end . Assuming that
no heat exits through the cylindrical outer surface , find the total amount of heat conducted
through the rod in two minutes
thick, and its ar巳a matches that of the po t. Assuming that heat is conducted into the water onl y through th巳 b o tto m ofth巳 pot ， find thew .tem.c
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perature at (a) the aluminllm- steel interfac巳 and (b) the steel
surface in contact with the heating e lement
w
Styrofoam whose area and thickness are 18 m2 and 0.10 m, respecw
ti ve. cly. Heat is lost via conduction through th巳 wall and the window.
.d o
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The temperature diffe rence between the inside and outs id巳 i s the
same for the wa ll and the window. Of the total heat lost by the wa ll
and the window, what is the percentage lost by the window?
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THE TRANSFER OF HEAT
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28. ( Sirius B is a whit巳阳阳t has a surface temperature (in
kelvins) that is four times that of our sun. Sirius B radiates only
0.040 times the power radiated by the sun . Our sun has a radius of
6.96 X 108 m. Assurning that Sirius B has the same ernissivity as the
sun , find the radius of Sirius B
29. ssm A car p缸ked in the sun absorbs energy at a rate of 560 watts
per square met巳r of surface area. The car reaches a temperature at
which it radiates energy at this same rate. Treating the car as a perfect
radiator (e = 1) ，自 nd the temperature.
Part (α) of the drawing shows a rectangular bar whose diw
Lo X 2Lo X 3乌. The bar is at the same constant tem.c
.d o
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perature as the room (not shown) in which it is located. The bar is
then cut , lengthwis巳 ， into two identical pieces, as shown in part (b) of
the drawing. The temperature of each piece is the same as that of the
original bar. (a) Wh at is the ratio of 由e power absorbed by 由etwo
bars in part (b) ofthe drawing t。由 si ngle bar in part (α)? (b) Suppose
that the temperature of the single bar in part (α) is 450.0 K. What
would th巳 temperature (i n kelvins) of the room and the two bars in
part (b) have to be so that the two bars absorb the same power as the
single 恼 in part (a)?
4.2 K in a spherical container (r = 0.30 m). The container is a perfect
blackbody radiator. The container is surrounded by a spherical shield
whose temperature is 77 K. A vacuum exists in the space between the
container and th巳 shield . The latent heat of vaporization for helium is
2.1 X 104 J/kg. What mass of 1i quid helium boils away through a
venting valve in one hour?
A~DD ! TIONAL
to
3Lo
LO
ø
均 31. ② Liqωhelium is stored at its boili暗point temperature of
k
(a)
(b)
ssm A solid cylinder is radiating power. 1t has a length that is ten
times its radius. 1t is cut into a number of smaller cylinders , each of
which has the same length. Each small cylinder has the same temperature as th巳 original cylinder. The total radiant power emitted by the
pieces is twice that emitted by the original cylinder. How many
smaller cylinders are there?
t. *34. One end of a 0.25-m copper rod with a cross-sectional area of
1.2 X 10- 4 m2 is driv巳 n into the center of a sphere of ice at 0 oC
(radius = 0.15 m). The portion of the rod that is embedded in the ice
is also at 0 oc. The rod is horizontal and its other end is fastened to a
wall in a room. Th巳 rod and the room are kept at a constant temperature of 24 oc. The emissivity of the ice is 0.90. What is the ratio of the
heat per second gained by the spher巳 through conduction to the net
heat per second gained by the ice due to radiation? Neglect any heat
gained through the sides of the rod.
r. *33.
PROBLEMS
35. ssm One end of an iron poker is placed in a fire where the temperature is 502 oC, and the oth巳r end is kept at a temperature of 26 oc.
The poker is 1.2 m long and has a radius of 5.0 x 10- 3 m. 19noring
the heat lost along the length of the poker, find the amount of heat
conducted from one end of the poker to the other in 5.0 s.
36. Concept Simulation 13.1 at www.wiley.comlcollege/cutnell
illustrates 由e concepts pertinent to this problem. A refrigerator has a
surface area of 5.3 m2. 1t is lined with 0.075-m-thick insulation whose
thermal conductivity is 0.030 J/(s . m . CO). The interior temperature is
kept at 5 oC , while the temperature at the outside surface is 25 oc.
How much heat per second is being removed from the unit?
of 0.700. AIso suppose that metabolic processes are producing energy
at a rate of 115 J/s. What is the temperature of the coldest room in
which this person could stand and not experience a drop in body temperature?
39. The concrete wall of a building is 0.10 m thick. The temperature
inside the building is 20.0 oC , while the temp巳rature outside is 0.0 oC
Heat is conducted through the wal l. When the building is unheated ,
the inside temperature falls to 0.0 oC, and heat conduction ceases
However, the wall do巳s emit radiant energy wh巳 n its temperature is
0.0 oc. The radiant energy ernitted per second per square meter is the
same as the heat lost per second per sqllare meter due to condllction.
What is the ernissivity of the wall?
37. The amount of radiant power produced by the sun is approximately 3.9 X 1026 W. Assuming the sun to be a perf巳ct blackbody
sphere with a radius of 6.96 x 10 8 m , find its surface temperature (in
kelvins).
*40. A solid sphere has a temperature of 773 K. The sphere is melted
down and recast into a cllbe that has the same ernissivity and emits
the same radiant power as th巳 sphere. What is the cube's temperature?
38. 咽F'
培 4 1.
Consult Multiple-Concept Example 8 to se巳 the concepts
that are pertinent here. A person's body is producing energy
internally due to metabo1i c processes. If the body loses more energy
than metabolic processes are generating , its temperature will drop. 1f
the drop is severe , it can be life-threatening. Suppose that a person is
unclothed and energy is being lost via radiation from a body surface
area of 1.40 m2, which has a temperature of 34 oC and an emissivity
~
1n a hOllse 出巳 temp巳rature at the surface of a window is
25 oC. The t巳mperature outside at the window sllrface is 5.0 oc. Heat
is lost through the window via condllction , and the heat lost per second has a certain vallle. The temperatllre olltside begins to fall , while
the conditions inside th巳 hOllse remain the same. As a reslllt ， 由 e heat
lost per second increases. What is the temp巳rature at the olltside window Sllrfac巳 when the heat lost per second doubles?
m
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30. 电P'
Multiple-Concept Example 8 discusses the ideas on which
this problem depends. Interactive LearningWare 13.2 at
www.wiley.comlcollege/cutnell reviews the concepts that are involved in this problem . Suppose the skin temperature of a naked person is 34 oC when the person is standing inside a room whose
temperature is 25 oc. The skin area of the individual is 1.5 m2
(a) Assurning the ernissivity is 0.80 , find the net loss of radiant power
from the body. (b) Deterrnine the number of food Calories of energy (1 food Calorie = 4186 J) that are lost in one hour due to the net
loss rate obtained in part (a). Metabolic conversion of food into energy replaces 出is loss
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problem.
Suppose the stove in 由 at example had a surface area of only
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2.00 m2 . What would its temperature (in kelvins) have to be so that it
still generated a net power of 7300 W?
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(a)
Problem 43
司位 43.
ssm www Two cy ]j ndrical rods are identical , except that one
has a thermaJ condllctivity k l and the other has a thermaJ condllctivity k2 . As the drawing shows , they are pJaced between two waJJs
that are maintained at different temperatllres Tw (warm巳r) and Tc
(cooJer). When the rods are arranged as in part a of the drawing , a
totaJ heat Q' ftow s from the warmer to the cooler wall , bllt when
the rods are arrang巳d as in part b , the total h巳at ftow is Q. Assllming
that the condllctivity k2 is twice as great as k l and that heat ftows
only along the lengths of the rods , determine the ratio Q' IQ.
A smaJJ sphere (emissivity = 0.90 , radillS = rl) is Jocated at the c巳n­
ter of a spherical asbestos sheJJ (thickness = 1.0 cm , ollter radillS = r2)'
The thickness of the shell is small compared to the inner and outer
radii of the sheJI. The temp巳ratllre of the small sphere is 800.0 oC,
while the temperature of the inner sllrface of the shell is 600.0 oC,
both temp巳ratures remaining constan t. Assllming that r21rl
10.0
and ignoring any 但r inside the sheJJ , find the temperature of the ollter
sllrface of the shel l.
在 * 44.
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of 由
t h巳 other iβs ∞
c opp 巳町r. Water in these pots
is boiling away at 100.0 oC at th巳 sam巳 rate. The temp巳rature of
the heating element on which the aluminum bottom is sitting is
155.0 oC. Assllme that heat enters the water only throllgh the bottoms of the pots and find the temp巳ratllre of the heating element on
which the copper bottom rests
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THE IDEAL GAS LAW
AND KINETIC THEORY
To the extent that the air in these
hot-air balloons behaves like an
ideal gas , its pressure , volume , and
temperature are related by the ideal
gas law, which is 0 门 e of the central
topics of th is chapter. (<<d Christophe
Karaba/Reuters/Corbis)
MOlECUlAR MASS , THE MOlE ,
AND AVOGADRO'S NUMBER
Often , we wish to compar巳 th e mass of on巳 atom w ith anothe r. To faciJitate th巳 com ­
parison , a mass scaJe known as the atomic mass scα le has be巳n estabJi shed. To s巳t up this
scale , a 1 巳f巳r巳 nce vaJue (aJong with a unit) is chosen fo 1' one of th巳巳l巳ments. Th巳 unit is
call巳d th巳 atomic mass unit (symbol: u). By internationaJ agre巳m 巳nt， th 巳 refe 1'ence e Jement
is chosen to b巳 th巳 most abundant type or isotop巳'" of carbon , which is call巳d carbon-12. Its
atomic mass T is defìn巳d to b巳 exactJy tweJve atornic mass units , or 12 u. Th巳 reJation s h j p
between th 巳 atomic mass unit and the kiJogram is
1 u = 1. 6605
Li
3 I Be
6 .941
4
I 9.01218
吧 982| 吧。:2
Figure 14.1 A portion of the periodic
table showing the atomic I1 umb巳r and
atomic mass of each e l em巳 n t. 1n the
periodic table it is customary to omit
the symbol "u" denoting the atomic
mass U I1I t
x 10… 27 kg
The atornic masses of all the 巳Jements are listed in the periodic tabJe , part of which is
shown in Fig ur巳 14. l. Th巳 complete periodic table is gi v巳 n on the in s id巳 of th巳 back cove r.
1n gen巳ral， the masses listed are averag巳 val ues and take into accollnt the variolls isotopes of
an el巳 me nt that exist natllrally. For brevity, the unit " u" is often omitted from the table. FOI
example , a magnesillm atom (Mg) has an average atomic mass of 24.305 u , while that for
th巳 lithium ato m (Li) is 6.941 u; thus , atornic magn巳sium is mor巳 massive than atomic
lithium by a factor of (24.305 u)/(6.941 u) = 3.502. 1n th巳 p巳riodic tabl巴 ， th巳 atornic mass of
carbon (C) is given as 12.011 u , rath巳r than 巳xactly 12 u , becaus巳 a small amollnt (abo ut 1 %)
of th巳 n aturally occurring material is an isotop巳 called carbon-13. The value of 12.011 1I is
an av巳rage that refl ects th巳 small contribution of carbon-13.
f'Isotopes are disCllssed in Section 3 1.1.
tln chemistry the expression "atomic weight" is freq u 巳ntly used in place
of 飞 tomic
mass."
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mass of a molecule is the sum of th巳 atomic masses of its atoms. For
the el巳ments hydrogen and oxygen hav巳 atomic mass巳s of 1. 007 94 u and
15.9994 u, resp巳ctively， so the molecular mass of a water molecule (H 20) is , therefOl飞
2( 1. 00794 u) + 15.9994 u = 18.0153 u.
Macroscopic arnounts of materials contain large numbers of atoms or molecules. Even in
a small volume of gas , 1 cm 3, for example，由巳 numb巳r is enormous. It is conveni巳nt to express such large numbers in t巳rms of a single unit, th巳 gram-mole， or simply the mole (symbol: mol). One gram-mole of a subsωnce contains as many particles (atoms or molecules)
as there are atoms in 12 grams of the isotope carbon-12. Experiment shows that 12 grams
of carbon-12 contain 6.022 X 10 23 atoms. Th巳 number of atoms per mole is known as
Avogadro's number NA , after the Italian scientist Amed巳o Avogadro (1 77ι1856):
NA
= 6.022
X 10 23 mol -
I
一一
n
N一凡
Thus , the number of moles n contained in any sample is th 巳 numb巳 r of particles N in the
sample divided by the number of particles per mole N A (Avogadro's numb巳 r):
in terms of carbon atoms , th巳 conc巳 pt of a mole can be applied to
of objects by noting that one mole contains Avogadro's nllmber of objects.
Thus ， on巳 mole of atornic slllfur contains 6.022 X 10 23 sulfur atoms , one mole of water
contains 6.022 X 10 23 H2 0 mol 巳cllles ， and one mol巳 of golf balls contains 6.022 X 10 23
golf balls. Th巳 mole is the SI base unit for expressing "the amount of a sllbstance."
Th巳 number n of moles contained in a sample can also be found from its mass. To see
how, multiply and divid巳 the right-hand side of th巳 previous equation by the mass mparlicle
of a singl巳 particle ， expressed in grams:
Although
any
d 巳fined
coll巳ction
m
饥
1
n
D 盯川t
一一一二一一一一一一
mparticleNA
一
Mass per mole
The num巳rator m parti出 N is the mass of a particle times the number of particles in th巳 S缸nple ，
which is the mass m of 由e sampl巳 expressed in grams. The denorninator mparticleNAis the mass
of a par甘c1 e times the number of particles per mole, which is 由巳 mass p巳r mole , expr巳ssed in
grarns per mole. The mass per mole (in g/mol) of any substance has the same numerical
value as the atomic or molecular mass ofthe substance (in atomic mass units). To under
stand this fact , consid巳r the carbon-12 and sodium atoms as examples. The mass p巳r mole of
carbon-12 is 12 g/mol , since , by definition , 12 grams of carbon-12 contain one mole of atoms.
On th巳 other hand , the mass p巳 r mole of sodium (Na) is 22.9898 g/mol for the following reason: as indicated in Figur巳 14 .1， a sodillm atom is more massive than a c缸 bon-12 atom by
th巳 ratio of their atornic masses , (22.9898 u)/ (1 2 u) = 1. 915 82. Therefore , the mass per
mole of sodillm is 1. 915 82 tÏmes as great as that of carbon-12 , which m巳ans equivalently
that (1. 915 82)(12 g/mol) = 22.9898 g/mo l. Thus , the nllmerical value of the mass per mole
of sodium (22.9898) is 由巳 same as the numerical value of its atornic mass.
Since one gram-mole of a substance contains Avogadro 's number of particles (atoms
or molecules) , the mass mp川 cl e of a particle (i n grams) can be obtained by dividing the
mass per mole (in g/mol) by Avogadro's number:
Ex创ηple
numb巳r
to
一­
m
Problem-solving insight
Mass per mole
NA
1 il1 ustrates how to use the conc巳pts of the mole , atornic mass , and Avogadro's
number of atoms and molecules pr巳sent in two famous g巳mstones.
deterrnin巳 the
Example 1
The Hope Oiamond and the Rosser Reeves Ruby
Figure 14 . 2αshows th 巳 Hope diamond (44.5 carats) , which is almost pure carbon. Figure 14.2b
shows th巳 Rosser Reeves ruby (1 38 c创'ats ) ， which is primarily aluminum oxid巳 (AI 2 0 3 ). One
carat is equivalent to a mass of 0.200 g. Deterrnine (a) the number of carbon atoms in the dia
mond and (b) the numb巳 r of AI 2 0 3 molecules in th巳 ruby.
The physics of
gemstones.
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The number of carbon atoms in the Hope diamond is
。
( )
m
8 90 g
一一一一一一----'='-12.011 g/mol
Mass per mole
N
= nNA = (0.741 mol)(6.022 x 10 23 atoms/mol) = I4 .4 6
X
10 23 atom s I
(b) The mass of the Rosser Re巳ves ruby is m = (1 38 carats)[(0.200 g)/ (1 carat)] = 27.6 g.
The moleα11ar mass of an alurninum oxide molecule (A1 20 3 ) is the sum of the atornic masses
of its atoms , which ar巳 26.9815 u for aluminum and 15.9994 u for oxygen (see the periodic
table on the insid巳 of the back cover):
Molecular mass
=
2(26.9815 u)
+ 3(1 5.9994 u)
=
10 1. 9612 u
、-一~一'-----v-一~
Mass of 2
aluminum atoms
Thus , th巳
mass
per
mol 巳 of Al 20 3
Mass of 3
oxygen atoms
is 10 1. 9612 g/m01. Calculations Ii ke those in part (a) reveal
that the Rosser Reev巳s r向 contains 0.271 m们rl 1. 63 X 10 23 molecl阳 of AI 2 0 3 1
~ CHECK Y 。
ωUR UNDER
附
STANDI川川
N唱
(The answers are given at the end of the book.)
1. Consider one mole of hydrogen (H 2 ) and one mole of oxygen (0 2 ) Which , if either, has
the greater number of molecules and which , if either, has the greater mass?
2. The molecules of substances A and B are composed of different atoms. However, the
two substances have the same mass densities. Consider the possibilities for the molecular
masses of the two types of molecules and decide whether 1 m3 of substance A contains
the same number of molecules as 1 m3 of substance B.
3. A gas mixture contains equal masses of the monatomic gases argon (atomic mass =
39.948 u) and neon (atomic mass = 20.179 u) . These two are the only gases present. Of the
total number of atoms in the mixture , what percentage is neon?
,
(b)
Figure 14.2 (α) The Hope diamond
surrounded by 16 smaIIer diamonds.
(Dane A. Penland/G巳 m & Mineral
Collection/Smithsonian Institution)
(b) The Ross巳r Reeves ruby. (Chip
Clark/ Smithsonian Institution) Both gems
ar巳 on display at th巳 Smithsonian
lnstitution in Washington , D.C.
叫 11
-守
THE IDEAl GAS lAW
An ideal gas is an idealized model for r巳 al gases that have suffìciently low densities. The condition of low density m巳 ans that the molecules of the gas are so far apart that
they do not interact (exc巳pt during collisions that are effectively 巳 lastic). The ideal gas
law 巳xpresses the relationship between the absolute pressur巳， th巳 Kelvin temperature , the
volum巳， and the number of moles of the gas
In discussing the constant-volume gas thermometer, S 巳ction 12.2 has alr巳 ady explained th 巳 relationship betw巳巳n the absolute pressur巳 and Kelvin t巳 mperature of a lowdensity gas . This th巳rmometer utilizes a small amount of gas (e.g., hydrogen or helium)
placed inside a bulb and kept at a constant volume. Since the density is low, the gas behaves as an id巳 al gas. Exp巳rim巳 nt reveals that a plot of gas pressure v巳 rsus temperature IS
a straight line , as in Figure 12 .4. This plot is redrawn in Figur巳 14.3 ， with th巳 chang巳 that
the temperature axis is now labeled in kelvins rather than in d巳grees Celsius. The graph
indicates that the absolut巳 pr巳 ssure P is directly proportional to the Kelvin temperature
T(P α T) ， for a 自 xed volum巳 and a fìxed number of molecules.
The relation b巳tw巳en absolute pressure and the number of molecules of an id巳 al gas
is simple. Experienc巳 indicates that it is possible to increase the pr巳 ssure of a gas by adding
mor巳 molecules ; this is exactly what happens when a tire is pumped up. When the volume
and t巳 mp巳rature of a low-density gas are kept constant, doubling the number of molecules
m
lic
Solution (a) Th巳 Hope diamond 's mass is m = (44.5 carats)[(0.200 g)/ (1 carat)] = 8.90 g
Since the average atomic mass of naturally occurring carbon is 12.011 u (see the p巳 riodic table
on the inside of the back cov巳 r) ， the mass per mole of this substanc巳 is 12.011 g/mol. The number of moles of carbon in th巳 Hop巳 diamond is
o
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Th巳 number N of atoms (or mol巳cules) in a sample is the number of moles n times
w
of atoms per mole N A (Avogadro 's number); N = nNA . We can det巳 rrnine the num.c
.d o
c u -tr a c k
ber of moles by dividing the mass of th巳 sample m by the mass per mole of the substance.
Reasoning
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THE IDEAL GAS LAW 417
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N
o
The absolute pressure P of an ideal gas is d让巳ctly proportional to the Kelvin tempera
ture T and the numb巳r of moles n of the gas and is inversely proportional to the volume
V of the gas: P = R(nT/V). In other words ,
O
PV= nRT
where R is the universal gas constant and has
th巳 value
(1 4.1)
of 8.31 J/(mol . K) .
Sometimes , it is convenient to express the id巳al gas law in terms of the total number
of parti c1es N , instead of the number of moles n. To obtain such an expression , we multiply
and divide by Avogadro's number NA 二 6.022 X 10 23 particles/mol* on 优 right in Equation
14.1 and recognize that 由巳 product nNA is equal to theωtal number N of particles:
PV =
.d o
m
o
m
w
c u -tr a c k
ω』3的的ω」且ωH3口
-的且〈
IDEAL GAS LAW
C
lic
k
to
bu
y
N
y
bu
to
k
lic
C
doubles the pressure. Thus , the absolute pressure of an ideal gas at constant temperature
.c
and
constant volume is proportional to the number of mol 巳cules or, equivalently, to th巳
c u -tr a c k
number of moles n of the gas (P α n).
To see how the absolute pressure of a gas depends on the volume of the gas , look at
the partially filled balloon in Figure 14 .4a. This balloon is "soft," because the pressur巳 of
the air is low. However, if all the air in the balloon is squeezed into a small巳r "bubble ," as
in p缸t b of the figure , the "bubble" has a tighter fee l. This tightness indicates that the pressure in the smaller volume is high enough to stretch the rubb巳 r substantially. Thus , it is
possible to increase the pressure of a gas by reducing its volume , and if the number of molecules and the t巳 mp巳rature are kept constant , the absolute pressure of an ideal gas is in
versely proportional to its volume V (P oc lI V).
The thr巳巳 relations just discussed for the absolute pressure of an ideal gas can be ex
pressed as a single proportionality, P α n Tl V. This proportionality can be written as an
equation by inserting a proportionality constant R , call巳d th巳 universal gas constant.
Experiments have shown that R = 8.3 1 J/(mol' K) for any real gas with a density suffi ci巳ntly low to ensure ideal gas b巳havior. The r巳 sulting equation is call巳d the ideal gas law.
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"
100
200
300
Temperature , K
Figure 14.3 The pressure inside a
constant-volume gas thermometer is
directly proportional to the Kelvin
temp巳rature ， a proportionality that is
charact巳ristic of an ideal gas
nRT 二 nN. 1 主 I T= NI 主一 IT
川\ NA /
\ NA /
The constant term R/NA is refe町巳d to as Boltzmann 云 constant， in honor of the Austrian
physicist Ludwig Boltzmann (1 84 4- 1906) , and is represented by th巳 symbol k:
R
k= 一一=
NA
831 J/(mol' K)
= 1.3 8 X 10- 23 J/K
6.022 X 10 23 mol - 1
(α)
*"Particles" is not an SI unit and is often omitted. Then , particles/mol
(b)
=
IImol
=
mol - I
Figure 14.4 (α) The air pr巳 ssure in the
partially filled balloon can be increased
by decreasing 由e volume of 出巳 balloon，
as illustrated in (b). (Andy Washnik)
.c
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N
y
to
bu
y
bu
k
lic
o
c u -tr
.
ack
c
E倒The physics
霄 oxygen
0'
in the lungs.
Exampl巳 2 pr巳 se nts
an application of the
0 E
白
xamp盯
帕
O
X均y归
明g 川阳 Lun
川川』汀川n
id巳 al
.d
(1 4.2) o c u - t r a c
PV = NkT
gas law.
In 由
t h 巳 lungs ， a thin respiratory membrane s巳parate s tiny sacs of air (absolute pr巳 ssure
5
1. 00 X 10 Pa) from the blood in the capillaries. Th巳se sacs are ca11ed alveoli , and it is from
them that oxygen enters 出巳 blood. Th巳 averag巳 radius of the alveoli is 0.125 mm , and the aiI
inside contains 14% oxygen. Assuming that the air b巳 have s as an ideaJ gas at body temp巳 raωm
(310 K) , fìnd th巳 number of oxygen mol 巳 cules in on 巳 of th 巳 sacs.
Reasoning Th巳 pr巳 ss ur巳 and temperature of th巳 air inside an alveolus are known , and its
volume can b巳 determined since we know the radius. Thus , the ideal gas law in the form
PV = NkTcan b巳 used directly to find the number N of air particles insid巳 on 巳 of the sacs. Th 巳
number of oxyg巳 n molecules is 14% of the numb巳 r of air particles.
Solution Th巳 volur盯阳
T
of 创
a ir par川tJ比
cle
巳 s队， w巳 hav巳
Problem -solving insight
In the ideal 日 as law , the temperature T must
be expressed on the Kelvin scale. The Celsius
and Fahrenheit scales cannot be used.
PV 一(l. 00 X 10 5 Pa)[ ~计(0 . 125 X 10 - 3 m)3
kT
(l.3 8 x 10- 23 J/K)(310 K)
N= 一一一
.
1.
n..
..
~
..
r\
1L1
9 X 10 '4
Th巳川ηber of oxygen mol巳cules is 14% of tt山 va l 盹 oro.14N=12 .7 x 10 13 1
With the aid of the id巳 al gas law, it can be shown that on 巳 mole of an id巳 al gas occupies a volume of 22 .4 liters at a temperature of 273 K (0 o c) and a pressure of one atmospher巳(1. 013 X 10) Pa). Th巳 se conditions of temp巳 rature and pressure are known as
standard temperature and pressure (STP). Conceptual Exampl 巳 3 discus s巳 s anoth巳r 10teresting application of the ideal gas law.
0'
The physics
rising beer bubbles.
Conceptual Example 3
Beer Bubbles on the Rise
If you look carefully at 由e bubbles rising in a glass of beer (see Figure
14. 匀 ， you'lls巳巳 them grow
in size as they mov巳 upward ， often doubling in volum 巳 by the time th巳y reach the s u rfac巳 .B 巴巴r
bubbles contain mostly carbon dioxide (C0 2) , a gas that is dissolved in the b巳巳r becaus巳 of the fermentation process. Whi ch variable describing 由 e gas is responsible for the growth of the tising
bubbles? (a) Th巳 Kelvin t巳 mp巳raωre T (b) The absolute PI它 ssure P (c) The numb巳r of moles n
Reasoning Th巳 variables T, P, and n are relat巳d to th巳 volum 巳 V of a bubble by the ideal gas
law (V = nR T.俨). We assume that this law applies and us巳 it to guide our thinking. According
to this law, an increas巳 in t巳 mp巳rature ， a d巳 crea s巳 in PI 巳 ss ure ， or an increase in th 巳 number of
moles could account for the growth in siz巳 of the upwa时 -moving bubbles.
Answers (a) and (b) are incorrect. Temp巳 rature can be eliminated immediately, sinc巳 It IS
constant throughout the be巳1'. Pressure cannot be dismissed so easily. As a bubble ris巳 s ， its depth
decreases , and so does the f1 uid pressure tllat a bubble experiences. Since volum巳 is invers巳 Iy
proportional to pr 巳ssure accor吐 ing to the ideal gas law, at least part of the bubble growth is du 巳
to the decreasing press lII巳 of the surrounding beer. How巳ver， som巳 bubbles double in volume
on th巳 way up. To account for the doub 1i ng , there wou ld need to be two atmosph 巳 res of pressure at the bottom of the glass , comp缸 ed to the one atmosph巳re at th 巳 top . The pressure increment due to d巳pth is pgh (see Equation 1 1. 4) , so an extra pressure of on 巳 atmosph巳re at th巳
bottom would mean 1.01 X 10 5 Pa = pgh. Solving for h with p equal to th巳 d巳nsity of water reveals that h = 10.3 m. Since most beer glasses ar它 only about 0.2 m ta11 , we can rule out a
change in pressl町巳 as the major cause of the change in volume.
Figure 14.5 The bubbles in a glass of
beer grow larger as
(阶
Cωour川
叫
I
th巳y
move upward.
Answer (c) is correct. The process of elimination brings us to the conclusion that the number
of moles of CO 2 in a bubble must som巳 how be incr巳 a s ing on the way up. This is , in fact , the
case. Each bubble acts as a nucleation site for CO 2 molecules dissolved in the s lIITounding b巴。1
so as a bubble moves upward , it accumulates carbon dioxid巳 and grows large r.
Related Homework: Problem 28
m
C
m
w
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.d o
o
to
C
lic
k
With this substitution , th巳 id巳 al gas law becomes
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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
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419
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T,
(14.3)
PiVi = PfVr
COllstallt 11
Figure 14.6 illustrates how pressure and volume change accordi ng to Boyle's law for
a fìxed nllmber of moles of an ideal gas at a constant t巳 mperature of 100 K. The gas begins with an initial pressur巳 and vo lum巳 of P i and Vi and is compressed. Th巳 pressur巳 111creases as the volume decr巳 ases ， accordi ng to P = nRTIV, lI ntil the fìnal pressur巳 and
volllm 巳 of Pr and Vf are reached. The curve that pass 巳 s through the initial and fìnal points
is called an isotherm, meaning "sa m 巳 temp 巳 rature." If the t巳 mperatur 巳 had b 巳巳 n 300 K,
rather than 100 K, th 巳 compression wOllld have occurred along the 300-K isotherm.
Different isotherms do not intersec t. Example 4 deals with an appli cation of Boyle's law
to sCllba diving
to
k
lic
、
、
m
w
.d o
o
m
o
COllstallt
、
1
c u -tr a c k
.c
、
‘‘
‘..
、
、
、
、
Pd-----
ω
isolherm
γ 300-K
C
k
lic
C
w
iE
bu
y
N
y
bu
to
lt1
w
Historically, the work of sev巳ral inv巳stigators led to th 巳 formulation of the ideal gas
.c
.d o
The Irish scientist Rob巳rt Boyl巳(1 627 -1691) discovered that at a constant temp巳ra­
law.
k
c u -tr a c
tur飞 the absolute pressure of a fìx巳d mass (fixed number of moles) of a low-density gas is
inversely proportional to its vo lum 巳 (P α1 / 盯 Thi s fact is often ca ll 巳d Boyle's law and
can b巳 derived from the ideal gas law by noting that P = nRT/V = constantlV when n
and T are constants. Alt忧巳rn川
l
u川
1口
mm
η1巳 (P i , 川
Vi)
tωo
a
fina
叫
创
a
tl
pr巳sSllr巳
and volll川
1mηme (P f , V山 l此t 比
川
11
i s possible tωo WI川it怡巳 PiV
川li = Fn1RT
i
and P f Vr = nR T. Since the right sides of these eq u ationsω·巳 eq u a l ， we may eqllate th 巳 left
sides to give the following conc ise way of expressing Boyle's law:
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14.2 THE IDEAL GAS L AW
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、
、
、.
、
、
E
Pi
卡 -----←-- -- --一---一
Vi
Vf
Volume
Figure 14.6 A pr巳 ssure-versus-volume
plot for a gas at a constant temperature
is called an isotherm. For an ideal gas ,
each isotherm is a plot of th 巳 equatlO n
P = nRT/V = constant/V
卢寸盯古古有ττ古节盯τ百币τπ7百古司了
……………·
Example 4
Scuba* Diving
The physics of
scuba diving.
When a sCllba diver descends to greater depths , the water pressure increases. The air pressure
inside th巳 body cavities (e.g. , lungs , sinuses) must be maintained at the same pressure as that of the
surrollnding water; otherwise th巳 cavities wOllld collaps巳 . A special valve automatically adj usts th巳 pressure of the air corning
from the scuba tank to ensure that the air pressure 巳 qllals th 巳 water pressure at all times. The scuba gear in Figure 14.7αcon­
sists of a 0.0150-m 3 tank fi lled with compr巳ssed air at an absolute pressllre of 2.02 x 107 Pa. Assume that th巳 diver consumes air at the rate of 0.0300 m3 per rninute and that th巳 t巳mp巳rature of th巳 air does not change as the div巳r goes deeper
into the water. How long (in mjnutes) can a diver stay lI nd 巳 r water at a d 巳pth of 10.0 m? Take the density of sea water to be
1025 kg/m 3 .
厂
Pl
广
f
=但zrrlC
η= Pres
(a)
陀
depth h
(b)
Figure 14.7 (α) The air pressure inside the body cavities of a scuba diver must be maintained at the sam巳 l 巳V巳 1 as the pressure of
the surrounding water. (Shirley Vanderbil tJlndex Stock) (b) The pressure P2 at a depth h is greater than the pressure P l at the surface
史
The
word is an acronym for s巳 I f- contai n ed underwater breathing apparatus
Continued
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k
to
bu
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N
y
bu
to
k
lic
o
Knowns and Unknowns The data for this problem are listed below:
Symbol
Value
Vj
Cph
Description
0.0150 m3
2.02 X 107 Pa
0.0300 m3/min
1025 kg/m 3
10.0 m
P1
1. 01 X 10 5 Pa
Comment
Explicit Data
Volum巳 of
川
p
air in tank
Pressur巳 of air in tank
Rat巳 of air consumption
Mass density of sea wat巳r
Depth of diver
Implicit Data
Air pressure at surface of water
Unknown Variable
Time that diver can remain at
1O .0-m depth
Atmospheric pressur巳
(see Section 11 .2).
at
sea level
7
Modeling the Problem
回国 Duration 川e 胁川ir insid巳 the scu川k has an in叫巳ssure
of Pj and a volume of Vj (the volume of the tank). A scuba diver does not breathe the a让
directly from the tank , because the tank pressure of 2.02 X 107 Pa is nearly 200 times
atmospheric pr巳 ssure and would cause his lungs to explode. Instead , a valve on th巳 tank
adjusts the pressure of the air bei吨 sent to th巳 diver so it 巳quals the surrounding water
pressure Pr. The time t (in minutes) that the diver can r巳 main under water is equal t。由
total volume of air consumed by the diver divided by the rate C (in cubic meters p巳r
minute) at which th巳 air is consumed:
份'b
一
一
t~r
，
The total volume of air consumed is th巳 volume Vr available at the breathing pres s ur巳
Pr minus th巳 volume Vj of the scuba tank , because this amount of air always r巳 mains
behind in the tank. Thus , we have Equation 1 at the right. The volume V and the rat巳
Car巳 known ， but the final volum巳 Vr is not, so w巳 turn to Step 2 to 巳valuate it
( 11
、 、，，
Total volume of air consum巳d
C
j
固盟 帅's Law 创
S川
m
仙灿阳叫
阳叫
nc
l比川叫
c臼阳阳
e创…
t阳
he
川巳创t
le
iv 号|
This 巳xpression
for Vr can be substituted into Equation 1, as indicated at the right. The
initial pressure Pj and volume Vj are given. However, w巳 stiU need to determine the
pressure Pr of the air inhaled by the div町， and we will evaluate it in the n巳xt step.
回111 Press…ω叩th in a Static Fl副 Figure 阳b shows the diver at a
depth h below the surface of the water. The absolute pr巳 ssure P2 at this depth is related
to the pressure P 1 at the surface of the water by Equation 11.4: P2 = P 1 + ρ'gh ， whereρ
，，.‘
、
volum巳 V
川r available to the diver at the pr巳ssur巳 Pr 凶
i s related tωo 由
th巳 initial pr巳 s sur巳 Pj
and volum巳 Vj of air in the tank by Boyle's law PjVj = PrVr (Equation 14.3). Solving for
Vr yields
m
C
m
o
C
.d o
w
w
Reasoning The time (in minutes) that a scuba diver can remain und巳r water is equal to the volume of air that is available
w
.c
.c
.
c u -tr a c k
divided
by the volume p巳r minute consumed by the diver. The volume of air avail able to the diver depends on the volume and d o c u - t r a c k
pr巳ssure of the air in the scuba tank , as well as th巳 pressure of the air inhaled by the diver, according to Boyle's law. The pressure of the air inhaled equals the water pressure that acts on th巳 diver. This pressure can be found from a knowledge of the
div町、 d巳pth ben巳ath the surface of the water.
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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
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(2)
N
y
substitut巳 this
.d o
(2)
+pgh l
expression for Pr into Equation 2, as indicated in the right column.
Solution Algebraically
combining 出 e
results of the
thr巳e
steps, we have
………
I
.
v: t
PjVj
,,
1
PjVj
+ ol! h
二十
审
V，
-
Pg'
t=-'一一一
c
The time that the diver can 1'emain at a
p，
ι
c
d巳pth
审
,
.I
-
c
of 10.0 m is
p.v:
一一---一一- V:
PJ
+ pgh
C
(2.02 X 107 Pa)(0.0150 m3)
+ (1 025 kg/m 3 )(9.80 rnIs2)( 1O .0 m)
5
1 01 X 10 Pa
0.0300 m3/min
一 0.0150 m3
=医王E
Note that at a fixed consumption rate C, greater values fo 1' h lead to smaller values for t.
In other words , a deeper dive must have a shorter duration.
Related Homework: Problems 24, 28
Another investigator whos巳 work contribut巳d to th巳 formulation of the ideal gas law
was the F1'enchman Jacques Cha.rl巳 s (1 746-1823). He discovered that at a constant pressure , the volume of a fixed mass (fixed number of moles) of a low-density gas is direct\y
proportiona\ to th巳 Ke\vin temperature (V α T). This re\ationship is known as Cha1'l 巳 s'
law and can be obtain巳 d from the ideal gas law by noting that V = nRTIP = (constant)T,
if n and P are constan t. Equivalently, wh巳 n an ideal gas changes from an initial volume
and temperature (町 ， T to a final volume and temperatur巳(町 ， Tr), it is possib\e to write
VJTj = nRIP and VrlTr = nRIP. Thus , one way of stating CharLes' Law is
j )
Constant P,
consωntn
Vj
Vr
贝
Tf
(1 4 .4)
,j" CHECK
YOUR UNDERSTANDING
(The answers are given at the end ofthe book.)
4. A tightly sealed house has a large ceiling fan that blows air out of the house and into
the attic. The owners tu rn the fan on and forget to open any windows or doors. What happens to the air pressure in the house after the fan has been on for a while , and does it become easier or harder for the fan to do its job?
5. Above the liquid in a can of hair spray is a gas at a relatively high pressure. The label
on the can includes the warning "DO NOT STORE AT HIGH TEMPERATURES." Why is the
warning given?
6. What happens to the pressure in a tightly sealed house when the electric furnace turns
on and runs for a while?
Continued
Problem-solving insight
When using the ideal gas law , either
directly or in the form of Boyle's law ,
remember that the pressure P must be the
absolute pressure , not the gauge pressure.
m
)
11
c u -tr a c k
o
m
o
We now
C
lic
k
to
bu
y
bu
to
k
C
lic
| Pr=PJ
w
，，飞 、
is. c the mass density of sea water and g is the magnitud巳 of the acc巳l巳 ration due to gravity.
Sinc巳 PJ is the air pressure at the surface of the water, it is atmospheric p1'essure. Recall
that the valve on the scuba tank adjusts the pressure Pr of the air inhaled by 出巳 diver to
b巳巳qual to th巳 pressure P2 of the surrounding water. Thus , P2 = Pr, and Equation 11 .4
becomes
c u -tr a c k
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8. Atmospheric pressure decreases with increasing a ltitude. Given this fact , explain why
helium-filled weather balloons are underinflated when they are launched from the ground
Assume that the temperature does not change much as the balloon rises.
9. A slippery cork is being pressed into an almost full (but not 100% ful l) bottle of wine
When released , the cork s lowly sl ides back ou t. However, if half the wine is removed from
the bottle before th e co rk is in serted , the cork does not s lide out. Exp lain .
10. Consider equa l masses of three monatomic gases: argon (atomic mass = 39.948 u) , kryp
ton (atomic mass = 83.80 u). and xenon (atomic mass = 131 .29 u). The pressure and volume
of each gas is the same. Which gas has the greatest and which the smallest te mpe rature?
(1)
11 KI NETI C TH EO RV 0F GAS ES
..
一寸 I~.
.-
I As lI seflll as it is , the ideal gas law provid巳s no insig ht as to how pressur巳 and temto properties of th巳 mol巳cu l es themselv町， such as their masses and
sp巳eds. To show how sllch microscopic properties ar巳 related to the pressure and t巳mper­
atur巳 of an id巳 al gas , this section examines the dynamics of moleclllar motion. The pres
sure that a gas 巳xerts on th 巳 walls of a container is due to the force ex 巳 rt巳 d by the gas
molecllles wh 巳 n th巳Y collid巳 w ith the walls. Th 巳refo 1'飞 we will b巳gin by combining the notion of collisional fo 1'ces exerted by a fl lI id (Section 1 1. 2) with Newton's second a nd thú'd
laws of motion (Sections 4.3 and 4.5). Th 巳 se concepts will allow lI S to obtain an expression for th巳 pl 巳SSUI 巳 in terms of microscopic properties. W，巳 will then 旦旦mbine this with
th 巳 id巳 al gas law to show that the av巳 rag巳 trans l ationa l kinetic e nergy KE of a particl 巳 In
an 阳1 gas is KE = ~kT， where k is B 阳nann's
tωur陀
e. 1n the process , we will also se巳 that th 巳 int巳 rnal e n 巳 rgy U of a monatomic ideal gas
is U = ~ nRT， wher巳 n is th巳 川
p巳 rature ar巳 related
THE DISTRIBUTl ON OF MOLECULAR SPEEDS
A macroscopic container fi lled with a gas at standard temp巳ratur巳 and pressure contain s a larg巳 numb巳r of particles (atoms or mol 巳cules) . These particles 缸·巳 in constant, random
motion , colliding with 巳ac h other and with the wall s of the co ntain巳r. 1n th巳 course of one sec
ond , a patticle undergoes many collisions , and each one chat1ges the particle's speed and di
rection of motion. As a resu lt, the atoms or molecllles hav巳 different speeds. It is possible ,
howev巳r， to sp巳ak about an av巳rag巳 patticl 巳 spe巳d. At any given instant , som巳 particles have
sp巳eds less than , some ne肌 and some greater than th巳 av巳rag巳. For conditions of low gas density, th巳 distriblltion of speeds within a larg巳 colJ 巳cti on ofmol巳cll les at a constant t巳mperature
was calculated by the Scottish physicist James Clerk Maxwell (1 831-1879). Figure 14.8
displays th巳 Maxwell sp巳巳d distriblltion curves fo 1' O2 gas at two diff，巳 rent t巳 mperature s
-2Uω」D
』E
。ω∞SEωU』ω止
UUE
的 ME3
-m〉LωH一
E它
且也
』ω
Figure 14.8 The Maxwell distribution
curves for molecular speeds in oxygen
gas at temperatures of 300 and 1200 K.
400
800
1200
Molecular speed , m/s
1600
m
When you climb a mountain , your eardrums 、 op" outward as the air press ure decreases. When you come down , theγpop inward as the pressure increases. At the
sea coast , you swim through a completely submerged passage and emerge into a pocket
of air trapped within a cave. As the tide comes in , the water level in the cave rises , and
your eardrums pOp. Is this popping analogous to what happens as you climb Up or climb
down a mountain?
在 节r
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CHAPTER 14 THE IDEA L GAS LAW AND KINETIC THEORY
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F-
c u -tr a c k
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Av 巳 rag巳 force
=
Final momentum - InitiaJ momentllm
...... . .
c u -tr a c k
' \
来/
. ..‘
(L
KINETIC THEORV
If a ball is thrown against a wall , it exerts a force on th巳 wal l. As Figure 14.9 suggests , gas particles do th 巳 sam巳 thing ， except that their masses ar它 smaller and their speeds
创它 greater. Th巳 number of p缸ticles is so great and they strike th巳 wall so often that th 巳巳f­
fect of their individual impacts appe创's as a continuous force. Dividing the magnitude of
thi s force by th巳 ar'ea of the wall gives the pressure exert巳 d by the gas.
To calculate the force , consider an ideal gas composed of N identical particles in a
cubical container whose sid巳 s have I 巳 ngth L. Exc巳 pt for 巳 l astic* collisions , these particles
do not interac t. Figure 14.10 focus巳 s attention on on巳 particle of mass m as it strikes th巳
right wall perpendicularly and rebounds elastically. While approac hing the wall , th巳 parti­
cle has a velocity +V and Ii near momentum +mv (see Section 7 .1 for a review of linear
mom 巳 ntum). The particle r 巳 bounds with a velocity -v and momentum -mv , travels to the
le仇 wall ， rebounds again , and heads back towar'd the righ t. The tim巳 t betw巳en collisions
with the right wall is the round-trip distance 2L divided by th 巳 s peed of the particle; that
is , t = 2 L1v. According to N巳wton 's second law of motion , in the form of the impulsemomentum theor巳 m ， th 巳 averag巳 force 巳xert巳d on th巳 particl 巳 by the wall is given by the
chang巳 in th巳 particle's momentum per unit time:
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--..，
C
m
J 将"-
---
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to
bu
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to
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lic
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When the t巳 mp巳rature is 300 K , the maximum in the curv巳 indicates that th巳 most proba. c speed is about 400 m/s. At a t巳mp巳rature of 1200 K , th巳 di stribution curv巳 shifts to the
ble
c u -tr a c k
right , and the most probable speed incr巳ases to about 800 m/s. One particularly useful type
of average s p巳ed ， known as th巳 rms speed and written as Vrms' is also shown in the drawing . When the temp巳rature of th巳 oxyg巳 n gas is 300 K the rms s pe巳 d is 484 m/s , and it increases to 967 m/s when th巳 temp巳rature rises to 1200 K. The meaning of th巳 rms spe巳d
and th 巳 reason why it is so important will be discussed shortly.
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Figure 14.9 The pressure that a gas
is caused by the collisions of its
molecllles with the walls of the
conta ll1 er.
巳xerts
L
L
(7.4)
Tim巳 b巳tw巳巳 n succ巳 ssiv巳 collisions
-mv 2
( mv) - (+mv)
2L1v
L
L
4
唔一一气
-u
According to Newton's law of action-reaction , the forc巳 applied to the wall by the particle is equal in magnitllde to 由 is value , bllt oppositely directed (i .e., +mv2/L). Th巳 magm­
tllde F of the total force exerted on th巳 right wall is equal to the nllmber of particles that
collid巳 with the wall during th 巳 time t mllltiplied by th巳 averag巳 force 巳X巳 rted by each
p 创ticle. Since the N particles move randomly in three dimensions , one-third of them on
th 巳 av巳rage strike the right wall during th巳 time t. Th 巳refore ， th巳 total forc巳 IS
F=(子)(子)
In this reslllt v 2 has b巳巳n [巳plac巳d by v 2 , the αverage value of the squared speed. The
collection of p缸 ticles possesses a Maxwell distribution of speeds , so an averag巳 value for
v 2 must b~ used , rather than a value for any individllal particle. The sqllar'e root of the
qllantity v 2 is called the root-mean-square speed, or, for short, the rms speed;
v rms = \j v 2 . With this substitution , the total force becomes
F=(子)(千)
Pressur巳 is force per unit 创·巳a， so the pressure P acting on a wall of ar'ea L 2 is
P
=..!..一 ( !i ì ( mV~ns ì
二一·一..
u
飞 3 ) 飞
---.
L3
J
Since the volum巳 of the box is V = L , the equation above can be written as
3
PV= i N(imuins)
(1 4.5)
''T he term "elastic" is used here to mean that 011 the average , in a large llu l11 ber of particles , there is no gai n or
loss of translational kinetic en巳rgy because of collisions
Figure 14.10 A gas particle is shown
colliding elastically with the right wall
ofth巳 container and rebollnding from it.
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bu
k
This result is similar to th巳 ideal gas law, PV = NkT (Equation 14 勾 Both 巳咀旦tions have
identical terms on the left, so the terms on the right must be equal: 沪(阻) = Nk T.
Ther巳for巳，
阻
2" 叫ns
=
~kT
(1 4.6)
Equation 14.6 is significant , because it a1 lows us to interpret temperature in terms of th巳
motion of gas particles. This equation indicates that the Kelvin temp巳rature is d让巳ctly proportional to th巳 average translational kin 巳tic energy p巳r particle in an ideal gas , no matt巳r
what the pressure and volume ar巳 On th巳 average ， the particles have greater kinetic
energies when the gas is hotter than when it is cooler. Conceptual Example 5 discusses a
common rrusconc巳ption about the r巳 lation b巳tween kinetic en 巳rgy and t巳mperature.
Conceptual Example 5
Does a Single Particle Have a Temperature?
Each particle in a gas has kin巳 tlC 巳e阳巳
ne r皂
gy. Fu巾巳町rrr
盯
mo
the re巳 lationship b巳创tw巳巳 n 由
t h巳 av巳rag巳 kin巳tlC energy p巳町r particle and 由
t h巳 t忧巳 mp巳ratur巳 of an
gas. Is it valid , then , to conclude that a single particle has a temperature?
id巳al
Reasoning and Solution We know that a gas contains an 巳normous number of particles
that are traveling with a distribution of speeds, such as those indicated by the graphs in Figure
14.8. Therefore , th巳 particles do not all have the same kinetic energy, but possess a distribution
ofkin巳tic en 巳rgies ranging from very nearly zero to extremely larg巳 values. If each particle had
a temperature that was associat巳 d with its kin巳 tlC en巳 rgy， there would be a whole range of diι
ferent temp巳 ratures within the gas. This is not so , for a gas at thermal equilibrium has only on巳
temperature (see Section 15.2) , a temp巳 rature that would be registered by a thermometer placed
in the gas. Thus , temperature is a property that characterizes the gas as a whole , a fact that is
inherent in the relation ~ mv~ns = ~ kT. The
览巳 t忧巳rm 叫
vJ斗ι口nrns
Th巳re巳for陀巳， ;mUιs is the 卢 verage kinetic energy per particle and is characteristic of the gas as
a whole. Since the Kelvin temperature is proportional to ~ mvιs' it is also a characteristic of the
gas as a whol巳 and cannot b巳 ascribed to each gas p缸ticle individually. Thus , a single gas
particle does not have a tetr.ψ eraω re.
If two id巳al gases have th巳 sam巳 temp巳削ure， the relation ~m哈s = ~ kT indicates that
the average kin巳tic energy of each kind of gas p缸ticle is th巳 sam巳 . 1n general , however, the
rms speeds of th巳 different particles are not the same , becaus巳 the masses may be diffl巳r­
en t. The n巳xt example illustrates these facts and shows how rapidly gas particles move at
normal temperatures.
Example 6
The Speed of Molecules in Air
Air is primarily a mixture of nitrogen N 2 (molecular mass
behaves like an
id巳a1
gas and
det巳rmine
the rms
=
28.0 u) and oxygen O2 (molecular mass = 32.0 u). Assum巳由 at 巳ach
and oxygen molecules when the air temperature is 293 K
sp巳巳d of 出巳 nitrogen
Reasoning As Figure 14.8 illustrates , the same type of molecules (巳 .g. ， O2 molecules) within a gas have different speeds ,
even though the gas itself has a constant temperature. The rms speed V rms of the mol巳cules is a kind ofαverage speed a吐 is
related to the kinetic energy (also an average) according to Equation 6.2. Thus , the average translational kinetic 巳nergy KE of a
molecule is 臣
KE = ; nm
肌
1仅Z 叫
t阳emp
巳阿erat
町阳
u町re
陀巳 Toft巾
h出巳 g
伊
a臼s 由山阳
rou咄
白
g
h 由耻巳 r附d
由巳
lat
圳
创t阳
ion
∞n E = i扫kT(
抑向
句qu川
E
阳
m
a创
∞
tiiωO叫
n川14.6
州
6仿) . 悦巳
WI c创
an find 配
th忱e nns
ηns 叩
sp巳创
ed 巾
0 f ea灿
ch 仰
ty叩
严巳
pe ofm
日
moω
时
伽
01l
t由
h巳n
民， from a knowledge of its mass and the temperature.
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C
m
C
lic
k
Equation 14.5 relates the macroscopic properties of the gas- its pressure and volumew
.c
.d o
to the microscopic properties of the constitu巳nt particles-their mass and speed. Sinc巳 th巳
c u -tr a c k
term ~mv.zm s is the average translational kinetic 巳I阳gy KE of an individual particle , it
follows that
PV = ~ N(阻)
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CHAPTER 14TH 巨 IDEAL GAS LAW AND KINETIC THEORY
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N
.c
Description
Symbol
Value
T
28.0 u
32.0 u
293 K
Molecular mass of nitrogen (N 2)
Mol巳cular mass of oxygen (0 2 )
Air temperature
Unknown Variable
Rms speed of nitrogen and oxygen molecules
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Knowns and Unknowns The data for this problem are:
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9
V rms
Modeling the Problem
E圃 叫倪创仇叫叫
d归阳
ωa
a川
缸
nE
沁
i sre
巳 lat旬
ed tωo
kin巳tic
energy KE by
(6.2)
U
where m is its mass. Solving for V rmS gives Equation 1 at
average kinetic 巳n巳rgy KE in Step 2.
th巳 righ t.
We will deal with the
…
回国 的er咿Kinet山 ergy and Temperature Since 阳 gas IS a
d to be
an id巳 al gas , the average kinetic energy of a molecule is directly proportional to the
Kelvin temperature T of the gas according to Equation 14.6:
| KE
= %kT I
=
F
f
E
'飞
=j 叫
E
l
( ''i)
甲川
where k is Boltzmann's constan t. By substituting this relation into Equation 1 at the right,
we can obtain an expression for th巳 rms speed.
Solution Algebraically combining the results
of 出巳 two
steps , we have
In this result , m is the mass of a gas molecule (in kilograms) . Since the molecular mass巳S
of nitrogen and oxygen are given in atomic mass units (28.0 u and 32.0 u, respectively) , we
must convert them to kilograms by using 由e conversion factor 1 u = 1. 6605 X 10- 27 kg
(see Section 14 .1). Thus ,
( 1.6605 X 10- 27 kg \
Nitrogen
mN2
= (28.0 u)\
Oxygen
mo ,
= (32.0 u)!\ 1 U
-
1
••
'"
\1 u
J
/
( 16605 X 10- 27 kg\
'" )
r
r
= 4.65
Á
"
•
^ _?f,
X lO- L Ò kg
= 5.31 X lO- l ò kg
The rms speed for each type of molecule is
店主旦 -
Nitrogen
V rmS
刊一飞
Oxygen
Vrms
砰=
V
mN,
13: 日(1.3 8
V
x 10-叫/K)(293 K)]
4.65 X 10- 26 kg
日t口
一 1511 m/sl
巳二二二二二
2 日 (1.3 8 X 10- 23 J/K)(293 K)l
币汇力
一
L = 1478 mJsl
5.3 1 X lO- L C> kg
L一一一一」
Note that th巳 nitrogen and oxygen molecules have the same average kinetic 巳nergy，
since the temperature is the same for both. The fact that nitrogen has the greater rms
speed is due to its smaller mass.
Related Homework: Problems 33, 58
、 ，，J
the average
r'
a gas
Problem-solving insight
The average translational kinetic energy is
the same for all ideal-gas molecules at the
same temperatu 陀， regardless 01 their
masses. The rms translational speed 01 the
molecules is not the same , however, because
it depends on the mass.
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I r川川川Y 川川日 MIC I 川AS
Chapter 15 deals with th 巳 science of thermodyna ll1 ics , in wh ich the concept of
energy plays an important role. Using the reslllts jllst dev巳 loped for the averag巳
translational kinetic 巳 nergy， we con c\ ude this s巳 ctio n by expressing th 巳 internal energy
of a monatomic ideal gas in a form that is suitable fo r use later on.
The int巳rnal energy of a sllbstance is th 巳 sum of the various kinds of 巳n 巳rgy that the
atoms or molecul巳s of th巳 substance possess. A monatomic ideal gas is cO ll1 posed of
single atoms. T hese atoms 创-巳 assum巳d to be so s ll1 a Il that th 巳 mass is conc巳 ntrated at a
point , with the r巳 s ul t that the mO ll1 ent of inertia 1 about the center of ll1 ass is negligible.
T hllS, the rotational kin etic 巳ne即 ~Iω2 is also 问ligibl巳. Vibrational kinetic and potential energi 巳 s are absent , b巳calls巳 th巳 atoms are not co nn 巳 ct巳 d by c h 巳 mica l bonds and ，巳x­
cept for elastic co lI isions , do not int巳 ract. As a r 巳 sult ， 由
t h 巳 int忧巳rna
剖l 巳 n陀阳巳创rgy U 臼
is 由
t h巳 total
int巳rnal
tran
丑础
创
a时
n1
i mU :Lnms = i kT a肌cc∞Oαr时.吐din吨g tωo Equ川a川t川ion 14札 t由h巳 internal 巳 nergy can be written in terms of
the Kelvin
te ll1 peratur巳 as
U = N( ~ kT)
Us ually , U is expressed in t巳 rms of the number of moles n , rath巳r than th巳 number of
atoms N. Using th 巳 fact that Boltz ll1 ann's constant is k = RINA , where R is the universal
gas constant and N A is Avogadro's nU ll1 ber, and realizing that NINA = n , w巳白 nd that
Monatomicidealgas
U = ~ nRT
(14 .7)
Thus , th巳 intern a1 energy depends on th巳 number of mol 巳s and the Kelvin temperature of the
gas. 1n fact , it can be shown that th 巳 int巳rnal en 巳rgy is proportional to the Kelvin temp巳 ra­
ture for any type of id巳 a l gas (巳 . g . ， monatomic , diatomic ，巳 tc.). For examp l巳， when hotair balloonists turn on th巳 burn巳 r， they incr巳 as巳 the temperature , and hence the internal
巳 nergy per mol巳 ， of the air inside the balloon (s巳e Figur 巳 14.11).
~ CHECK Y 。
ωUR 川
U胁
NDωDERSTANDI川川'酌
N、唱a
(The answers are
Figure 14.11
Sinc巳 air b巳 haves
approximat巳 Iy
as an ideal gas , th巳
per mol 巳 insid巳 a hot-air
balloon increases as the t巳 mperatur巳l
ríses. (Erik Lam/Alamy Im ages)
int巳rnal 巳nergy
伊
9 IV
阳
en
at the end of the book.)
11. The kinetic theory of gases assumes that , for a given collision time , a gas molecule
re bounds with the same speed after colliding with the wall of a containe r. If the speed after
th e collision were less than the speed before the co llision , the duration of the collisio n
remaining the same , would the pressure of the gas be greater than , equal to , or less than
the pressure predicted by kinetic theory?
12. If the temperature of an ideal gas were doubled from 50 to 100 oC, would the average
translational kinetic energy of the gas particles also double?
13. Th e pressure of a monatomic ideal gas doubles , while the volume decreases to
one-half its initial value. Does the internal energy of the gas increase , decrease , or remain
unchanged?
冒 4. The atoms in a container of helium (He) have the same translational rms speed as the
atoms in a container of argon (Ar) . Treating eac h gas as an ideal gas , decide which , if eithe r,
has the greater temperature.
15. The pressure of a monatomic idea l gas is doubled , while its volume is reduced bya
fa cto r of four. What is the ratio of th e new rms speed of the atoms to the initial rms speed?
o
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k
to
The 巳quation KE = ~ kT has also been applied to particles m 川 l吨er than atoms or
w
molecules. The Engli sh botanist Rob巳rt Brown (1 773-1858) observed through a micro.c
.d o
c u -tr a c k
scope that pollen grains SllSp巳 nded in water move on very irr巳gular， zigzag paths. This
Brownian motion can also be observed with other p 臼ticle S ll Sp巳 nsions ， such as 自 ne smoke
parti c\ es in air. 1n 1905 , A lb巳 rt Einstein (1879-1955) showed that Brow nian motion cOllld
b巳巳xplain巳d as a response of the large susp巳 nd巳 d particl巳 S to impacts from the moving
molecules of the f1 uid medium (巳 . g . ， water or air) . As a reslllt of th 巳 impacts ， the susp巳nd巳d
parti c\ es J!旦v e the sam巳 average translational kinetic energy as the f1 uid moleculesnamely，阻 = ~ kT. Unl ike 阳 mo l ecll l es ， however， 阳 pωticles are large 巳nough to b巳
seen through a microscop巳 and ， b巳cause of their relatively 1缸~g巳 mass ， have a cO ll1 paratively
s ll1 all av巳rag巳 speed .
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You can smell the fragrance of a perfume at some di s tan c巳 fro m an o p巳 n bottle because perfume molecules leave th 巳 s p ace above the Ii quid in the bottle , where they are relatively conce ntrated , and spread o ut into the air, wh 巳 r巳 th ey ω'e I 巳 ss concentrated. During
their j ourney, they collid巳 with o the r m o l巳c ul es ， so their paths rese mbl e the zigzag paths
ch aract巳ri s tic of Brownian motio n. Th巳 proc巳 ss in which 1110lecul es m o v巳 from a region of
higher conce ntration to one of lowe r con c巳ntration is called d.旷usion. Diffusion al so occurs in liquids and solids , and Fig ur巳 14 .1 2 illu s trat巳 s ink diffusing through wa t巳1'.
However, co mp ar巳d to th巳 rate of diffusio n in gases , th巳 rat巳 I S g巳 n 巳rall y smaller in Ii quids
and 巳ve n smaller in solids. T he host medium , such as th 巳 a i r o r water in the 巳xampl es
a b ov巴， is referred to as the solvent, while the diffusing substance , li ke the p巳 rfume m o l巳­
cules or the ink in Figure 14.12 , is known as the solute. Relati vely speaki ng , diffusion is a
slow process , even in a gas. Co n c巳 ptu a l Example 7 illu strat巳s why.
Conceptual Example 7
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f牛土豆\
*DIFFUSION
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Inilially
Laler
Figure 14.12 A drop of ink placed in
water eventllally becomes completely
dispersed beca ll S巳 of diffllsion.
Why Diffusion Is Relatively Slow
The fragrance from an open bottle of perfllme takes several seconds or sometimes even minlltes to reach the other side of a room by th巳 proc巳 ss of diffll sion. Which of the fo llowing accounts for the fact that diffllsion is relatively slow? (a) The natllre of Browni an motion (b) The
relati vely slow translational rms speeds that charac teri ze gas m o l 巳 c lll e s at room te mp巳ratllre
@
@
e
Reasoning The important charac teri stic of the paths fo llowed by objects in Brownian motion
is their zigzag s h ap巳 s. We have calclllated typical t:ranslational rms speeds fo r gas molecllles
near roo m te mp巳rature in Example 6, and those results will gllide our reaso ning here.
Answer (b) is incorrect. 1n Example 6 we have seen that a gas m o l eC lll 巳 n ear roo m t巳 mp era­
ture has a translational rms speed of hllnd reds of meters per second . Sllch speed s 缸·巳 n o t slow.
It wOllld take a moleCllle traveling at sllch a s pe巳 d jllSt a frac ti on of a s巳co n d to cross an ordinary roo l11
Answer (a) is correct. When a p巳rfllm e moleCllle di仔u ses thro llgh air, it makes millions of collisions each second with air moleCllles. As Figllre 14.13 illu s tra t巳 s ， th巳 V巳 l oc ity of th巳 m o l 巳c lll e
changes abruptly b巳ca ll s巳 of each collision , bllt b e tw巳巳 n co llisions , it m ov巳s in a straight lin巳.
Althollgh it does move very fas t between colli sions , a pe rfllm巳 m o l ec lll e wanders only slowly
away fro m the bottle becallse of the zigzag path . It wOllld t ak巳 a long tim巳 ind 臼 d to diffuse in
this manner across a room. Usually, h owev凹， convection Cllrrents are present and calTy th巳 fra­
gra nc巳 to the other sid巳 of th巳 roo m in a matter of seconds or minlltes .
@
@
Figure 14.13 A P巳rfllme molecllle
collides with mjJj ions of air molecllles
dllring its jOllrney, so 由巳 p ath has a
zigzag s h ap巳 A ltholl gh the air
molecllles ar巳 s h own as stationary, they
are also moving.
Related Homework: Problems 49, 51
Diffusion is th巳 basi s for drug delivery sy st巳 ms that bypass th巳 n巳ed to admini s t巳r
medicatio n orally or via injecti o ns. Fi g ur巳 1 4 .1 4 shows o ne such syste m , th巳 tran sd e rm al
patch. The word " tran s d巳rm al " m 巳 a n s "across the skin ." Such patc hes , for 巳 x ampl巴 ， are
Conlrol
membrane
....
\
The physics of
drug delivery systems.
Figure 14.14 Using diffusion, a
transdermal patch deli v巳rs a drllg
directly into the skin , where it 巳 nte rs
blood vessels. The backing contains the
dru g within the reservoir, and the
control m巳 mbran e limits the rate of
diffllsion into the skin
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(α)
Cross-sectional area = A
Figure 14.15 (α) Solute diffuses
through the channel from the region
of high巳 r concentration to th巳 region
of lower conc巳ntration . (b) Heat is
conduct巳d along a bar whose ends are
maintained at diff，巳rent temperatures.
(b)
、，
，-
φ'' -
h、
T 一
'κ­
A 一
used to deliver nicotine in programs d巳 slgn巳d to help you stop smoking. Th巳 patch is attach 巳d to the skin using an adh巳 sive ， and the backing of the patch contains the drug within
a reservoir. Th 巳 concentration of the drug in the reservoir is relatively high , just lik:e th巳
concentration of p巳tfume molecul 巳 s above the liquid in a bottle. Th 巳 drug diffus巳 s slowly
through a controI membrane and directly into th巳 skin ， where its conc巳 ntration is relatively
low. Diffusion carries it into th巳 blood v巳 ssels present in the skin. The purpose of the control m巳 mbrane is to limit the rate of diffusion , which can also b巳 adjusted in the r巳 servoir
by dissolving th巳 drug in a neutral mat巳rial to low 巳 r its initial concentration. Another
diffusion-controlled drug deliv巳ry system utilizes capsules that ar巳 insert巳 d surgically
beneath the skin. Contraceptiv巳 Sal 巳 adrninist巳r巳 d in this fashion , for instance. The drug in
the capsule diffuses slowly into the bloodstream over extended p巳riods that can b巳 as long
as a y巳ar.
The diffusion proc巳 ss can b巳 described in terms of the arrangement in Figure 14.15a.
A hollow channel of length L and cross-s巳 ctiona l area A is filled with a f1 uid. Th巳 I巳ft end
of the channel is conn 巳cted to a contain巳r in which the solute conc巳 ntration C2 is relatively
high , while the right 巳 nd is conn巳cted to a container in which the solut巳 conc巳 ntration C 1
is lower. These conc巳 ntrations are defined as the total mass of th巳 solute molecules divided
by the volume of the solution (e.g. , 0.1 kg/m 3). B 巳cause of the difference in conc巳ntratlO n
between the ends of the channel , D. C = C2 - C 1, th巳re is a net diffusion of the solute from
th巳 left end to the right 巳 nd.
Figure 14.15αis sirnilar to Figure 13.7 for the conduction of heat along a bar, which ,
for convenience , is reprodllced in Figure 14.15b. When the ends of the bar are maintained
at differ巳nt temperatures , T2 and T 1, the h巳at Q conducted along the bar in a time t is
A-L
(
Q (13.1)
where D. T = T 2 - T" and k is the thermal condllctivity. Wher巳 as conduction is the flow of
heat from a region of higher temperature to a region of low巳r temperature , diffusion is th巳
mass flow of solute from a region of high巳r conc巳 ntration to a region of lower concentration. By analogy with Equation 13.1 , it is possible to write an equation for diffusion:
(1) replac巳 Q by the mass m of solute that is diffusing through th 巳 channel ， (2) replac巳
D. T = T2 - T 1 by the diff，巳r巳nce in concentrations D. C = C2 - C 1, and (3) replac巳 k by a
constant known as the diffusion constant D . The resulting 巳 quation ， first formulated
by the German physiologist Adolf Fick (1829-190 日， is referred to as Fick's law of
d旷usion.
m
o
o
c u -tr a c k
C
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k
to
Cross-sectional area = A
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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
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The mass m of solute that diffuses in a time t through a solvent contained in a channel
of length L and cross-sectional ar巳aA is*
(DA ð. C)t
.d o
m
o
m
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o
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C
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to
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FICK'S LAW OF DIFFUSION
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DIFFUSION F 429
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(1 4. 8)
m= ----一
L
where ð. C is th巳 concentrati o n differe nce be tween th e 巳 nds of the channel and D is the
diffusion constan t.
51 Unit for the Diffusion Constant: m 2/s
It can be verified fro m Equation 14.8 that th巳 diffu sio n constant has units of m 2 /s, the exact valu巳 de p巳 ndin g o n th巳 n ature of the solute and th巳 sol ve n t. For example , th巳 diffu sio n
constant for ink in w at巳r is differe nt from that for ink in be nzene . Example 8 illustrates an
important application of F ick's law.
Example 8
Water Given Off by Plant Leaves
Large amounts of water can b巳 gi ve n off by plants. It has been esti m at巳d ， for in s tanc巳， that a
single sunflower plant can lose up to a pint of water a day during th巳 growing season. F igure
14.16 shows a cross-sectional view of a leaf. Inside the leaf, water passes fro m the liquid phase
to the vapor phase at the walls of the mesophyll cells. T he water vapor then diffuses through
the intercellular air spaces and eventuall y ex its the leaf through small openings , call 巳d stomatal
pores. The diffusion constant for water vapor in air is D = 2 .4 x 10 m 2/s. A stomatal pore has
a cross-sectional area of about A = 8.0 X 10- 11 m2 and a length of about L = 2.5 x 10- 5 m. The
conc巳ntrati on of wat巳r vapor on the interi or side of a pore is roughly C2 = 0.022 kg/m 3 , while
the concentration on the outsid巳 i s approximately C 1 = 0.011 k g/m 3 . Det巳rmine the mass of
water vapor that passes through a stomatal pore in one hour.
The physics of
water loss from plant leaves.
•
Reasoning and Solution Fick's law of diffusion shows that
(DA ð. C)t
112
(14.8)
L
112 =
(2 .4 X 10- 5 m 2/s)(8.0 X 10- 11 m 2 )(0.022 kg/m 3
2.5 X 10 m
-
0.0 11 kg /m 3)(3600 s)
•
Mesophyll
cells
= 13.0 X 10- 9 时
This amount of water may not seem significan t. However, a single leaf may have as many as a
mi llion stomatal pores , so the w at巳r lost by a n 巳 n t:ire pl ant can be substantial.
。
y" CHECK
epidermis
Y。
ωUR UNDERS τ A 酌NDI川川
E酌N、唱a
(The answers aιre gl 阳
V，en at the end of the book.)
16 . τ~ In the lungs , oxygen in very small sacs called alveoli diffuses into the blood . The
曹 diffusion occurs directly t hrough the walls of the sacs , which have a thic kness L
The total effective area A ac ross which diffusion occurs is th e sum of the individual areas
(each quite sma ll) of the various sac walls. Considering the fact that the mass m of oxygen
that enters the blood per second needs to be large and referring to Fick's law of diffusion ,
what can you deduce about L and about the total number of sacs present in the lungs?
'在 The same solute is diffusing through th e 户
same solvent in each of three cases . For
Case
Length
Cross-Sectional Area
(a)
~L
A
each case , the table gives the length and
cross-sectional area of the diffusion channe l.
(b)
L
~A
The concentration difference between the
ends of the diffusion channel is the same i n~
~L
2A
each case. Rank the diffusion rates (in kg/s)
in descending order (Iargest first).
law assumes that th巳 te m peratu re of the solvent is constant throughout the channel. Experiments
indicate that the d i仔u s i o n constant depends strongly on the temperature
不 F i ck ' s
} LOWEr
Figure 14.16 A cross-sectional view of
a leaf. Water vapor di旺ü ses out of the
leaf through a stomatal pore.
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This chapt巳r introduces th巳 id巳 al gas law, which is a r巳 lation b巳 twe巳 n the pressure ,
volume , temperature , and number of moles of an ideal gas. In Section 10.1 , we examined
how the compression of a spring dep巳 nds on the force applied to it. Example 9 revi巳ws
how a gas produces a force and why an id巳 al gas at different temperatures causes a spring
to compress by different amounts.
(α)
Concepts & Calculations Example 9
The ldeal Gas Law and Springs
Gas 1
(b)
Figure 14.17 shows thre巳 id巳 ntical chambers containing a piston and a spring whose spring
constant is k = 5.8 X 10 4 N/m. The chamber in p缸.t a is completely evacuat巳d ， and the
piston just touches its left end. 1n thi s position , the spring is unstrained . 1n part b of the
drawing , 0.75 mol of ideal gas 1 is introduc巳d into the chamb町， and the spring compresses by
XI
15 cm. 1n part c , 0.75 mol of ideal gas 2 is introduced into the chamber, and the spring
compr巳 sses by X2 = 24 cm. Find the temperature of each gas
Concept Questions and Answers Which gas exerts the greater
Answer
amoun t.
W，巳 know
Th 巳 refore ，
(c)
Fi gure 1 4.17 Two ideal gases at
different temperatures compress the
spring by different amounts
the piston ?
that a great巳 r force is r巳quired to compress a spring by a greater
gas 2 巳xerts the greater forc巳.
How is the force required to compress a spring related to
its un strain巳d position?
Gas 2
forc巳 on
th巳 displac巳 ment
of th巳
spring
from
Answer According to Equation 10.1 , the applied forc巳 Fr^pplied required to compress
a spring is directly proportional to th巳 displac巳 ment X of the spring from its un strain巳d
position; F/ pplied = kx , wher巳 k is the spring constant of the spring.
Which gas exerts the greater pressure on the piston ?
Answer According to Equation 1 1. 3, pressure is d巳白 ned as th巳 mag nitude F of the force
acting perpendicular to the surface of the piston divided by the area A of the piston;
P = F/A. Since gas 2 ex巳 rts the greater force , and th巳 area of the piston is the sa m巳 for
both gas巳s ， gas 2 exerts the gr 巳 at巳r pressur巳.
Which gas has the greater temperature?
Answer According to 由 e id巳 al gas law, T = PV/(nR) , gas 2 has t h 巳 greater temperature.
Both gases contain the same numb巳 r n of moles. How巳V巳r， gas 2 has both a greater pressure P and greater vo lum巳 V. Thus, it has the great巳 r temperatur巳
Solution W巳 can use the ideal gas law in the form T = PV/(nR) to determine th巳 temp e rature
T of each gas. First , how巳ver， w巳 must find the pressure. According to Equation 11 .3 , th巳
pressure is th巳 magnitude F of the forc巳 that the gas ex巳rts on th巳 piston divided by the 盯ea
A of the piston , so P = F/A. R巳call that the forc巳 F/pplied applied to the piston to compress
the spring is related to the displacement X of the spring by F/ pplied = kx (Equation 10.1). Thus ,
F = F/pplied = kx , and the pressure can be written as P = kx/A. Using this expression for the
pressure in th巳 ideal gas law gives
T =
PV
一
nR
(子)
V
-
nR
However, the cylindrical vo lurn巳 V of the gas is equal to th巳 produ c t of the distance X and
缸ea A , so V = xA. With this substitution , the expression for the temperature becornes
(子) V (子)俐
Th巳 temperaωres
Gas 1
kx 2
of the gases are
Ta
kXl
..." 2
nR
(58
lVN/m)(15
日百曰
,_ .- ×.. ... ....".- × 10-2 m) 2 = 1210
K1
(0.75 mol)[8.31 J/(rnol' K)]
'-一-'
th巳
m
w
Unstrained spring
o
m
o
.c
lic
k
GONGEPTS & GALGULATIONS
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c u -tr a c k
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THE IDEAL GAS LAW AND KINETIC THEORY
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As anticipated , gas 2 , which compresses the spring more , has
1
'-一一一」
w
th 巳 high巳r t巳mp巳rature.
。
The kinetic theory of gases is important because it allows us to understand the re1ation betw巳en the macroscopic properties of a gas , such as pressure and temp巳ratur巳， and
th巳 microscopic prop巳rties of its parti c1es , such as sp巳巳d and mass. Th巳 following example
reviews th巳巳 ssential features of this th巳ory.
Concepts &
Ca lculatïons 庭xample 冒 O
O Hydrog川toms in Outer SP哇
In outer spac巳 the d巳nsity of matter is 巳xtrem巳 ly low , about on巳 atom per cm 3. Th巳 matt巳r IS
mainly hydrog巳 n atoms (m = 1. 67 X lO - 27 kg) whos巳 rm s speed is 260 mls. A cubical box ,
2.0 m on a si 白 ， is placed in out巳 r space , and 出巳 hydrogen atoms are allowed to enter. (a) What
is the magnitud巳 of the forc 巳 that the atoms 巳X巳 rt on on 巳 wall of th 巳 box? (b) Determin 巳 th 巳
pressure that the atoms 巳X巳rt. (c) Does outer space hav巳 a temperatur飞 and ， if so , what is it?
Concept Questions and An swers Why do hydrog巳 n atomsexe此 a forc巳 on 出巳 walls ofthe box?
Answer Every tim巳 an atom col1id巳 s with a wall and rebounds , th巳 atom exerts a force
on th巳 waU. lmagine op巳 ning your hand so it is Aat and having som巳on巳 throw a ball
straight at it. Your hand is like the wall , and as th巳 ball rebounds , you can feel th巳 force
Intuitively, you wou ld expect the force to be∞ me greater as the speed and mass of th巳 ball
become greater. This is indeed th巳 cas巳
Do the atoms generate a pressure on the walls of th巳
box?
Answer Yes. Pressure is defìned as the magnitude of the force exerted perpendicularly on
a wall divided by th巳 area of the wall . Since the atoms 巳X巳rt a forc巳， th巳y also produc巳 a
pressure.
Do
hydrog巳 n
atoms in outer space have a
properties of the atoms?
temperatUJ 巳?
If so , how is the temperature
r巳 lat丑d
to
th 巳 microscopic
Answer Yes. The Kelvin temperatul 巳 is proportional to the averag巳 kin 巳tic energy of an
atom . The averag巳 kin巳 tic energy, in turn , is proportional to the mass of an atom and th 巳
square of the rms speed. Since we know both these quantities , we can d巳t巳 I币un 巳 the temp巳 rature of the gas
Solution (a)
Th巳 magnitud巳 F
of the force
exert巳d
on a wall is given by
(se巳 S 巳ction
14.3)
F=(子)(千)
where N is th巳 number of atoms in th巳 box ， m is th巳 mass of a single atom , V rmS is the rms sp巳巳d
of the atoms , and L is th巳 l巳 nσth
of one side of th 巳 box. The volume of th巳 cubical box is
D
(2 .0 X 10 2 cm)3 = 8.0 X 10 6 cm 3. The number N of atoms is equal to the numb巳r of atoms
p巳 r cubic centimeter times th巳 volum巳 of the box in cubic centimet巳rs
N =(非主) (8.0 X 川3) =
8.0
X川
The magnitude of the force acting on one wall is
叶子) (千) = (乓旦) [Jl67 × 12)问
(b) The pressure is the magnitude of the force divided by
F
A
p=~=
1. 5 X lO -
16
(2.0 m) L
N
th巳 area
= 13.8 X lO -
A 0f a wall:
川
11
Pa
l
1
.d o
m
1
to
日石Kl
540 K
k
=
o
.
ack
bu
y
o
m
C
c u -tr
w
lic
lic
k
to
To吨=一二一= _k_X2_2 = ....:(_5._8_X_l_04_N_/n--'l)~(2_4_X_l.-:...0→
m)2
- . ...,,--,
nR
(0.75 mo1)[8.31 J/(mo1.K)]
Gas2
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14.5 CONCEPTS & CALCULATIONS
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THE IDEAL GAS LAW AND KINETIC THEORY
N
14
to
k
T = mV~ms = ( 1. 67 X \0 - 27 kg)(2ωm/s)2日. 7 KI
= 一一一一=
一
= 12.7 KI
3k
3( 1.38 x 10 一 υJ/K )
L一一」
This is a frigid 2.7
k巳 lvins
above absolute zero.
。
nON'CEPT SUMMARV
If you need more help with a concept, use the Learning Aids noted next to the discussion or equation. Examples (Ex.) are in the text
of this chapter. Go to www.wiley.com/college/cutnell for the fo llowing Learnin g Aids:
Interactive LearnlngWare (l LW) - Additional examples solved in a fi ve-step
Concept 51mulations (CS) Interactlve 5olutlons
(15) 一-
Animat巳 d
forma t.
text fi gures or animations of important concepts.
Models for certain types of problems in
th巳 chapter
homework. Th巳
calculations
are carried out
int巳 ractively
learning Aids
Discussion
Topic
Atomic mass unit
Molecular mass
int巳 racti ve
14.1 MOLECULAR MASS , THE MOLE , AND AVOGADRO'S NUMBER Each element in the periodic
table is assigned an atomic mass. One atomic mass uni t (u) is exactly one-twelfth the mass of an
atom of carbon-12. The mol巳cul ar mass of a molecule is the sum of the atomic mass巳s of its atoms.
The number of moles n contain巳d in a sample is equal to the number of particles N (atoms or molecu les) in the sample divided by the number of particles per mole N A ,
N
n= - -
Number 01 moles
Avogadro's number
Number 01 moles
Mass per mole
NA
NA is called Avogadro's number and has a value of NA
= 6.022
X 1023 partic\ es p巳 r mol 巳.
The number of moles is also equal to th巳 m ass m of the sample (expressed in grams) divided by
the mass per mole (express巳d in grams per mole):
nz
Ex.
n
Mass per mole
wh巳re
The mass per mol 巳(i n g/mol) of a s ub s tan c巳 h as the salη巳
lecular Ill ass of one of its parti c\ es (i n ato mic mass units).
nUlllerical
1
value as the atolllic or mo-
The mass mpart i出 of a particle (in grams) can be obtai ned by dividing the mass per mole (i n g/mol)
by Avogadro's number:
Mass per mole
Mass 01 a particle
lηparti cJc
A. l
" A
14.2
THE IDEAL GAS LAW Th巳 i d ea l gas law relates the absolute pressure P, the volume V, the
n of mol郎， and the Kelvin temperature T of an ideal gas accord ing to
numb巳r
PV
Ideal gas law
Universal gas constant
where R
=
8.31 J/(mol ' K) is th巳
unive rsa l
Boltzmann's constant
where N is the number of parti c\ es and k
id巳al
( 14.1 )
gas constan t. An alternative fonn of the ideal gas law is
PV
Ideal gas law
nRT
=
R
= 一一-
NA
NkT
=
(14.2) Ex. 2, 3 , 9
is Boltzmann's constan t. A real gas behaves as an 15 14.25, 14.29
gas when its density is low enough that its p创ticles do not interact , except via elastic collisions.
A form of the ideal gas law that applies when the number of 1ll 01es and the temperature are constant is known as Boyle's law. Using the subscripts "i" and "f" to denote , re spectiv巳 I y， initial and
final conditions , we can write Boyle's law as
Pi 只 =
Boyle's law
PrVr
(1 4.3) EX.4
A form of the ideal gas law that applies when the number of moles and the pressure are constant
is called Charles' law
Charles' law
14.3
Maxwell speed distribution
只
Vr
盯
Tr
(1 4 .4)
KINETIC THEORY OF GASES The distriblltion of particle speeds in an id巳al gas at constant
is the Maxwell speed distribution (see Figllre 14.8). The kinetic theory of gases
temp巳 ra ture
o
c
k.
m
lic
o
c u -tr a c
C
m
C
lic
k
to
(c) According ωEquation 14.6 , the Kelvin t巳 mperature T of the hydro g巳 n atoms is relat巳 d to
th巳 av巳r吨巳 kin巳tic energy of an atom by ~ kT = tmv~ms ' where k is Bo !tzmann 's constant
w
.c
.d o
c u -tr a c k
Solving thi s equation for the temperature gives
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」n
『
.
.t6
@司
、
ea
aE
"u
『、
Average Iranslalional kinelic
energy
nnnuBt man" uea
E
"
"
"Mr aE
indicates that the Kelvin temperature T of an ideal gas is related to the
energy KE of a particle according to
averag巳 translational
阻 =jmuι = ~ kT
where v rms is the
root-mean-squar巳 speed
kinetic
(14.6)
.d o
m
o
m
o
c u -tr a c k
C
k
lic
C
to
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y
to
bu
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Learning Aids
Oiscussion
Topic
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EX.5.6.10
.......,
IS 14.39, 14.61
n
of the particles
The internal energy U of n moles of a monatomic ideal gas is
U = ~ nRT
Inlernal energy
The internal energy of any type of
Kelvin temperature
id巳al
(1 4.7)
gas (e.g. , monatomic , diatomic) is proportional to its
14.4 DIFFUSION Diffusion is the process wher巳by solute molecules move through a solvent
from a region of high巳r solute concentration to a region of lower solute concentration . Fick's law
of diffusion states that the mass m of solute that diffuses in a tim巳 t through the solvent in a channel of length L and cross-sectional area A is
(DA /::"C)t
Fick's law 01 diHusion
(14.8) Ex. 7, 8
m= 一一一一一一
L
wh巳re /::,. C is the solute concentration
diffusion constan t.
di旺er巳 nc巳 betwe巳n
Note to 11I structors: All of the questions shown here are available for
or WebAssign
Section 14.1 Molecular Mass, the Mole ,
and Avogadro's Number
1. All but one of the followin!!: statements 缸巳 true. Which one is not
true? (a) A mass (in grams) equal to the molecular mass (in atomic
mass units) of a pure substance contains the same number of molecules , no matter what the substance is. (b) One mole of any pure
substance contains the same number of molecules. (c) Ten grams of
a pure substance contains twic巳 as many molecules as five grams of
the substance. (d) Ten grams of a pure substance contains th巳 sam巳
number of molecules , no matter what the substance is. (e) Avogadro's
number of molecules of a pure substance and one mole of the substance
have the same mass.
ass号nment
the ends of the
chann巳1
and D is the
via an online homework management program such as WileyPLUS
initial value , while the temperature is held constan t. How many moles
/::"n of the gas must be allowed to escape through the valve , so that the
pressure of the gas does not change?
。
2. A mixture of ethyl alcohol (molecular mass = 46.1 u) and water
(molecular mass = 18.0 u) contains one mole of molecules. The
mixture contains 20.0 g of ethyl alcohol. What mass m of water does
it contain?
Section 14.2
The Ideal Gas Law
3. For an ideal gas , each of th巳 following unquestionably leads to
an increas巳 in the pressure of the gas , except one. Which one is it?
(a) Increasing the temp巳rature and decreasing the volume , while
keeping the number of moles of the gas constant (b) Increasing
the temperature , the volume , and the number of mol巳 s of the gas
(c) Increasing the temperature , while keeping the volume and the
number of moles of the gas constant (d) Increasing the number of
moles of th巳 gas ， while keeping the t巳 mperature and the volume
constant (e) Decreasing the volum巴 ， while keeping the temperature and the number of moles of the gas constant
4. The cylinder in the drawing contains 3.00 mol of an ideal gas. By
moving the piston , the volume of the gas is reduced to one-fourth its
5. Carbon monoxide is a gas at 0 oC and a pressure of 1.01 x 10 5 Pa
It is a diatomic gas , each of its molecules consisting of one carbon
atom (atomic mass = 12.0 u) and one oxygen atom (atomic mass =
16.0 u). Assuming that carbon monoxide is an ideal gas , calculate its
density p
Section 14.3
Kinetic Theory of Gases
6. If the speed of every atom in a monatomic ideal gas were doubled ,
by what factor would the Kelvin temperature of the gas be mu 1t iplied?
(a) 4
(b) 2
(c) 1
(d) ~
(e) ~
7. Th巳 atomic mass of a nitrogen atom (N) is 14.0 u, while that of an
oxygen atom (0) is 16.0 u. Three diatomic gases have the same temperature: nitrog巳n (N2) , oxyg巳n (0 2) , and nitric oxide (NO). Rank these
gas巳 s in asc巳nding ord巳r(small巳st 白 rst) ， according to the values of their
translational rms speeds: (a) O 2, N 2, NO (b) NO , N 2, O2 (c) N2,
NO , O2 (d) O2, NO , N 2 (e) N 2, O 2, NO
8. The pressure of a monatomic ideal gas is doubled , while the volume
is cut in half. By what factor is the internal energy of the gas mu 1tiplied? (a) ~ (b) ~ (c) 1 (d) 2 (e) 4
.c
THE IDEAL GAS LAW AND KINETIC THEORY
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of 0.25 cm. Wh at wOllld be the diffu sion rate 111. /1 in a channe l with a
cross-secti onal ru 巳a of 0.30 cm2 and a length of 0. 10 cm?
PROp LEMS
Note to Instructors: Mosl of Ihe homework p lV blems in Ihis chapter are available fo r ass ig ll lll en t
as Wil 巳y P LUS
or WebAssign , a l1 d those marked
ùZleraclivity. See p， 呐ce fo r additional details.
l\I
ilh Ihe ico l1
NOle: The pre
Molecular Mass, the Mole, and Avogadro's Number
1. ssm Hemoglobin has a molecul ar mass of 64500 u. Find
(in kg) of one m o l 巳c llle of hemoglobin .
th巳 m ass
啕害".
Manllfacturers of headache re medi es rOlltinely c1 aim that
th e ir ow n brands are more potent pain re li ev巳rs than th巳 co m
peting brands. The ir way of makin g th巳 co mp ari s on is to compare the
nllmber of mo lecllles in th巳 s ta nd a rd do s ag巳 Ty l e n o l lI ses 325 mg of
acetamin ophen (C 8H9N0 2) as the standard d os巳， while Advil lI ses
2.00 X 10 2 mg of ibllprofen (C I3H 180 2)' Find the nllmb巳r of mo lecllles
of pain re li巳ve r in the standard doses o f (a) Tyleno l and (b) Advi l
2.
自
3. The arti 白 c i a l sweete ner Nu traS weet is a chemi cal called aspartame (CI4H I8N20 5)' What is (a) its mo leclllar mass (i n atomic mass
units) and (b ) the mass (in kg) of an aspartame mo lecul e?
4. ③ A 削ain 巳l巳m削 h as a mass per m由 of 196.967 g/mol
What is the mass of a single atom in (a) atomic mass units and
(b) kil ograms? (c) How many moles of atoms ru 巳 in a 285-g sampl e?
5. 唱伊 ssm
The active ingredient in the aJl ergy medication Claritin
contains carbon (C), hydrogen (H) , chl orine (CI) , nitrogen (N) ,
and oxygen (0 ). Its moleclllru' formul a is C22H23CIN20 2' The standard
adll lt dosage utili z巳s 1. 572 X 10 19 molecllles of thi s species. Determine
the mass (i n grams) of the active ingredient in the standru'd dosage.
~
6. A mass of 135 g of a certain element is known to contain
30.1 X 10 23 atoms. What is the e l巳 m ent ?
*7. A runner weighs 580 N (abollt 130 Ib), and 71 % of thls weight
is water. (a) How many m o l 巳s of water are in the runner's body?
(b) How many water mol ecules (H 2 0 ) 创毛 th el 巳?
*咱8. ③ Cωon ω
巾
s l耐d臼阳
E创
阳r a mi x
( m o l eclll 创
a r mass = 39.948 g/mη0 1 ) ， 2.60 g of neon ( m o le
巳C川
ll1让la
缸r mass =
20.1 80 g/mη 0 1 ) ， and 3.20 g of he Lillm ( mη01 巳cαωu 阳
l创
a r mηIaSS = 4.0026 g/mo l).
For this mi xture , determine the p巳 rc巳 ntag巳 of th巳 tota l number of atoms
that corresponds to each of the components
*9. ssm A cylindrical g lass of water (H20 ) has a radiu s of 4.50 cm and
a height of 12.0 cm. The density of water is 1. 00 g/cm3 How many
moles of water molecllles are contained in the glass?
*10.
唱铲 T he
preparati on of ho meopathi c " remedies" invo lves the repeated dillltio n of sollltions containing an acti ve ingredient
sllch as arseni c tri ox id巳 ( A S2 0 3 ) . S lI ppose one b巳g in s with 18.0 g of
arseni c tri ox ide di ssolved in water, and repeatedly dillltes the sollltion
widl pure water, each di lll tion redllcing the amolln t of 创 seni c tri oxid
~
an onlille
h Olll e wo rk 川allagelllenl
prog ram
SlI ch
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T叽l川n 巾
ssm Solution is in the Student Solutions Manua l.
www Solution is available online at www.
Section 14.1
νia
tml are presenled in Wiley PL US l/S川g a guided Iulorial forma l Ihat p lV vides enhanced
b川
ion
刷佣删川…
111阳阳叫
n阳E四e<阳
re maining in the sollltion by a fac tor of 100. Assumin g perFect mi xing at each diluti on, what is the max imum nllmber of diluti ons one
may p巳 rfo rm so th at at leas t one m o l 巳c ul e o f arseni c tri ox ide remains
in the diluted solution? For co mpari son, h o m巳o p a thi c " remedi es" are
co mm onl y dilllted 15 0 1' even 30 times
Section 14.2
T he Ideal Gas L aw
11. ssm 1t takes 0. 16 g of he lillm (He) to fì ll a ball oon. How many
grams of nitrogen (N 2) would b巳 reqllired to fìll the ball oo n to the
sa me pressure , vo lum 巴 ， and te mperatllre?
12. An empty oven w ith a vo lum巳 of 0. 150 m3 and a temperature of
296 K is tll rn巳d on. The oven is vented so th at the air p res s llr巳 in s id e
it is always the srun e as th e air pressllre of the enviro nmen t. lniti all y,
the air pressure is 1. 00 X 10 5 Pa , bll t after th巳 ove n has warmed up
to a fì nal temperatllre of 453 K , th巳 atm os ph eri c air pressllre has decreased to 9.50 X 10 4 Pa due to a c h a n bσe in weather co nditi ons. How
many m o l 巳s of air leave the ov巳 n whil e it is heating IIp?
13. An ideal gas at 15.5 oC and a pressllre of 1. 72 X 10 5 Pa occupi es
a volllme of 2.8 1 m3 (a) How Ill any moles of gas are present?
(b) lf th e volume is raised to 4. 16 m3 and th巳 temp巳rature raised to
28.2 oC , what will be the p l 巳ss ure of the gas?
14. (
Four c1 0sed tar盹 A ，川， and D ，巳川 contain an i出a l gas
The table gives the abso lute press llr巳 a n d vo lllme of the gas in each
tank. 1n each case , there is 0.10 11101of gas. Using this number and the
data in the table , compute the te mperatllre o f the gas in eac h tank.
Absolute
press LII 巳 ( Pa)
Volum 巳 ( m 3 )
A
B
C
25 .0
30.0
20.0
4 .0
5 .0
5.0
15. 电p' O xyg巳n
D
2.0
75
for hospital pati巳nts is kept in special tanks , w here 由巳
oxygen has a pressllre of 65.0 atmospheres ruld a tem peratur巳
of 288 K. Tl回 国1I<:s are stored in a separat巳 roO Ill ， and 由e oxygen is
pllmped to the pati巳nt ' s room, where it is ad ministered at a pressure of
1.00 atlllosphere and a temperatllr巳 of 297 K. What volume does 1.00 m3
of oxygen in the tanks occupy at the conditions in the patient's room?
16. A Goodyear blimp typicall y contains 5400 m3 of helillm (H巳) at
an absolute press ure of 1.1 X 10 5 Pa. The t巳 mp巳ra饥Ire of the he lillm
is 280 K. What is the mass (in kg) of the he lilllll in
th巳 blimp ?
o
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9. The. c fo llowing statements concern how to increase the rate of difc u -tr a c k
fusion (in kg/s). A lI but o n巳 s tatem e nt ar巳 a l ways tru e. Whi ch o n巳 I S
not necessaril y tru巳? (a) lncrease the cross-secti onal area of the diffusion channe l, keeping constant its length and the di 仔巳 ren ce 111
solllte concentrations between its ends. (b) lncrease the difference
in solllte concentrati ons between th e 巳nd s of the di ffu sion ch a nn巳 1 ，
keeping constant its cross-sectional area and its length. (c) D巳c rease
the length of the di ffu sion channe l, ke巳pin g constant its cross-sectional
area and the di ffere n c巳 in solute co n c巳ntra ti o n s between its 巳 nd s.
w
w
.d o
(d ) [ncrease the cross-sectional area of the di ffusion channel, and dew
crease its length, keeping co nstant the d i仔巳 re n ce in solllte co ncentra.c
.d o
c u -tr a c k
tions between its ends. (e) lncrease the cross-secti onal area of the
diffusion channel, increase the d i 仔ere n ce in solute concentrations between its ends , and increase its length.
10. The diffll sion rate for a solllte is 4 .0 X 10- 11 kg/s in a solventfìll ed channel th at has a cross-secti onal area of 0.50 cm2 and a length
k
Di ffusion
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Section 14.4
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14
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CHAPTER
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434
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19. ssm ln a diesel 巳ngin 巳 ， the piston compl 巳 sses air at 305 K to a
vo lume that is one-sixteenth of th 巳 original volllme and a press ure
that is 48 .5 times the original press ur巳. What is the temperature of the
air after the co mpression ?
20. A 0.030-m 3 contain巳 r is initial1 y evacuated. The口 ， 4.0 g of water is
placed in the container, and , aft巳r some time , a1l th巳 water evaporates.
lf the t巳 mperature of the water vapor is 388 K , what is its pressure?
21. On th巳 s llnlit s urfac巳 of Venus , the atmosph巳I1 C pressure IS
9.0 X 10 6 Pa , and the temperature is 740 K. On the earth's surface th巳
atmospheric press ure is 1.0 X 10 5 Pa , while the surface temperatur巳
can reach 320 K. These data imply that VenllS has a "thicker" atmospher 巳 at its surface than do巳 s th巳巳 arth ， which means that the numb巳r of molecllles p巳 r unit vo lum巳 (NIV) is greater on the sllrface of
V巳 nll s than on the earth. Find the ratio (N/ V)venu，/(NIVh川'
22. When you push down on the handle of a bicycle pump , a piston
in the pump cylinder co mpre ss巳s th巳 air inside th巳 cy linder. When the
pl 巳 ss ure in the cy linder is grea t巳r th an th巳 pre ss ure inside the innel
tube to which the pump is attached , air begins to flo w from th巳 pump
to the inner tube. As a biker slowly begins to push down the handle
of a bicycle pllmp , the pl 巳ss ur巳 in s id巳 the cylinder is 1.0 X 10 ) Pa ,
and the piston in the pllmp is 0.55 m above th巳 bottom of the cylin
d巳r. The press ure in s id巳 th巳 inn e r tub巳 i s 2 .4 X 10 ) Pa. How far down
mllst the biker push the handl e before air b巳gin s to flow from th巳
pump to th巳 inn e r tube? Ignore the air in the hose conn巳cting th巳
pump to the inn 巳 r tube , and assume that the temperatllre of the air in
th巳 pump cylinder does not change
*23. ssm www The drawing shows two thermally
insulated tanks. They are
conn巳cted by a valve that
is initially clos巳d. Each
tank contains n巳on gas at
the pl巳ss ure， temp巳rature，产， = 5.0 x 10 5 Pa
5
P2 = 2.0 x 10 Pa
and volume indicated in
T~ = 220 K
T 2 = 580 K
the drawing. When th巳时= 2.0 m3
V2 = 5 .8 m ~
valve is opened , the contents of the two tanks nllX, and the pressure becomes constant through
ou t. (a) What is th巳 fi nal tempel矶Ire? Ignor巳 any chang巳 111
temperature of th巳 tank s the m s巳 lves. (Hint: The heat gained by the gas
in one tank is equal to that lost by the otha) (b) What is th巳
fìnal press UI毛7
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' 26. (
The dl 川19 s how s 川 deal gas confined 川咐毗r by
a massless pi ston that is attached to an ideal spring. Olltside th巳
cylinder is a vac uum. The cross-sectional area of the piston is
A = 2.50 X 10- 3 m2 The initial pr巳 ss ur巳 ，
vo lume , and temperature of the gas are ,
res pecti ve ly , PO, Vo = 6.00 X 10- 4 m 3, and
几= 273 K , a nd the spring is initially
Piston
stretched by an amount Xo = 0.0800 m with
res p巳 ct to its un strained length . Th巳 ga s IS
he a t时 ， so that its final preSS llre , volllm巳， and
temperature are Pr, Vr, and Tfo and the spring is
stretched by an amount x r = 0.1000 m with
res pect to its un strained length. What is the
到 nal temp巳 ratlll 巳 of the gas?
~ 27.
ssm The relativ巳 hllmidity is 55 % on a day when th巳 t巳 mperature
is 30.0 o Using the graph that accompanies Problem 75 in Chapter
12 , d巳 termine th 巳 number of moles of water vapor per cubic meter
of aiI
c.
司 28. M lI ltipl巳-Conce pt
Example 4 and Conc巳 ptual Example 3 are
pertinent to thi s problem. A bubble , located 0.200 m ben巳ath th巳 s ur­
face in a glass of b巳er， ri s巳 S to the top. The air press ur巳 at th巳 top IS
1.01 X 10 5 Pa. Assum巳 that the density of beer is the same as that of
fresh wat巳r. lf the te mperature and number of moles of CO 2 in the
bubble 1 巳 m a in co nstant as the bubbl巳 n ses ， 自 nd the ratio of th巳 bub
ble's vo lume at the top to its vo lllme at the bottom.
1' 29. Interactlve Solutlon 14.29 at www.wiley.comlco l1 ege/cutnel1
off，巳rs one approach to this problem. One assumption of the ideal gas law
is that 出e atoms or molecules themselves occupy a n巳gligible volum巳.
V巳门fy that this assumption is reasonable by considering gaseous xenon
(Xe). Xenon has an atomic radius of 2.0 X 10- IO m. For STP conditions ,
calculate th巳 percentage of the total volume occupi巳d by th巳 atom s.
川 30.
A s ph巳 rical balloon is made from a material whose mass is
3.00 kg. The thickness of the material is n 巳gligibl巳 compared to the
1.50-m radius of th巳 balloon. The balloon is filled with helium (He)
at a t巳 mperature of 305 K and ju st f] oats in air, neither rising nor
falling. The dens ity of th 巳 s urrounding air is 1. 19 kg/m 3 . Find the
absolut巳 pr巳 ssure 0 1' th e helium gas.
出 * 3 1.
ssm A cylindrical glass beaker of height 1.520 m rests on a table.
half of the beaker is filled with a gas , and th巳 top half is
fìlled with liguid mercury that is ex pos巳 d to the atmosphere. The gas
and mercury do not mix because they aI巳 separated by a frictionless
movable piston of negligibl 巳 m ass and thickness. The initial tempel
ature is 273 K. Th巳 t巳 mperature is increased lI ntil a valu巳 is reached
when one-half of the m巳rcury has spilled Oll t. Ignore the thermal
expansion of the glass and the mercury , and fìnd this temperature.
Th巳 bottom
**32. A gas fìll s th e ri ght portion of a horizontal cylinder whos巳 radius
_. theσas
þ
is 5.00 cm. The initi al oressure of
is 1.01 X 10 5 Pa. A frictionless movable pi sto n separates
th巳 gas from the left portion of the
|
sprint Jln
cylinder, which is evacuated and
contains an ideal spring , as the
drawing shows. The pisto n is 1l1 1tiall y he ld in place by a pin. The
spring is initi all y un strained , and
the length of th巳 gas- fìll ed portion is 20.0 cm. wh巳 n the pin is remov巳d
and the gas is all owed to ex pand , the length of the gas-fì1led charnbeI
doubles. The initial and fìnal temperatures ar巳 equal. Determine the
spring constant of the spring.
~._ UV
!' 24. 孔iJultipl e-Con c巳 pt
Example 4 reviews the principles that play rol 巳S
in this problem. A primitive diving bell consists of a cylindrical tank
with one end open and one end closed. The tank is lowered into a freshwater lake , open end downward. Water rises into the tar虫， compress
ing the trapped air, whose temperature remains constant during the
d巳scen t. The tank is brought to a halt wh巳 n the di stanc巳 between the
surface of the water in the tank and the s UIfac巳 of the lak巳 i s 40.0 m.
Atmospheric pressure at the surface of th巳 lake is 1.01 X 10 5 Pa. Find
th巳 fraction of the tank's volume that is fìlled with ai r.
习' 25. Ref，巳r
to Interactlve Solutlon 14.25 at www.wiley.comlcolI ege/
cutnelI for help with problems like thi s on 巴. An apartment has a room
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18. (
The volllme of an ideal gas is held constan t. Detemu削he ratio P2 1P 1 of 由e fìnal preSSllre to the initial preSSllre when th巳 temperature
ofth巳 gas ri ses (a) 台 om 35.0 to 70.0 K 创ld (b) from 35.0 to 70.0 oC.
lic
k
lic
C
w
whos巳 dimensions are 2.5 m X 4.0 m X 5.0 m. Assume that th巳 air in
the room is co mpo s巳d of 79% nitrogen (N 2) and 21 % oxygen w(0
.c
.d o 2 )
c u -tr a c k
At a t巳m p巳rature of 22 oC and a pressure of 1.01 X 10 5 Pa , what is
the mass ( in grams) of th 巳 air?
w
17. A clown at a birthday party has brought along a helillm cylinder,
with
. c which he intends to fill balloons. When full , each balloon con.d o
c u -tr a c k
tains 0.034 m3 of he lillm at an absolllte pressur巳 of 1.2 x 10 5 Pa. Th巳
cylinder contains helillm at an absolute pl 巳ssure of 1.6 x 10 7 Pa and
has a volllm巳 ofO.0031 m3 • Th巳 temp巳 ratllre of th巳 h e lillm in the tank
and in tbe balloons is the sa m巳 a nd remains constan t. What is the
maximllm number of balloons that can b巳 自 lled ?
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PROBLEMS
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Ki netic Theory of Gases
33. Consult Multiple-Concept Example 6 to review the principles
invo lved in this problem. Near the surface of Venus, th巳 rm s speed
of carbon di ox ide molecules (C0 2) is 650 mJ s. What i s 由e temperature (in kelvins) of th巳 atmosph巳 re at that point?
c u -tr
a nd D 盯
ar陀巳 白阳11巳e d 川
w it由h mona
削
a创t 阳
34. ③ Four tanks A , B , C , 削
ga s巳s. For each ta nk， 由巳 mass of an individual atω
0口
m1 η1 and 由
th巳 rn
口
mη1 S s p巳巳d
of 阳
th
忱
1e
巳 ato m s are expressed in terms of m and V忖
nl1
阳
】
1
!
斗ιnl1
h
2
tabl 巳吩). Suppose 由
tha
刽tm = 3.32 X 10 - 药
6 kg , and Vrms = 1223 m/s. Find
the t巳mpe ratllre of 由e gas in each tank.
A
Mass
月1
Rms speed
Vnns
B
C
D
m
2v rm s
2m
2m
V rm s
2v rm s
35. (ãi) Suppose a tank contains 680 m3 of neo n (Ne) at an absolute
pressure of 1. 01 x 10 5 Pa. The temperature is changed from 293.2 to
294.3 K. What is the increase in the internal energy of the neon?
36. l f the translational rms speed of the water vapor molecules (H20 )
in air is 648 mJs, what is the translational rms speed of the carbon
dioxide molecules (C0 2) in the same air? Both gases are at the same
temperature.
37. ssm Initially, the translational rms speed of a molecule of an
ideal gas is 463 m/s. The pressure and volume of this gas are kept
constant, while the number of molecules is doubled. What is the fìnal
translati onal rms speed of the molecules?
38. Tw o gas cylinders ar巳 identic al. One contains the monatomic
gas argon (Ar) , and the other contains an equal mass of the
monatomic gas krypton (Kr). The pressures in the cylinders are th巳
same , but the temperatures are differe n t. Determjne th巳 ratlO
粤皿 of
A、.&.....1
Argon
the
…
ge kinetic energy of a krypton atom to the
average ki netic energy of an argon atom
39. Consult Interactive Solution 14.39 at www.wiley.comlcollege/
cutnell to see how tills problem can b巳 so lved . Y，巳ry fìne s m ok巳 p ar­
ticles are sllspended in ai r. The translati onal rms speed of a s mok，巳
parti cle is 2.8 x 10 mJ s, and the t巳mp巳rature is 301 K. Find the
mass of a p articl巳 .
•
40. A contai ner holds 2.0 mol of gas. The total average kinetic energy
of the gas molecules in the contruner is equal to the kin eti c 巳 n ergy of
an 8.0 X 10 - 3句 bullet with a speed of 770 m/s. What is the Kelvin
temp巳ra tur巳 of the gas?
拉 41. ssm
average rate (in atoms/s) at which neon ato l11 s collide with one side
.c
.d o
of the co ntain巳r? Th巳 m ass of a single neo n atom is 3.35 X 10- 26 wkg.
c u -tr a c k
Section 14.4
Diffusion
45. 哼，. lnsects do not have lungs as we do , nor do they breathe
~ through their mouths. Instead , th巳y have a system of tiny
tllbes , called tracheae , through w h.i ch oxygen diffuses into their bodies.
The trach ea巳 beg in at the surface of an insect's body and pe n巳trate
into th巳 int巳rior. S lI ppose that a trachea is 1.9 mm long with a crosssectional area of 2.1 X 10- 9 m2. The concentration of oxygen in the
air outs id巳 the insect is 0.28 kg/m 3, and the diffusion constant is
1. 1 X 10- 5 m2/s. If the mass per second of oxyg巳n diffusing through
a trachea is 1. 7 X 10- 12 kg/s, fìnd the oxygen concentration at the inteno l 巳nd of the tube.
46. A large tank is filled with methane gas at a concentration of
0.650 kg/m 3 . The valve of a 1.50-m pipe connecting the tank to the atm os ph巳 re is inadvertently left open for twelve hours. During thi s
tim巳， 9. 00 X 10- 4 kg of meth ane diffuses out of the tank , leaving the
concentrati on of methane in the tank essenti ally unchanged. Th巳 dif­
fusion constant for methane in ru r is 2. 10 X 10 m2/s. What is the
cross-sectional area of the pipe? Assume that th巳 co n ce ntrati o n of
methane in the atmosphere is zero.
•
47. ssm The diffusion constant fo r the alcohol ethanol in water is
12 .4 X 10- 10 m2/s. A cylinder has a cross-sectional area of 4.00 cm2
and a length of 2.00 cm. A diff巳re n c巳 in ethanol concentration of
1.50 kg/m 3 is mruntained between the ends of th巳 cylinder. In one
hour, what mass of ethanol di旺u ses through the cylinder?
48. ( When a gas is diffusing through air in a diffusion chann巳1 ，
the di仔U si o n rate is the number of gas atoms per second diffusing
fro m one end of th巳 c h a nn el to th巳 o th er end. The faster the atoms
move, the greater is the diffusion rate , so the diffusion rate is proporti onal to the rms speed of the atol11 s. The atom.i c mass of ideal gas A
is 1. 0 1I , and that of ideal gas B is 2.0 u. For diffusion through the
same channel under the same conditions, fìnd the rati o of the diffusion rate of gas A to the diffusion rate of gas B
49. Review Conceptllal Exa mple 7 before wo rki n g 由i s problem .
For wa t巳 r vapor in air at 293 K, the di ffusion constant is
D = 2 .4 X 10- 5 m2/s. As outlined in Problem 51 (纱， the time required for th巳自 rs t solute molec ules to traverse a channel of length
L is t = U/(2D ), according to Fick's law. (a) Find the time t fo r
L = 0.010 m. (b) For comparison, how long would a water molecule tak巳 to travel L = 0.010 m at the translati onal rms speed of
water 1110lecules (assul11 ed to be an ideal gas) at a temperatu re of
2
www The pressure of sulfur dioxide (S02) is 2. 12 X 10 4 Pa.
There al毛 42 1 moles of tills Dσas in a vo lume of 50.0 m3 Find the
translati onal rms speed of the sulfur dioxide molecules.
Hel.i ul11 (剧 ， a 1110natomic gas ， 创 I s a 0.0 1O -m 3 container
Th巳 press ure of the gas is 6.2 X 10 5 Pa. How long would a 0.25-hp
engine have to run (1 hp = 746 W) to produce an amount of energy
eqllal to the intern al energy of this gas?
14 2.
(
忡43.
ssm www ln 10.0 s, 200 bllllets s trik巳 a nd embed th巳 mse l ves in
a wall. The bullets strike the wall perpendicularly. Each bullet has a
mass of 5.0 X 10- 3 kg and a speed of 1200 m/s. (a) What is the
average change in momentllm per second fo r th e bullets?
(b) Determine the average force exerted on the wall. (c) Assu m.i ng
the bullets are spread out over an area of 3.0 X 10- 4 m2, obtain the
average pressllre th巳y ex巳rt on this region of the wall.
才 * 44.
A cllbical box with each sid巳 of length 0.300 m contains
1.000 moles of neon gas at rool11 t巳 mperature (293 K). What is the
5.00 x 10- 3 m
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Section 14.3
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CHAPTER 14 THE IDEAL GAS LAW AND KINETIC THEORY
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一一一一-
Dry
air
**52. The drawing shows a container that is partially filled with
2.0 grams of water. The temperature is maintained at a constant 20 oC.
The space above the bquid contains air that is completely saturated
with water vapo r. A tube of length 0.15 m and cross-sectional area
53. At the start of a trip , a driver adj usts th巳 absolut巳 pressure in her
tires to b巳 2 . 81 X 10 5 Pa when the outdoor temperature is 284 K. At
the end of the trip she measures the pressure to be 3.01 X 10 5 Pa
Ignoring the expansion of the tires , find the air temperature inside the
tires at the end of the t目 p.
54. 电V A young male adult takes in about 5.0 X 10- 4 m 3 of fresh
air during a normal breath. Fresh air contains approximately
21 % oxygen. Assuming that the pressure in the lungs is l. 0 X 10 5 Pa
and that air is an ideal gas at a temperature of 310 K, find the number
of oxygen molecules in a normal br 刨出 .
55. ssm Wh at is 出e density (i n kglm 3) of nitrogen gas (molecular mass
= 28 u) at a pressure of 2.0 atmospheres and a tempera阳re of3lO K?
忱
1e di迂仙
f
56. ③ The
2
has a va川lu巳 of l. 06 X 10- 9 m
口1 勺
/s . In a 2.0-cm-long tube with a cross4
2
s巳ction
阳
1旧
al area of l. 5 X 10- rπm 气
， the mass rate of dωif旺
fu
山
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I览
盯iωon 臼
1 s m/μt =
1
4
4.2 X 10- kg/s , becaus巳 t由h巳 g剖lycωIII 巳 concentration is maintained
at a value of 8.3 X 10 kg/m 3 at one end of the tube and at a lower
value at the other end. What is the lower concentration?
•
57. ssm The average value of the squared speed v 2 does not equal
the square of the average speed (百 ) 2 . To verify this fact , consider
three particles with the following speeds: V I = 3.0 m/s , V 2 =
7.0 m/s , and V 3 = 9.0 m/s. Calcu late (a) v 2 = ~ (v? + v l + v } )
and (b) (v? = [ ~ (VI + v2 十 问 )] 2.
58. Refer to Multiple-Concept Example 6 for insight into the concepts
used in this problem. An oxygen molecule is moving near the earth 's
surface. Another oxygen molecule is moving in the ionosphere (the uppermost part of the earth's atmosphere) where the Kelvin temperature
is three times greater. Deterrrune th巳 ratio of the translational rrns speed
in the ionosphere to the translation aI rms sp巳巳d near the earth' s surface.
59. ssm A tank contains 0.85 mol of molecular nitrogen (N 2).
Determine the mass (i n grams) of nitrogen that must be removed from
the tank in order to lower 由 e pressure from 38 to 25 atm. Assume that
the volume and temperature of the nitrogen in the tank do not change.
习 60.
Estimate the spacing between the centers of neighboring atoms
in a piece of solid aluminum , based on a knowledge of the density
(2700 kg/m 3) and atomic mass (26.98 15 u) of aluminum. (Hint:
Assume that the volume of the solid is filled with many small cubes,
with one atom at the center of each.)
世 6 1.
Intera c1 ive Solution 14.61 at www.wiley.com/college/cutnell
a model for problems of this type. Th巳 temperature near th巳
sllrface of the earth is 291 K. A xenon ato m (atomic mass = 13 1. 29 ll)
has a kinetic energy equal to the average translational kinetic 巳 nergy
and is moving straight up. If th巳 atom do巳s not collide with any other
atoms or molecllles , how high IIp will it go before coming to rest?
Assume that 由e acceleration dlle to gravity is constant throUghOllt the
ascen t.
provid巳s
*62. Compressed air can be pumped llnderground into huge cavems as
a form of energy storage. The vol ll me of a cav巳m is 5.6 X 10 5 m3,
and the pressure of the air in it is 7.7 X 10 6 Pa. Assume that air is a
diatomic id巳al gas whose intemal energy U 比 given by U = ~ I1RT. If
one home llses 30.0 kW . h of en巳rgy per day , how many homes cOllld
th is intemal e n巳rgy serve for one day?
-* 63.
电害，.
When perspiration on the human body absorbs heat , some
of the perspiration tums into water vapor. The latent heat of
vaporization at body temperature (37 oc) is 2 .42 X 10 6 J/kg. The heat
absorbed is approximately equal to the average energy E given to a
single water molecllle (H 20) tÏ mes the nllmber of water molecules that
are vaporized. What is E ?
3
l'* 64. The mass of a hot-air balloon and its occupants is 320 kg (excluding the hot air inside the balloon). The air outside the balloon has a
pressure of l.01 X 10 5 Pa and a density of l.29 kg/m 3 . To lift off, the
air insid巳 the balloon is heated. The volllme of th巳 h巳ated balloon is
650 m3 . The pressure of the heated air remains the sam巳 as the pressure of the olltside ai r. To what temperature (in kelvins) must th巳 air
be heated so that the balloon jllSt lifts 0旺? The moleclllar mass of air
is 29 II
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3.0 X 10- 4 m2 connects the water vapor at one end t。但r 出at remains
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completely dry at th巳 other end . The diffusion constant for water
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por in air is 2 .4 X 10- 5 m2/s. How long does it take for the water in
the container to evaporate completely? (Hint: R吃fe r to Problem 75 in
Chapter 12 tofind the pressure ofthe water vapo r.)
w
ssm www Review Conceptual Exar口ple 7 as background for this
problem.
It is possible to convert Fick's law into a forrn that is us巳ful
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c u -tr a c
when the concentration is zero at one end of the diffusion channel
(C 1 = 0 in Figure 14.15α) . (a) Noting that AL is 出e volum巳 Vofthe
channel and that m/V is the average concentration of solute in the channel , show that Fick's law becomes t = C/(2D). This form of Fick's law
can be used to estimate 由etim巳 requ ired for the first solute molecules
to traverse th巳 channel. (b) A bottle of perfum巳 is opened in a room
where convection currents are absen t. Assuming that the diffusion constant for perfume in a让 is l. 0 X 10- 5 m2/s, estimate th巳 minimum time
required for 由巳 perfume to be smelled 2.5 cm away.
C
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ADDITIONAL PROBLEMS 437
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丁 HERMODYNAMICS
A high-performance Funny Car
dragster accelerates down the track.
Such dragsters use a heat engine for
propulsion that can develop 8000 or
more horsepower. Note the "header
flames" from the exhaust system
The laws that govern the use of
heat and work form the basis of
thermodynamics , the subject of this
chapter. (@ Oavid Allio/lco 门 SM I/Corbis)
THERMOOVNAMIC SVSTEMS
ANO THEIR SURROUNDINGS
Figure 15.1 The air in thjs colorful
hot-air balloon js one example of a
thermodynarnk system
(@ William Panzer/Alamy Images)
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We have studied heat (Chapter 12) and work (Chapter 6) as separate topics. Often ,
however, th巳y occur simultaneously. 1n an automobile engine , for in sta nc巳 ， fuel is burned
at a relatively high temperature , some of its int巳 rnal 巳 nergy is used for doing th 巳 work of
driving the pistons lI P and down , and th巳 excess h 巳 at IS 1 巳 mov巳 d by th 巳 cooling system
to pr巳vent overheating. Thermodynamics is th 巳 branch of physics that is built upon the
fundam 巳 ntal laws that heat and work obey.
1n thermodynamjcs the collection of objects on which attention is b巳ing focus巳 d is
called the system, whil巳巳V巳rything els巳 in the environment is called th 巳 surroundings. For
巳xample ， the system in an alltomobile engine could b巳 th巳 burning gasoline , while the surroundings would th巳 n inclllde the pistons , the 巳 xhaust system , th 巳 radiator， and th 巳 outs ide
a让， The system and its surroundings are se p创'ated by walls of some kjnd. Wall s that permit h 巳 at to flow through them , such as those of th 巳 engine block , are called diathermal
walls. P巳rfectly insulating walls that do not permit h巳at to flow between the system and its
surroundings are known as adiabatic walls.
To understand what the laws of thermodynamics have to say about the relation ship
betw 巳巳 n heat and work , it is n巳 ce ssary to describ巳 the physical condition or state of a
system. We might be int巳 rested ， for instance , in the hot air within the balloon in Figure
15. 1. The hot air itself would b巳 th 巳 system ， and th 巳 skin of th 巳 balloon provides the
walls that separate this syst巳 m from th 巳 s urrounding cooler ai r. The state of the system
would b巳 sp巳cifi巳d by giving values for th巳 pressur巳 ， volum巴 ， temperature , and mass of
th 巳 hot ai r.
As this chapter discusses , th巳re are four laws of thermodynamics. W巳 begin with the
one known as the zeroth law and th巳 n consider the remaining three.
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The zeroth law of thermodynanucs deal s with th 巳 conc巳 pt of thermal equilibrium.
Two systems ar巳 said to be in thermal equi librium if there is no net 饲 ow ofh 巳 at b巳 tween
them when th巳y are brought into thermal contac t. For instance , you ar巳 d巳finitely not in
thermal equilibrium with the wat巳r in Lake Michigan in January. Just dive into it , and you
will find out how quickly your body l os巳 sh巳 at to the frigid water. To help 巳xplain the c巳 n­
tral idea of the zeroth law of thermodynamics , Figure 15 .2αs hows two s yst巳 ms labeled A
and B. Each is within a container whose adiabatic wall s are made from insulation that prevents the flow of heat, and each has the same t巳 mp巳rature ， as indicated by a thermometer.
1n pωt b , one wall of 巳 ach container is replaced by a thin silver sheet, and th巳 two sheets
are touched together. Silver has a larg巳 thermal conductivity, so heat flow s through it r巳 ad­
ily and the s ilv巳r she巳ts b巳hav巳 a s diathermal walls. Even though the diathermal walls
would pern1i t it , no net flow of heat occurs in part b , indicating that th 巳 two systems are in
thermal equilibrium. Th巳re is no net flow of h巳 at b巳cau se the two systems have the sa m 巳
temperature. We see , th巳 n ， that temperature is the indicator 01 thermal equilibrium in the
sense that there is no net .flow 01 heat between two systems in thermal contact that have
the same temperature.
In Figure 15.2 the thermometer plays an important role. System A is in equilibrium
with the thermometer, and so is system B. ln each cas巳， the th巳 rmometer regi sters the sam 巳
temperatul 巳， th 巳reby indicating that the two sys t巳 ms ar巳 equally ho t. Consequently, systems
A and B are found to be in thermal equilibrium with 巳ach other. ln effect , th巳 thermometer
is a third syst巳 m. The fact that system A and system B are 巳ach in thermal equilibrium with
this third syst巳 m at th巳 same temperature m巳 an s that th巳y aI巳 in thennal equilibrium w ith
each other. Thi s finding is an example of the zeroth law 01 thermodynamics.
Diathermal walls
(silver)
(b)
Figure 15.2 (α) Systems A and B aI巳
surrounded by adiabatic walls and
register the same temperature on a
thermometer. (b) Wh en A is put into
thermal contact with B through
diathermal wa lI s, no net flow of heat
occurs b巳 tween the syst巳 ms .
Tw o systems individually in thermal equilibrium with a third system* are in tbermal
equilibrium with eacb otb巳E
The zeroth law 巳 stablis be s temperature as the indicator of thermal equilibrium a nd
implies that all parts of a system mu st be in th 巳 rmal eq uilibrium if the system is to have
a definable single t巳 mperature. 1n oth巳 r words , ther巳 can b巳 no tlow of heat within a system that is in thermal equilibrium.
THE FIRST LAW OF THERMODVNAMICS
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Problem-solving insight
""" """" 'b
*The state of the third system is the same when it is in th巳rmal 巳quilibriu l1l with either of the two systems. 1n
Figllre 15.2 , for examp l e ， 由巳l1lercll ry level is th巳 same in the ther l1l ometer in either syste l1l.
c u -tr a c k
(α)
THE ZEROTH LAW OF THERMODYNAMICS
Th巳 atoms and molecul 巳 s of a s ub sta nc巳 have kin 巳tic and potential 巳 nergy. These
and other kinds of molecular energy constitute the internal e nergy of a s ub sta nc巳 .Wh 巳 na
substance participates in a process involving energy in the fo rm of work and h 巳 at ， th 巳 111ternal en巳 rgy of the substance can chang巳. Th巳 relationship between work , heat , and
changes in the int巳rnal energy is known as the first law of thermodynamics. W，巳 will now
see that the first law of thermodynamics is an 巳 xpression of the cons巳 rvation of 巳 nergy.
Suppose that a system gains heat Q and that this is th巳 only effect occurring. Consist巳 nt
with the law of conservation of energy, th巳 internal energy of the syst巳 m incl 巳ases from an
initial value of Uj to a fìnal value of Ur, th巳 chang巳 b巳 ing t;" U = Uf - Uj = Q. ln writing thi s
equation , w巳 us巳 th巳 convention that heαt Q is positive when the system gains heat and
negative when the system loses heat. The int巳rnal en巳 rgy of a system can also change because of work. If a system does work W on its sLmoundings and there is no heat flow，巳 n巳rgy conservation indicates that th巳 interna l energy of the system decreases from Uj to Ur,
the change now being t;" U = U
"'"'1, - U
"-'1 = - W. The minus s iσn is included b巳cau se we follow
the convention that work is positive when it is done by the system and negative when it is
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NII THE ZEROTH LAW OF THERMODVNAMICS
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15.3 THE FIRST lAW O F THER MODYNAMICS
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THERMODYNAMICS
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THE FIRST LAW OF THERMODYNAMICS
Work
Heat
The int巳mal energy of a system changes from an initial value Uj to a fin a1 value of Uf
due to heat Q and work W:
6.. U = U f
(a)
-
Uj = Q - W
(1 5.1)
Q is positive when the system gains heat and negative when it loses heat. W is positive
when work is done by the system and
乒任，问
negativ巳 when
work is done on the system.
Surroundings,
Example 1 illustrates the use of Equation 15.1 and the sign conventions for Q and W.
。 Exarn回e 1
Wørk
Heat
(b)
Figure 15.3 (a) The system gains
energy in the form of heat but loses
energy because work is done by the
system. (b) Th巳 syst巳m gains energy in
the form ofh巳at and also gains energy
because work is done on the system
Problem-solving insight
When using the first law of
thermodynamics , as expressed by
Equation 15.1 , be careful to follow the
proper sign conventions for the heat Q
and the work W.
Positive and Negative Work
Figure 15.3 illustrates a system and its surroundings. In part a , th巳 system gains 1500 J of h巳 at
from its surroundings , and 2200 J of work is done by the system on the surroundings. In part b ,
the system also gains 1500 J of heat, but 2200 J of work is done on th巳 system by the surroundings. In each case , determine the change in the internal energy of the system
Reasoning In Figure 15.3a the system loses more energy in doing work than it gains in the
forrn of heat , so th 巳 internal energy of th巳 system decreases. Thus , we expect the change in the
int巳 rnal energy, !1 U = Uf - Uj , to be negative. In p盯t b of the drawing , the system gains 巳n­
ergy in the form of both heat and work. The internal energy of the system increases , and we expect !1 U to be positive.
Solution (a) Th巳 heat is positive , Q = + 1500 J, since it is gained by the system. Th巳
work is positive , W = +2200 J , since it is done by the syst巳 m. According to the first law of
thermodynamics
!1 U= Q - W= (+1500 J) 一 (+2200 J)
The minus sign for !1U indicates that
th巳 internal
=
1-700 J 1
(15 .1)
energy has decreased , as expected.
(b) Th巳 heat is positive , Q = 十 1500 J, since it is gained by the system. But th巳
tive , W = -2200 J, since it is done on the system. Thus ,
work
!1 U= Q - W= (+1500 J) 一 (-2200 J) = 1+ 3700 J 1
The plus sign for !1 U indicates that the
internal 巳nergy
has
increas巳d ，
is nega-
(15.1)
as expected.
In the first law of thermodynamics , the internal energy U , heat Q, and work W are enquantities , and each is expressed in energy units such as joules. However, th巳r巳 is a
fundamental difference between U , on th巳 one hand , and Q and W on the other. The n巳xt
example sets the stag巳 for explaining this di旺erence .
巳rgy
O
Exarnple 2
An Ideal Gas
Th巳 temp巳rature
of three moles of a monatomic ideal gas is reduced from T j = 540 K to
T f = 350 K by two different methods. In the first method 5500 J of heat flows into the gas ,
whereas in the second , 1500 J of heat flows into it. In 巳 ach case find (a) th巳 change in the internal energy and (b) the work
don 巳 by
the gas.
Reasoning Sinc巳优 internal ene韧 of a monatomic i伽I gas is U = ~nRT (Equation 14.7)
and sinc巳 the number of moles n is fixed , only a change in temperature T can alt巳r the internal
en巳rgy. Because the change in T is the same in both methods , the change in U is also the same.
From the given t巳mperatures ， the change !1 U in internal energy can be determined. Then ，由e
first law of thermodynamics can be used with I1 U and the giv巳nh巳at values to caIc ulate the
work for each of the methods.
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done on the system. A system can gain or lose 巳nergy simultaneously in the form of heat Q
and work W. The change in internal energy due to both factors is given by Equation 15.w1.
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Thus , th巳 first law of thermodynamics is just the conservation-of-energy principle applied
to heat , work, and the change in th巳 intemal 巳nergy.
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of a monatomic ideal gas , we find
= ~nR(Tf - T = ~(3.0 mol)[8.31
j)
(b) Since ð. U is now known
to deterrnine th巳 work :
and 由巳 heat
J/(mol' K)](350 K - 540 K)
= 1-
ð. U= 目00 J 一(一 7100 J ) =
W= Q -
2ndmethod
W=Q 一 ð. U= 臼00 J 一 (斗 100 J) = 18600 J I
m
c u -tr a c k
7100 J 1
is given in each method , Equation 15.1 can
1st method
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internal 巳nergy
m
Solution (a) Using Equation 14.7 for the
for
. c each method of adding heat that
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15 .4 THER MAL PROCESSES 441
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b巳 u sed
11 2 600 J I
In each method the gas does work , but it does more in the first method.
To und巳rstand th巳 difference between U and eith巳r Q or W, consider the value for !:::.U
in Example 2. In both methods !:::.U is th巳 same. Its value is deterrnin巳d once the initial and
final t巳 mperatures are specified because the internal energy of an id巳 al gas d巳pend s only
on the temperature. Temperature is one of the variables (along with pressure and volume)
that d巳自 n巳 the stat巳 of a system . The internal energy depends on秒 on the state of a system, not on the method by which the system arrives at a given state. In r巳cognition of thi s
characteristic , int巳rnal energy is referred to as a function of state. * In contrast, heat and
work are not functions of state b巳cau se they have different values for each diff，巳rent method
used to make the system change from one state to another, as in Example 2.
Problem-solving insight
~ CHECK YOUR UNDERSTANDING
(The answer is given at the end of the book.)
1. A gas is enclosed within a chamber that is fitted with a frictionless piston . The piston is
then pushed in , thereby compressing the gas. Which statement below regarding this process
is consistent with the first law of thermodynamics? (a) The internal energy of the gas will
increase. (b) The internal energy of the gas will decrease. (c) The internal energy of the
gas will not change. (d) The internal energy of the gas may increase , decrease , or remain
the same , depending on the amount of heat that the gas gains or loses.
守 11
uHl.
.-
THERMAL PROCESSES
..I A system can interact with its surro undings in many ways , and the heat and work
that come into play always obey th巳白 rs t law of thermodynamics. This section introduces
four common thermal processes. In each case , the process is assumed to be quasi-static,
which means that it occurs slowly enough that a uniform pressure and te mperature exist
throughout a11 regions of the syst巳 m at all tim巳 s.
An isobaric process is one that occurs at consωnt pressure. For instance , Figure 15 .4
shows a substance (solid , liquid , or gas) contained in a chamber fitted with a frictionless
piston. The pressure P experi巳 nc巳 d by the substance is always the same , b巳cau se it is deterrnined by the external atmosphere and th巳 weight of the piston and the block resting on
it. Heating the substance makes it expand and do work W in lifting the piston and block
through the displacement 言 The work can b巳 calcul ated from W = Fs (Equation 6 .1) ,
where F is the magnitude of the forc巳 and s is the mag nitllde of the di splacemen t. The
forc巳 i s generated by the pressure P acting on the bottom sllrface of the piston (area = A) ,
according ω F = PA (Equation 10.19) . 认rith this substitlltion for F, the work becomes
W = (PA)s . But the product A . s is the change in volllme of the material, !:::. V = Vf 一町，
where Vf and Vj are the final and initial volumes , respectively. Thus , the relation is
lsobaric process
W= P !:::.V= P(Vr - Vj )
(1 5.2)
Consistent with our sign convention , this reslllt predicts a positive vallle for the work don巳
by a system when it expands isobarically (Vf exceeds Vj ) . Equation 15.2 also applies to an
*The fact that an ideal gas is used in Example 2 does not restrict our conclusion. Had a real (nonideal) gas or
other material been used , the onl y di 仔'eren ce would have been that the expression for the internal energy would
have been more compl icated. 1t might have involved the volume V, as well as the temperature T , for instance
F= PA
F= PA
kk
Figure 15.4 The substance in the
chamber is expanding isobarically
because the pressure is held constant
by the external atmosphere and the
weight of the piston and the block
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Example 3
Iso b a ri c Exp ansi o n o f Water
One gram of water is placed in the cylinder in Figur巳 15 .4， and the pressure is maintained at 2.0 X 105 Pa. The temperature
of the water is raised by 31 C O • In one cas巳 ， the water is in the liquid phase， 巳xpands by the small amount of 1. 0 x 10- 8 m 3,
and has a specifìc heat capacity of 4186 J/(kg . C O ). In another case , the water is in the gas phase , expands by the much
greater amount of 7.1 X 10- 5 m3, and has a sp巳cifìc h巳at capaci ty of 2020 万(kg . C O ) . Detennin巳 the chang巳 in th巳 internal
energy of the wat巳r in each case.
Reasoning The chang巳 ~U in the intern a1 energy is given by the fìrst law of thermodynamics as ~U = Q - W (Equation
15.1). The h巳at Q may be evaluated as Q = cm ~T (Equation 12.4). Fin a1 1y, since the process occurs at a constant pr巳 ssur巳
(isobaric) , the work W may b巳 found using W = P ~ V (Equation 15 .2).
Knowns and Unknowns The following table summarizes the
Description
giv巳n
Symbol
Value
m
data:
Comment
Mass of water
Pressur巳 on water
Increase in t巳 mperature
Increase in volume of liquid
Specifìc h巳 at capacity of liquid
Increa s巳 in volume of gas
Specifìc h巳 at capacity of gas
~VI叩 id
Cgas
1. 0 g
2.0 X 105 Pa
31 C。
1. 0 X 10- 8 m3
4186 J/(kg' C O )
7 .1 X 10- 5 m3
2020 J/(kg . C O )
Unknown Variables
Change in internal energy of liquid
Change in internal energy of gas
~Uliquid
7
~Ugas
7
P
~T
Cliquid
~Vgas
0.0010 kg
Pressure is constan t.
Expansion occurs.
Expansion occurs.
Modeling the Problem
E圃 T、heF
川
山
eF
First 山川啊
Law
川r咱叫
w
of
旷川le
阳阳町…
e町
e叫
口
rrm
m
m
is given by 出
t h巳 first law of 仙
t h巳町rmodyna
缸mη
旧
ùcs趴， as shown at the righ t. In Equation 15.1 ,
neither the h巳 at Q nor the work W is known , so we turn to Steps 2 and 3 to evaluate
them.
M
E圃I Heat and Specific Heat Ca川ty Acc叫ng to Equation 山， 川削 Q
needed to raise
th巳 t巳mperature
of a mass m of material by an amount
~T is
|Q= cm ~T I
••
I Q=cm 飞
(1 2.4)
where c is the material' s specifìc heat capacity. Data 缸巳 available for all of the terms on
出 e right side of this expression , which can be substituted into Equation 15.1, as shown
at the righ t. The remaining unknown variable in Equation 15.1 is the work W, and we
巳valuate it in Step 3.
m固 W
响or此k】ω{Done 剖
a tCωon削
sta删
nt Pressure
町e Und
毗
1缸 c∞onst
川
s创tan叫m盹 Oωr i osωob阳
缸
an
∞
c
ondi凶
tlOns趴，
the work W don巳
i臼s
given by Equation 15.2 as
| W= P ~V I
where P is 出e pressure acting on the mat巳rial and ~ V is the chang巳 in the volume of th巳
materia l. Substitution of this expression into Equation 15.1 is shown at the righ t.
(15.1)
(1
2 .4)
(15.2)
m
lic
o
c u -tr
.
ack
C
m
C
lic
k
isobar北 compression (Vf \ess than VJ Th巳 n ， the work is negative , since work must b巳 done
w ,
on the system to compress it. Example 3 emphasizes that W = P ~ V applies to any system
.c
.d o
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solid , liquid , or gas , as long as the pressure remains constant while the vo\ume changes.
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442 CHAPTE R 15 TH ERMODYNAMICS
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Applying this
A【U
叫
/1叩 jd
r巳 sult
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m
o
m
w
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c u -tr a c k
C
lic
k
to
bu
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to
k
lic
C
Solution
Combining the results of each step algebraically, we find that
c
w
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-X
15.4 THERMAL PROCESSES F 443
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to the Ii quid and to the gaseous water gives
= C1 叩j q耶甲叩uωA
= [4186 J/尺(kg'C。勺) ](0.0010 kg)(31 C。可)
-
(ο2. 0
x
10
伊5 Paω)(ο1. 0
x
10一 8 m3巧)
= 130J 一 0.0020 J = 1130 JI
!1 Ugas
= cgasm ÂT -
P !1 V:
= [2020 J/(kg' C )](O.OO lO kg)(31 C ) 一 (2 . 0 X 105 Pa)(7.1 X 10- 5 m 3)
O
= 63
O
J 一川=囚
For the Ii quid , virtually all the 130 J of h 巳 at s巳rves to chang巳 the int巳rnal energy, sinc巳
the volum巳 chang巳 and the corresponding work of expansion ar巳 so small. 1n contrast ,
a significant fraction of the 63 J of heat added to the gas causes work of expansion to
be done , so that only 49 J is left to change th巳 int巳rnal energy.
Related Homework: Problem 14
'
P
UL
3的的ω』
ι
O
Vf
Vj
Volume
Figure 15.5 For an isobaric process ,
a pressure-versus-volume plot is a
hori zontal straight line, and the work
done [W = P(Vr - Vj )] is the colored
reclangular area under the graph.
才llIAIl-t ?
Pf
U』3町的UL
ι
Pj
11
·V
1t is often convenÏent to display thermal processes graphically. For in s tanc巳， Figur巳
15.5 shows a plot of PI 巳 ssure versus volum 巳 for an i so b aric 巳 x p a n s ion . Since th 巳 pressllre
is constant , th巳 graph is a horizontal straight line , beginning at the initial volume V j a nd
巳 ndin g at the final volume Vf . 1n terms of such a plot , the work W = P(Vr - VJ is th 巳 are a
und巳 r the graph , which is the shaded rectangle of height P and width Vr - Vj •
Another common thermal process is an isochoric process, one that occurs at constant
volume. Figure 15 .6α illu strates an isochoric process in which a sllbstance (solid , liquid ,
or gas) is heated. The substance would expand if it could , but the rigid container k巳eps the
vo lum 巳 co n stant ， so the pressure-volum 巳 plot shown in Figure 15 .6b is a vertical straight
line. Because the volume is constant, the pressure inside ri ses , and the substance exerts
more and more force on the walls. Althouσh
enormous forc巳 s can b巳 σ巳 nerated in the
D
c1 0sed container, no work is done (W = 0 J) , sinc巳 th巳 walls do not move. Consistent with
zero work being done , theω'ea under th 巳 vertical straight lin 巳 in Figur巳 15.6b is z巳ro . Sinc巳
no work is done , the first law of thermodynamics indicates that th 巳 h巳 at in an isochoric
process serves only to chang巳 the int巳rnal energy: !1 U = Q - W = Q.
A third important thermal process is an isothermal process, one that takes place at
constant temperature. The next s巳ctio n illustrates the importa nt features of an isothermal
process when th巳 sys tem is a n ideal gas.
Last , there is the adiabαtic proce叫 one that occurs without the transfer of
heat. Since there is no heat transfer, Q equals zero , and th 巳白 rs t law indicates that
!1 U = Q - W = - W. Thus , wh巳n work is done by a system ad iabatically, W is pos itive
and th 巳 int巳 rna l 巳 nergy decreases by exactly th 巳 amount of the work done. When work
is don e o n a system adiabatically , W is negative and the intern a l 巳 n ergy in c reas巳 s corr巳 s pondin g l y. Th 巳 n 巳 xt s 巳 c tion di sc usses an adiabatic proc巳 ss for an id 巳 al gas
A process may be complex e nough that it is not recognizable as one of the four just
discussed. For instance , Fi g ure 15 .7 shows a proc巳ss for a gas in which the press ur巳， volume , and te mperature ar 巳 changed along the straight line from X to Y. With th巳 aid of
integral calculus , th巳 following can be proved . Problem -solving insight: The area under a
pressure-volume graph is the work for any kind of process. Thus , the area representII1 g th巳 work has b e巳 n color
Vol ume
(b)
(a)
Figure 15.6 (α) The substance in th巳
chamber is being heated isochorically
b巳cau se the rigid chamber keeps the volume constan t. (b) The pressure-volume
plot for an isochoric process is a vertical
straight line. The area under the graph is
zero , indicating that no work is done.
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THERMODYNAMICS
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to
bu
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to
c
Det巳rmine
.d o
the work for the process in which the pr巳 ssure ， volume , and temperature of a gas are
changed along the straight Ii n巳 from X to Y in Figure 15.7
Reasoning The work is given by the area (i n color) und巳r the straight line between X and Y.
Since the volum巳 increases ， work is done by the gas on the surroundings , so the work is positive. The area can b巳 found by counting squares in Figure 15.7 and multiplying by th巳 area per
square.
Solution We estimate that there are 8.9 color巳d sqllares in th巳 drawing. The area of one
is (2.0 X 10 5 Pa)( l. O X 10- 4 m3) = 2.0 X 10 1 J , so th巳 work is
sq ll ar巳
w= +(8.9 叩ares)(2.0 X 川 J/叩则= 1+180 J I
。
ω』3的的ω』
ι
~ CHECK
YOUR UNDERSTANDING
(The answers are given at the end of the book.)
2. The drawing shows a pressllre-versus-volume plot for a three-step process: A • 8 ,
B • C, and C •A. For each step , the work can be positive , negative , or zero. Which
answer in the table co rrectly describes the work for the three steps?
Wo rk Done by the System
A
•
labede
l)
{
)
{
)
ll
B
Positive
Positive
Negative
Positive
Negative
B
•
C
Negative
Positive
Negative
Negative
Positive
B
C
•
A
Negative
Negative
Positive
Zero
Zero
ωL2国的ω』且
Volume
Figure 15.7 The colored area gives the
work done by 出巳 gas for the process
from X to Y.
A
C
Volume
3. Is it possible for the temperature of a substance to rise without heat flowing into the
substance? (a) Yes , provided that the volume of the substance does not change. (b) Yes ,
provided that the substance expands and does positive work. (c) Yes , provided that work
is done on the substance and it contracts.
E3 的的ω』
ι
4. The drawing shows a pressure-versus-volume graph
C
in which a gas expands at constant pressure from A to 8 ,
and then goes from 8 to C at constant volume. Complete
the table by deciding
Q
ð. U
W
whether each of the four
unspecified quantities
?
+
7
is positive (+), negative A • B
A
B
?
?
B
•
C
+
(一)， or zero (0)
Volu me
5. When a solid melts at
constant pressure , the volume of the resulting liqu id does not differ much from the volume
of the solid. According to the first law of thermodynamics , how does the interna l ene rgy
of the liquid compare to the internal energy of the solid? The internal energy of the liquid
is (a) greater than , (b) the same as , (c) less than the internal energy of the solid.
Hot water at temperature T
(α)
ωL3
的的ω』且
THERMAL PROCESSES USING AN IDEAL GAS
ISOTHERMAl EXPANSION OR COMPRESSION
Volume
(b)
Figure 15.8 (a) The ideal gas in the
cylinder is expanding isothermally at
temperature T. Th巳 force holding th巳
piston in place is reduced slowly, so
th巳 expansion occurs quasi-statically.
(b) The work done by the gas is given
by the colored area.
wh巳n a system performs work isothermally, the temperature remains constan t. In
Figure 15.8α ， for instanc巳 ， am巳tal cylind巳 r contains n mol 巳 s of an id巳 al gas , and the large
mass of hot water maintains the cylind巳r and gas at a constant Kelvin temperature T. The
piston is held in p\ac巳 initially so the vo\um巳 of the gas is V i • As th巳 externa\ force appli巳d
to th巳 piston is reduc巳d quasi-statically, the pressure decreas巳 s as th巳 gas 巳xpands to the
fìnal volum巳 Vf • Figure 15 .8b gives a plot of pressur巳 (P = nRT/V) v巳rsus vo\um巳 for the
process . Th巳 solid red line in th巳 graph is called an isotherm (m巳 aning "constant temperature") and represents the relation betw巳en press ur巳 and vol ume wh巳 n the t巳 mperature is
m
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Work and the Area Under a Pressure-Volume Graph
m
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Example4
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nR叫
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m
w
o
m
Isothermal
expanS lO n or
compression of
an ideal gas
C
lic
k
held constan t. The work W done by the gas is not given by W = P ß V = P (Vf - VJ because
the pressure is not constan t. Nevertheless , the work is eqllal to the area under the
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graph. Th巳 techniques of integral calculus lead to the following result* for W:
o
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15.5 THERMAL PROCESSES USING AN IDEAL GAS
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(15.3)
Where-does the 巳 nergy for this work originate? Since th 巳 internal energy of any ideal
gas is proportional to the Kelvin temperature (U = ~ nRT for a monatomic ideal gas , for
巳xample) ， the int巳rnal energy remains constant throughollt an isothermal process , and the
change in internal energy is zero. As a result , the first law of thermodynamics becomes
ßU = 0 = Q - W. ln other words , Q = W, and the en巳 rgy for the work originates in the
hot water. Heat ftows into the gas from the water, as Figure 15.8a illustrates. If the gas is
compressed isoth巳rmally， Equation 15.3 still appli臼， and heat ftows Ollt of the gas into the
wat巳r. Example 5 deals with the isoth巳rmal expansion of an id巳 al gas.
Exan冒 ple5
Isothermal Expansion of an Ideal Gas
Two moles of the monatomic gas argon expand isothermally at 298 K , from an initial volum巳
of Vj = 0.025 m3 to a final volllme ofV f = 0.050 m3 . Assllming that argon is an ide aJ gas , find
(a) the work done by the gas , (b) the change in the internal en巳 rgy of th巳 gas ， and (c) the heat
supplied to the gas.
Reasoning and Solution (a) The work done by
W=n川
R盯T 叫丘旦)
飞
V, I
=
由
t h巳 gas
(ω2
川川
O叫
m州
mo
01仰)[阳
1) 8幻3川
1门Jν阳/
…
(b) The internal energy of a monatornic ideal gas is U
change when
(c)
Th巳 heat
can be fOllnd from Eqllation
飞 U.UL己3 町1γ" I
=
15.3
丘:
4
一一一一一一-一
~ nRT (Eqllation 14.7) and does not
perature is constan t. Therefo战 1 ßU = 0 JI
Q 'supplied can be deterrnined from the first law of thermodynarnics:
Q = ßU
+ W = 0 J + 3400 J
=
~豆豆E
(1 5 .1 )
ADIABATl C EXPANSION OR COMPRESSION
Adiabatic
expans lO n or
compression of
amonatomic
idealgas
Insu lating material
(a)
Pj
ω』2町的ωL
且
When a system perfonns work adiabatically, no heat ftows into or out of the system. Figure 15.9a shows an arrangem巳 nt in which n moles of an ideal gas do work under
adiabatic conditions ，巳xpanding quasi-statically from an initial volum巳 Vj to a final volume
Vr. The arrang巳 ment is similar to that in Figure 15 .8 for isothermal 巳xpansion. However, a
diffl巳rent amount of work is done her巳 ， b巳cause th巳 cylinder is now surrounded by insulating material that pr巳vents the ftow of heat from occurring , so Q = 0 J. According to the first
law of th巳rmodynarnics ， the change in internal 巳 nergy is ßU = Q - W = - W. Since the internal energy of an ide aJ monatornic gas is U = ~ nRT (Equation 14凡 it follows directly that
ßU = Uf - Uj 二 ~nR(冉一引)， where 贝 and 冉的 the initial and final Kelvin temperatures.
With this substitution , the relation ßU = - W becomes
PtI- --→一-~
问
Volume
Vf
(b)
W = ~nR(引鸟)
(1 5 .4)
When an ideal gas expands adiabatically, it does positive work , so W is positive in
Equation 15 .4. Therefore , the te口口 Tj - Tf is also positive , so 由e final temperature of the
gas must be less 出 an the initial temperature. The internal energy of the gas is reduced to
*In this result , "In" denotes the naturallogarithm to the base e = 2.71828. The naturallogarithm is related to
the common logarithm to the base ten by In(Vf/V;) = 2.303 10g(Vf/V;).
Figure 15.9 (α) The ideal gas in the
cylinder is expanding adiabatically. The
force holding the piston in place is
redllced slowly, so the expansion occurs
qllasi-statically. (b) A plot of pressure
verslls volllme yields the adiabatic
cllrve shown in red , which intersects
由e isotherms (bllle) at 由e initial
temperature Tj and the final temperature
Tf . The work don巳 by the gas is given
by the colored area.
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THERMODYNAMICS
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Table 15.1 Summary of Thermal Processes
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First Law of Thermodynarnics
Typ巳 of Thermal
Process
Isobaric
(constant
pr巳ssur巳)
Isochoric
(constant
volum巳)
Work
(ð. U
Don 巳
W= P(Vr - Vj)
Q - W)
=
ð. U = Q - P(Vr - V j)
、-r-----------'
W
W=OJ
Isoth巳rmal
(constant temperature)
W= nRT In
{ V, \
I 一ι|
飞V:/
oJ
=
Q-
」一-..--'
飞
ð. U for an
W = ~ nR(Tj - T r)
(for a monatornic
ideal gas)
~r )
V .
J
一一斗千二二J
ideal gas
Adiabatic
(no h巳at f1 ow)
nRTl n (
W
AU =OJ- j nR何一 Tr)
、---v--'
、--一一一~一一一~
Q
W
provide the nec巳 ssary 巳 n巳rgy to do the work , and because th 巳 internal energy is proportional to the KeJvin temp巳 rature ， th巳 t巳 mpe rature decreases. F igure 15.9b shows a plot of
pressure versus voJum 巳 for an adiabatic proc巳 ss . The ad iabatic curve (red) inte创rs巳ct岱s th巳
i臼sot由
h沱
m
巳创r丁
-m
fìnal 忧t 巳 mp巳町ra创tu
盯
I江
r巳 [盯乓f = 冉
Pr 只
Vjr 尺
/ (nR)刀]. The ∞
c 010ωI 巳
ed area un
md巳创r 由
盯
th巳 adiaba
创仙
t l C cur
凹
V巳 r陀巳pr纪巳s巳 nts
the work don巳.
Th巳 r巳verse of an adiabatic 巳xpansion is an adiabatic compression (W is n 巳gat l v时， and
Eq uati on 15 .4 indicates that the final t巳 mperature 巳 xceeds th 巳 initial temperature. The
en 巳 rgy provid巳d by th 巳 ag巳 nt doing th巳 work increases the interna J 巳 nergy of the gas. As a
result, th巳 gas becomes hotter
The 巳quation that gives the adiabatic Cllrve (red) betw巳巳 n the initi al pressur巳 and volume (Pj , V;) and th巳 fìnal pressure and voJume (Pr , V r) in Figure 15.9b can b巳 derived
llsing int巳gral caIc ulus. The result is
Adiabatic
expanswn or
compressioll of
an ideal gas
PjVj "Y = PrV/
(1 5.5)
where the exponentγ(Greek gamma) is the ratio of the sp巳 cifi c heat capacities at constant pr巳 ssur巳 and constant volume ， γ = cp/cv. Equatio n 15.5 applies in conjunction
with the id 巳 al gas law, becaus巳 e ach point on the adiabatic curve satisfies the relation
PV = nR T.
Table 15.1 sllmmarizes the 飞Ñ' ork don巳 in th巳 four typ巳 s of 由
t h巳
αr画: m
h av巳 b巳巳 n considering. For each proc巳 ss it also shows how th 巳 fìrst law of thermodynamics depends on th巳 work and other variables.
~ CHECK VOUR 川
U NDERSτANDU
川
N
(The answers are given at the end of the book.)
6. One hundred joules of heat is added to a gas , and the gas expands at constant pressure. Is it possible that the internal e nergy increases bγ100 J? (a) Yes (b) No; the
increase in the internal energy is less than 100 J , since work is done by the gas. Icl No;
the increase in the internal energy is greater than 100 J , since work is done by the gas.
在 A gas is compressed isothermally, and its internal e nergy increases. Is the gas an ideal
gas? (a) No , because if the temperature of an idea l gas remains constant , its internal
energy must also remai n constant. (bl No , because if the temperature of an idea l gas
remains constant , its internal energy must decrease. (c) Yes , because if the temperature
of an ideal gas remains constant , its interna l energy must increase
8. A material undergoes an isochoric process that is also adiabatic. Is the internal energy
of the material at the end of the process (a) greater than , (b) less than , or (c) the same
as it was at the start?
c
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11 SPEGIFIG HEAT GAPAGITIES
IÔ II
... ,.
I In this section the 自rst law of thermodynamics is us巳d to gain an understanding of
the factors that determine the sp巳cific heat capacity of a material. R巳 m 巳 mber， when th巳
temp巳rature of a substance changes as a result of heat flow , th巳 chang巳 in temperature !::.T
and 出e amount of h巳at Q are relat巳d according to Q = cm !::.T (Equation 12 .4). In this expression c d巳 notes the specific h巳 at capacity in units of J/(kg. C O ) , and m is th 巳 mass in
kilograms. Now, however, it is more convenient to express th巳 amount of material as the
number of moles n , rather than the number of kilograms. Therefore , we replace the expression Q = cm !::.T with the following analogous expression:
Q = Cn!::.T
(1 5.6)
where the capital letter C (as opposed to the lowercase c) refers to the molar specifìc heat
capacity in units of J/(mol. K). In addition , the unit for measuring the temp巳 rature change
!::.T is th巳 kelvin (K) rath巳r than the Celsius d巳gr巳巳 (C )， and !::.T = TI' - T , where Tf and T
are th巳 final and initial temp巳ratures. For gases it is necessary to distinguish between the
molar specific heat capacities C p and C v , which apply, respectively, to conditions of constant pressure and constant volume. With the help of the first law of thermodynarnics and
an ideal gas as an example , it is possible to see why Cp and Cv diffl巳r.
To determine the molar specific heat capacities , we must first calculate th巳 heat Q
n巳eded to rais巳 the t巳mperature of an ideal gas from T to Tf • According to the first law ,
Q = !::.U + W. We also know that the internal energy of a monatomic ideal gas is
U = ~ nRT (Equation 14.7). As a result , !::.U = Uf - Uj = ~ nR(冉一只). When the heating process occurs at constant pressure , the work done is given by Equation 15.2:
w= P !::. V = P(Vf - V;). For an ideal gas, PV = nRT, so th巳 workbecom巳s W = nR(Tf - Tj ).
On th巳 oth巳r hand , when the volum巳 is constant, !::. V = 0 m3, and th巳 work done is zero.
The calculation of the heat is summarized below:
O
j
j
j
Q =!::.U
+W
Qc川mlp…c = j nR(骂一引)
+ nR(冉一只)
= ~nR(冉一盯
η)
η) + 0
R( 乓 - 贝
Qc∞OI1S川 v叨01阳u川阳
1川llC
11咀c = j n础
Th巳 molar specific heat capacities can now
that C = Q/[n(Tf - T;)]:
Constant pressure
for a monatomic
idealgas
COllstant volume
for a monatomic
idealgas
con 阳11 volu I11 e _ 3 ,.,
= .2:.
C"v-n(Tf
TI)- EH
the specific heats is
一一
一一
叩'
一一
Monatomic
ideal gas
口
n(Tf-Ti)- 2H
'主抖
ratioγof
Qconsl川 pressure
户
up
户
均G
-
The
b 巳 determined ， sinc 巳 Equation
15.6 indicates
(1 5.7)
(15 .8)
573
(15.9)
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9. The drawing shows an arrangement for an adiabatic free expansion or "thro创 ing "
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process.
The process is adiabatic because the entire arrangement is contained within
perfectly insu lating walls. The gas in chamber A rushes
suddenly into chamber B through a hole in the partition. Chamber B is initially evacuated , so the gas expands there under zero external pressure and the work
(w = P ß V) it does is zero. Assume that the gas is an
ideal gas. How does the final temperature of the gas
after expansion compare to its initial temperature? Th e
final temperature is (a) greater than , (b) less than ,
(c) the same as the initial temperature.
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15.6 SPEC IFI C HEAT CAPACITIES F
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THERMODYNAI\i\ ICS
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Cp
-
In fact , it can be shown that Equation 15.10
diatornic , etc.
Cv = R
(15.10)
appli 巳 s
to any kind of ideal gas-monatomic ,
~ CHECK Y 。
ωUR 川
U ND
、
(The answers are given at the end of the book.)
10. Suppose that a material contracts when it is heated. Following the same line of reasoning used in the text to reach Equations 15.7 and 15.8 , deduce the relationship between
the specific heat capacity at constant pressure (Cp ) and the specific heat capacity at constant volume (C v). Which ofthe following describes the relationship? (a) Cp =Cv (b) Cp
is greater than Cv (c) Cp is less than Cν
11. You want to heat a gas so that its temperature will be as high as possible. Should you
heat the gas under conditions of (a) constant pressure or (b) constant volume? (c) It
does not matter what the conditions are.
THE SECONO lAW OF THERMODVNAMICS
Ic巳 cream melts wh 巳 n left out on a warm day. A cold can of soda warms up on a
hot day at a picnic. Ice cream and soda n巳ver b巳com巳 colder when left in a hot environment , for heat always 刊 ows spontaneously from hot to cold , and never from cold to ho t.
The spontaneous flow of heat is the focus of one of th 巳 most profound laws in a l1 of
science , the second law of thermodynamics.
THE SECOND LAW OF THERMODYNAMICS: THE HEAT FLOW STATEMENT
Heat flows spontaneously 仕om a substance at a higber temp巳rature to a substance at a
lower temp巳rature and does not flow spontaneously in the reverse d让ection.
It is important to realize that th 巳 second law of thermodynamics deals with a diffl巳 rent
of nature than does the fìrst law of thermodynamics. The second law is a statem 巳 nt
about the natural t巳nd 巳 ncy of heat to f1 0w from hot to cold , wher巳 as the fìrst law d巳 als with
energy conservation and focuses on both h 巳 at and work. A number of important devices
dep巳 nd on heat and work in their operation , and to understand such devices both laws are
ne巳 ded. For instance , an automobile engine is a type of heat 巳 ngine because it us巳 sh巳 at to
produc巳 work. In discussing h巳at 巳 ngines ， Sections 15.8 and 15.9 wiU bring togeth 巳 r the
fìrst and s巳cond laws to analyze 巳 ngin 巳巳ffìciency. Then , in Section 15.10 we will s巳e that
refrig巳rators ， air condition 巳l飞 and heat pumps also utiliz巳 heat and work and are c1 0sely
related to heat engin巳 s. The way in which thes巳 thre巳 appliances operate also depends on
both the fìrst and second laws of th 巳 rmodynamics.
asp巳ct
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C
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For real monatomic gases near room temp巳 ratur巳， expenm巳 ntal values of Cp and Cv give
w
ratios very c10s巳 to th巳 theor巳tical value of ~
.c
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Many gases are not monatomic. In stead , th巳y consist of molecules formed from more than
one atom. The oxygen in our atmosphere , for example, is a diatornic gas , b巳cause It conslsts
of mol巳cules formed from two oxygen atoms. Sirnilarly, atmospheric nitrogen is a diatomic
gas consisting of molecules form刨 食om two nitrogen atoms. wh巳r巳as the individual atoms in
a monatornic ideal gas can exhibit only translational motion ，出 e molecules in a diatomic id巳al
gascan 巳xhibit translational and rotational motion , as well as vibrational motion at suffìciently
high temperaωr巳s. Th巳 r，巳sult of such additional motions is that Equations 15.7 -15.9 do not
apply to a diatomic ideal gas. In stead , if the t巳 mp巳rature is suffìciently moderate that the
diatornic molecules do not vibrat巳， the mol 盯 specific_ heat capacities of a diatomic ideal gas
are C p = ~R and C v = ~ R， with 由巳 result thatγ= 艺 =i
The difference between C p and Cvaris巳 s becaus巳 work is done when th 巳 gas 巳xpands
in respons巳 to the addition of heat under conditions of constant pressur巳 ， wher巳as no work
is done under conditions of constant volume. For a monatomic ideal gas , Cp 巳 xce巳ds C v
by an amount equal to R , th巳 id巳al gas constan t:
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A heat engine is any device that uses heat to perform work. It has thre巳 essential
features:
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15.8 HEAT EN G INES
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The physics of
a heat engine.
1. Heat is supplied to the engi ne at a relatively high input temperature from a place
call巳d the hot reservoir:
2. Part of th巳 input heat is used to perform work by th巳 working substance of th巳巳 n­
gin 巴 ， which is th 巳 mat巳 rial within the engine that actually do巳 s th 巳 work (e.g. , the
gasoline- air mixture in an automobile engine).
3. The remai nder of the input heat is r巳j 巳cted to a place ca ll 巳d the cold reservo Ìl;
which has a temperature low巳 r than the input temp巳 rature.
Figur巳 15.10 illustrates these features . Th巳 symbol QH ref，巳 rs to the input heat , and th巳
subscript H indicates th巳 hot reservoir. Similarly, th巳 symbol Qc stands for the r巳Ject巳d
h巳 at ， and the subscript C denotes the cold res巳 rvoil". T he symbol W ref，巳 rs to the work
done. The vertical bars enclosing each of these thr巳巳 symbols in th巳 drawing are incllld 巳d
to emphasize that w巳 are co n c巳 rn ed here with th 巳 absolute vallles , or magnitudes , of th巳
symbols. Th盹 IQHI indicates 由巳 magnitud巳 of the input heat, IQcI denotes the magnitude of
由e rej巳cted heat, and 1wl stands for 伽e magnitude of the work don巳 Problem - solving insight:
Since IQHI , IQd , and Iwl r，φrω magnitudes only, they never have negative values ass机ed
to them when they appear in equations.
To be highly efficient , a heat 巳ngin e must produce a r巳 l ativ巳 l y large amount of work
位 om as little input heat as possible. Th巳refore ， th巳 efficiency e of a heat engine is defìned
as the ratio of the magnitud巳 of the work Iwl done by th巳 engine to the magnitud巳 of the
inpllt h巳at IQHI:
Iwl
(15.11)
e = IQ可
If th巳 input heat w巳r巳 co nv erted 巳 ntirel y into work , the engine wou ld
of 1. 00, 剑
s in
肌ce叫|阳
圳1 = 1ωQHI ; such an eng
W
♂lI1 e
hav巳 an
Figure 15.10 This schematic
representation of a heat eng in巳 shows
th巳 input heat (magnitude = IQHI) that
ori ginat巳 s from th巳 hot reservo 日， the
work (magnitude = IWI) that the
巳ngine does , and the heat (magnitude
= IQ cI) that th巳 engme r句 ects to th巳
cold reservoi r.
efficiency
quoted αωsp
严er，
陀
.cenωge臼s obtained 均
b'y mult句切圳秒刷
1)y川
lin
咔
g the ratio Iw l/IQHI by a factor of 100
Thus , an efficiency of 68 % would m巳 an that a valu 巳 of 0.68 is used for the effìciency in
Equation 15.11
An engine , lik巳 an y device , must ob巳y the principle of co n s巳 rvation of energy. Som 巳
of the engine 's input h巳at IQHI is converted into w。此 Iw l, and th巳陀阳maηa创ind巳r IQ cI is 陀
j扣巳cαt巳
ed tωo the cold r巳servoir. If th巳re are no other losses in th巳 eng in e， the principle of energy cons巳rvation requires that
1QH 1 = 1wl
+
(15.12)
1Qc 1
Solving this 巳quation for Iwl and substituting the 1 巳sult into Equation 15.11 leads to the
fo llowing a1ternative expression for the effic iency e of a heat engine:
e = JQHI 一 I Qc L =
IQHI
Example 6 illustrates how the concepts
to a h巳at 巳 ngin巳.
1 _
1Qcl
IQHI
of 巳ffìc i ency and 巳 n ergy co n servat i on 创 e
A NALVZING MULTIPLEExample 6
(15. 13)
applied
ONCEPT PROBLEMS
An Automobile Engine
An automobile engine has an effìciency of 22.0% and produces 2510 J of work. How much heat is rejected by the engine?
Reasoning Energy conservation indicates that th巳 arnount of heat 1可 ected to t l1巳 cold reservoir is the part of the input heat that
is not converted into work. The work is given, and the inpllt heat can be obtain巳d since the efficiency of the 巳ngine is also given.
Continued
.c
w
Symbol
Value
Efficiency of engine
E
Magnitud巳 ofwork
Iwl
22.0% (0.220)
2510 J
Unknown
m
.d o
l旬riable
Magnitude of rej 巳cted heat
?
IQd
Modeling the Problem
11:11且画.
-i
、1'/
r，，、
The Conservation of Energy According ωthe energy-conservation
principle , the magnitudes of the input h巳at IQHI , the work done Iwl , and the rejected heat
IQd are related according ωIQHI = Iwl + IQd (Equation 15.12). Solving for IQd gives
Equation 1 at the righ t. In 出is result , Iwl is known , but IQHI is not , al由ough it will be
巳valuated in Step 2
3
E圃 问阳
EnE咆g刽昭盼…
i白Er
=平 |
which can be substituted into Equation 1 as shown in the right colurnn.
| IQHI
'EA
、‘，，，
，
| IQHI
rs-、
e = Iwl/IQHI . Solving for IQHI , we find that
=平 |
Solution Combining the resu Its of each step algebraically, we find that
甲甲wl
IQc l =
IW1(+ 一 1) = (2510 怕
Related Homework: Problems 45, 84
Problem-solving insight
When efficiency is stated as a percentage
(e.g. , 22.0%) , it must be converted to a
decimal fraction (e.g. , 0.220) before being
used in an equation.
In Example 6 , less than one-quarter of the input heat is conv巳 rted into work because
the efficiency of the automobil 巳巳 ngine is only 22.0 %. If th巳 engine were 100% efficient ,
all th巳 input heat would be converted into work. Unfortunately, nature do巳 s not permlt
100%-efficient heat engines to exi st, as the next section discusses.
。 11
-: 10
CARNO T' S PRINCIPlE AND THE CARNOT ENGINE
What is it that allows a heat engine to operate with maximum efficiency? The
French engineer Sadi Carnot (1796-1832) proposed that a heat engin巳 h as maximum efficiency when the process巳 s within the engine are reversible. A reversible process is one in
which both the system and its environment can be returned to exactly the states they
were in before the process occurred.
In a reversible process , both th巳 sy s tem and its environment can be return 巳d to their
initial states. Therefore , a process that involves an en巳rgy-dissipating mechanism , such as
friction , cannot be r巳versible because the energy wasted due to friction would alter th巳 sy s ­
tem or th巳巳nvironment or both. There are also r巳 asons other than friction why a process
may not be r巳versible . For instance , the spontaneous flow of heat from a hot substance to
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Description
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Knowns
and Unknowns The following data are available:
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THERMODYNAMICS
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a cold substance is 盯eversible， even though friction is not present. For heat to flow in the re.c
vers巳
d让巳ction ， work must be done , as we will se巳 in S巳ction 15.10. The agent doing such
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work must be located in the environm巳nt of the hot and cold substances , and , therefore , the
environment must chang巳 while the heat is moved back from cold to hot. Since th巳 system
and 由巳 environment cannot both be returned to their initial states , the process of spontaneous
heat flow is irreversible. In fact , all spontaneous proc巳ss巳s ar，巳 irrev巳rsible， such as th巳巳xplo­
sion of an unstable chemical or the bursting of a bubble. When the word "reversibl巳" is used
in connection with engines , it does not just mean a gear that allows the engine to operate a
device in reverse. All cars have a reverse gear, for instance , but no automobile engine is thermodynarnically reversible , since friction exists no matter which way the car moves
Today, the idea that the efficiency of a heat engin巳 is a maximum when the engine
operates reversibly is referred to as Carnot 亏 principle.
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15.9 CARNO T' S PRINCIPLE AND THE CARNOT ENGINE
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CARNOT'S PRINCIPLE: AN ALTERNATIVE STATEMENT
OF THE SECOND LAW OF THERMODYNAMICS
No irreversible engine operating between two reservoirs at constant temperatures can
have a greater efficiency than a reversibl 巳 engine op巳 rating between the same temp巳阳，
tures. Furthermore , all reversible engines operating between the same temperatures
have the same efficiency.
Carno t' s principle is quite remarkabl 巴， for no mention is mad巳 of the working subthe engine. It does not matter whether the working substance is a gas , a liquid ,
or a solid. As long as the process is reversible , the efficiency of the 巳 ngine is a maximum.
How巳ver， Carno t' s principle does not stat巳， or 巳ven imply , that a reversible engin巳 has an
efficiency of 100%.
It can be shown that if Carno t' s principle wer巳 not valid , it would be possible for h巳 at
to flow spontaneously from a cold substance to a hot substanc巳， in violation of the s巳cond
law of thermodynamics. In 巳ffect， then , Carnot's principle is another way of expressing the
S巳cond law.
No real engin巳 operates reversibly. Nonetheless , the idea of a r巳versible engine provides a useful standard for judging the performanc巳 of real engines. Figure 15.11 shows a
r巳versible engin巳， call巳d a Carnot engine, that is p缸 ticularly us巳ful as an idealized mode l.
An important feature of a Carnot engine is that all input heat (magnitude = IQHI) originates
from a hot reservoir at a single temperature T H and all r巳:jected heat (magnitude = IQcI)
goes into a cold reservoir at a single temperature Tc . This important feature is emphasized
in Problem 61 , which focuses on a pressure-versus-volume plot for a Carnot engine that
utilizes an ideal gas as its working substance.
Carno t' s principle implies 由at the efficiency of a reversible engin巳 is independent of the
working substance of 由巳 engine ， and therefore can depend only on the temperatures of the hot
and cold res巳rvoirs. Since efficiency is e = 1 一 IQcI /IQHI according to Eq uation 15.13 , the
ratio IQcI /IQHI can d巳严nd o n1 y on 由巳 reservoir t巳mperatures. This observation led Lord
Kelvin to propos巳 a thermodynamic temperaωre scale. H巳 proposed that 出e thermodynamic
也mp巳raturl臼 of the cold and hot res巳rvoirs be defined such that their ratio is equal to IQcI/ IQHI
Thus， 由e thermodynarnic temperature s叫e is rela忧d to the heats absorbed and 叫ected by a
Carnot engine , and is independent of the working substanc巳 . If a reference t巳 mp巳rature is properly chosen , it can be shown 由 at the thermodynamic t巳 mp巳rature scale is identical to the Kelvin
scale introduced in Section 12.2 and used in the ideal gas law. As a result, the ratio of the magnitude of the rej巳cted heat IQcI to 出巳 magn山de of the input heat IQHI is
stanc巳 of
IQcl
IQHI
wh巳re
Tc
(15.14)
TH
temperatur巳s
Tc and TH must be expressed in kelvins.
a Carnot engine can b巳 written in a
substituting Equation 15.14 into Equation 15.13 for the efficie
fdn
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mvDE
-A
plvc
C‘
a o
the
The 巳fficiencye
跄
Ca盯r口n】刚 of
pa
创rticωu
川la
缸1忖 useful
way by
主几
一-
河，啤
-MM
例M
如m
(1 5.15)
Temperature = T H
Temperature =Tc
Figure 15.11 A Carnot engine is a
reversible engine in which all input
heat IQHI originates from a hot reservoir
at a single temperature TH , and all
rejected heat IQcI goes into a cold
reservoir at a single temperature Tc .
The work done by the engine is Iwl.
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T hi s relation gives the maximum possible e.庐ciency for a heat e ngi ne operating between
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two Kelvin temp巳ratures Tc and TH , and the next 巳xa mpl e illu strat巳 s its application.
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The physics of
extracting work from a warm ocean.
Problem-solving insight
When determining the efficiency of a
Carnot engine , be sure the temperatures
Tc and T H of the cold and hot reservoirs
are expressed in kelvins; degrees Celsius or
degrees Fahrenheit wil l not do
Example 7
A Tropical Ocean as a Heat Engine
Water near the surface of a tropical ocean has a temp巳 ra tur 巳 of 298.2 K (25.0 o c) , whel 巳 as
water 700 m beneath th 巳 surface has a temp巳rature of 280.2 K (7.0 o c). 1t has been proposed
that th 巳 warm water be used as the hot res巳rvoir and the cool water as the cold reservoir of a
heat engine. Find the maximum possibl巳巳fficiency for such an engine.
Reasoning The maximum possible efficiency is th巳巳ffìci巳 ncy that a Carnot engine wou ld have
(Equatio n 15.15) operating between temp巳ratures of T H = 298.2 K and Tc = 280.2 K.
Solution Using TH = 298.2 K and Tc = 280.2 K in Equatio n 15.15 ,
er:l rnnl
山川
Tr
一 ~==
TH
we 且 nd
that
280.2 K
1 _ _
1
一一一一一一 = 10 .060 (6.0 %) 1
298.2 K
.1
u
•
。
In Example 7 the maximum possible effìciency is o nl y 6.0%. The small efficiency
of the hot and cold reservoirs are so close. A greater
巳ffìcie n cy is possible on1y when there is a greater difference betwe巳 n the [ 巳s巳 rvoir temperatures. However, there 创·巳 li mits on how large the effìciency of a heat engine can be , as
Conceptual Exampl巳 8 discusses.
创'i ses becallS巳 th巳 Kelvin temp巳ratures
Conceptual Example 8
Natural Limits
o o巾日ficiency 们阳t Engine
Consid巳 r
a hypothetical eng in 巳 th at receives 1000 J of heat as input fro m a hot reservoir and
1000 J of work , rejecting no heat to a cold r巳 servoir whose temperature is above 0 K.
Which law of th巳 rmodynamics does this 巳ng ll1巳 violate? (a) The fìrst law (b) The second law
(c) Both the fìrst and second laws
d巳 livers
Reasoning Th巳 first law of th巳rmody n ami cs is a n 巳xpression of 巳 nergy conservation. The
second law states that no irreversible 巳 ngine operating between two r巳S 巳r voirs at constant
temperatures can have a great巳I 巳fficien cy than a reversible engi ne operating between the
same temperatures. The efficiency of sllch a 1 巳versible engine is eCarn ot> th 巳 efficiency of a
Carnot 巳n gi n e.
Answers (a) and (c) are incorrect. From 由 e point of view of en 巳 rgy co n s巳 rvation ， nothing is
wrong with an 巳 n g lll 巳 th at conv巳rts 1000 J of h巳 at into 1000 J of work. Energy has been n 巳 1 th 巳 r c [ 巳 ated nor destroyed; it has only been transformed fro lll one form ( h 巳 at) into another form
(WOl北). TherefOl巳， this 巳 ngine does not vio l at巳 th e fìrst law of therlllodynamics.
Answer (b) is correct. Si nce a Ll of the input heat is converted into work , the efficiency of the
engi ne is 1, or 100%. But Equation 15 .15 , which is based on the second law of therlllodynalllics , indi cates that th巳 lllaXillllllll possibl巳巳ffìciency is e CarnOl = 1 - TcfTH , where Tc and T H are
the telllp巳ratures of th巳 Cοld and hot reservoirs , resp巳ctively. Sinc巳 we 缸?巳 told that Tc is above
o K , it is clear that 由巳 ratio Tc /TH is greater than zero , so 由 e Ill axilllulll possible effìciency is
l ess 由 a n 1 (or less than 100%). The e n gir眩 ， therefore , violates the second law of therlllodynalllics , which Ii lllits the effìci 巳 nc i es of h eat 巳ngin es to va lu 巳 s less than 100%.
。
Example 8 has e mphasized that even a perfect heat engine has an efficiency that is
reg创"d， we note that th巳 maximum possible effìcie ncy, as
g lv巳 n by Equatio n 15 .15 , approach巳 s 1. 0 w h巳n Tc approach巳 s absolllte z巳ro (0 K).
However，巳x pen m 巳 nts h av巳 shown that it is not possible to cool a substanc巳 to absolute
zero (se巳 Sect ion 15.12) , so nature do巳 s not allow a 100%-effìcient heat e ngine to exist. As
a result , there will always be h巳at rejected to a cold reservoir w h 巳 never a h eat 巳 n gin 巳 I S
llsed to do W Ol飞 even if friction a nd other ÌlTeversible process巳S 缸"e elimi n at巳d completely.
This rej 巳 cted h巳 at is a form of the rmal poll11tion. The s巳 co nd law of thermodynamics reqllires that at least some thermal poll11tion be g巳 n erated wh 巳n 巳ver heat 巳 ngines aJ巳 u sed to
p巳 rform work. This kind of thermal polllltion can b巳 redllced o nly if society reduces its
dependenc巳 on h 巳 at 巳 ng in es to do work
less than 1. 0 or 100%. 1n this
The physics of
thermal pollution.
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(The answers are given at the end of the book.)
12. The second law of thermodynamics , in the form of Carnot's principle , indicates that
the most efficient heat engine operating between two temperatures is a reve rsibl e one.
Does this mean that a reversi ble engine operating between the temperatures of 600 and
400 K must be more efficient than an irreversible engine operating between 700 and 300 K?
13. Concept Simulation 15.1 at www.wiley.com/college/cutnell allows you to explore the
concepts th at relate to this question. Three reversible engines, A, B, and C, use th e sa me cold
reservoir for their exhaust heats. However, they use different hot reservoirs that have the following temperatures: (A) 1000 K，但) 1100 K, and (C) 900 K. Rank these engines in order of increasing efficiency (smallest efficiency first). (a) A, C, B (b) C, 8, A (c) 8, A, C (d) C, A, 8
14. In Concept Simulation 15.1 at www.wiley.com/college/cutnell you can explore th e
concepts that are important in this question. Suppose that you wish to improve the efficiency of a Carnot engine. Which answer describes the best way? (a) Lower the Kelvin
temperature of the cold reservoir by a factor of four. (b) Raise the Kelvin temperature of
the hot reservoir by a fa cto r of fou r. (c) Cut the Kelvin temperature of the co ld reservoir in
half and double the Kelvin temperature of th e hot reservoir. (d) AII three choices give th e
same improvement in efficiency.
15. Consider a hypothetica l device that takes 10000 J of heat from a hot rese rvoir and
5000 J of heat from a cold reservoir (whose temperature is greater than 0 K) and produces
15000 J of work. What can be said about this device? (a) It viol ates the first law of thermodynamics but not the second law. (b) It violates the seco nd law of th ermodynamics
but not the first law. (c) It violates both the first and second laws of thermodynamics.
(d) It does not violate either the first or the second law of t hermodyn am ics.
~ il
m
REFRIGERATORS , AIR CONDITIONERS , AND HEAT PUMPS
Th巳 natural tendency of heat is to flow from hot to cold , as indicated by th巳 sec ­
ond l aw of thermody namks . However, if work is u s巳d ， heat can be mα de to flow fro m
cold to hot, against its natural tendency. Refrig巳rators ， air condition巳 rs ， and heat pumps
are, in fact, devices that do just tha t. As Figure 15 .1 2 illu strates, th巳s巳 devices use work
(magnitude = IWI) to extract heat (magnitude = IQ
fro m the cold res巳rvoir and deposit
heat (m agni tude = IQHI ) into the hot res巳rvoir. Generally sp巳akin g ， such a process is called
a r，价igeration process. A compari son of the l eft and right sides of thi s drawing shows that
the dir 巳ction s of the arrows symbolizing heat and work in a refJ培巳ration process are oppo-
cI)
site to those in an engine process. Non巳thel巳ss ， en巳rgy i s conserved during a refri g巳 rati on
cI.
proc巳ss ， just as it is in an engine process, so IQHI = Iwl + IQ Moreover, if the process occurs r巳versi bly， we have ideal d巳vices that are called Carnot refri g巳 rato rs ， Carnot air co nditioner飞 and Carnot h巳at pumps. For th巳S巳 id巳al d巳vic町， th巳 relation
(Equation 15.14) applies , just as it does for a Carnot e n g in 巳 .
IQcll IQH I = TcIT ,
J•
The physics of refrigerators. ln a r，刷gerator， th巳 interior of the unit is the cold reservoir,
while the warmer exterior is th巳 hot reservoir. As Fiσure
15 .13 illustrates , the refri gerator
b
takes h巳 at fro m the food inside and deposits it into th巳 kitc h巳 n ， along w ith the energy
needed to do th e work of making
the h 巳 a t flow fro m cold to ho t. For thi s r巳 aso n ， the
b
Refrigeration Process
Engine Process
Figure 15.12 In the r巳fri gerati on
process on the left, work I 叫 i s used
to remove heat IQd from 由巳 cold
m巳川 ir and deposit heat IQHI into
th巳 hot reservoir. Compare this with
the en gi l1巳 process 011 the righ t.
Hot reservoir
(outside
refrigeratorJ
•
Figure 15.13 A refri gerator.
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ωUR UNDERSTAND It川
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15.10 REFRIGERATORS , AIR COND ITIONERS , AND HEAT PUMPS
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outside surfaces (usually th巳 sides and back) of most refrigerators 缸巳 warm to the touch
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c u -tr a c k
An air conditioner is like a refrigerator, except that the room itself is the cold res巳 r­
voir and th巳 outdoors is the hot reservoir. Figure 15.14 shows a window unÜ, which cools
a room by removing heat and depositing it outside , along with the work used to make th巳
heat flow from cold to ho t. Conceptual Example 9 considers a common IlÚ sconception
about refrigerators and air conditioners.
Tlle physics of
air conditioners.
Conceptual Example 9
You Can't Beat the Second Law of Thermodynamics
Is it possible (A) to cool your kitchen by leaving the refrigerator door open or (B) to cool your
b巳droom by putting a window air condition巳r on the floor by th巳 bed? (a) Only A is possible.
(b) Only B is possible. (c) Both are possible. (d) Neith巳 r is possible.
Re <l soning During a refrigeration proc巳ss (b巳 it in a refrigerator or in an air condition 时， heat
(magnitude = IQ c!) is remov巳d from a cold r巳servoir and h巳at (magnüude = IQHI ) is deposited into a hot r巳S巳rvoir. Moreover, according to the second law of thermodynarnics , work
(magnitude = IWI) is 叫 uired to move 出is heat from the cold r巳阳voir to th巳 hot res巳创I川 I忧r..
Th巳 princiψple
巳 of ∞
c on阳vat1刚
10nαn of 巳n巳
we will
Figure 15.14 A window air conditioner
removes heat from a room , which is
the cold reservoir, and deposits heat
outdoors , which is the hot reservoi r.
us
优巳
this as a
guid巳 111 ass巳sS ll1 g 由
th巳 pos
岱si油
bi凶
liti巳臼s.
Answers (剖， (时， and (c) are incorrect. If you wanted to cool your kitchen by leaving the refrigerator door open , th巳 refrig巳 rator would have to take heat from directly in front of th 巳 open
door and pump less heat out the back of th巳 unit and into 由巳 kitch巳 n (since the refrigerator is
supposed to b巳 cooling the entire kitchen). Lik巳wis巳， if you tried to cool your entire bedroom by
placing the air conditioner on the f1 00r by th巳 bed ， the air conditioner would have to take heat
(magnitude = IQ c!) from directly in front of the unit
th巳 back. According to the second law of thermodynamics this cannot happen , since IQHI =
out 由
Iwl 十 IQc!;由at 队 IQHI is greater than (not less than) IQc! because I
is greater than zero
wl
A可swer (d) is correct. The heat (magnitude = IQ c!) removed from the air directly in front of
the open refrig巳rator is deposited back into th巳 kitch巳n at the rear of the uni t. Moreover, according to 出e second law of tI阳modynarnics ， work (magnitud巳= IWI) is ne巳ded to move that heat
from cold to hot, and 由 e energy from this work is also deposited into 由巳 kitch巳 n as additional
hea t. Thus ，出巳 open refrigerator puts into th巳 kitch巳n an amount of heat IQHI = Iwl + IQc!,
which is more 由 an it removes from in front of the open refrigerator. Thus , rather than cooling the
kitchen , the op巳nr巳frig巳rator warms it up. Putting an air conditioner on the floor to cool your bedroom is sirnilarly a no-win game. The heat pump巳d out the back of the air condition巳 r and into
the bedroom is g1巳ater than the heat pulled into th巳仕ont of the uni t. Consequ巳 ntly，出 e air condi
tioner actually warms th巳 bedroom.
Relat~d
Homework: Problem 71
。
Th巳 quality of a refrigerator or air conditioner is rat巳d according to its co巳fficient of performance. Such appliances perform well wh巳nth巳y remov巳 a relatively large amount of heat
(magnitud巳= IQc!)仕om a cold reservoir by using as sm a1 1an amou川 of work (magniωde = IWI)
as possible. Therefore , th巳 coefficient of performance is d巳fìned as 由e 则o of IQc! to Iwl , and the
greater this ratio is , th巳 b巳tter the performance is:
Refrigerator or
air conditioner
Figure 15.15 In a heat pump the cold
reservoir is the wintry outdoors , and the
hot reservoir is the inside of the house.
Coefficient of
performance
I Qc I
IwI
(15.16)
Commercially available refrigerators and air conditioners have coeffici 巳 nts of performance in th巳 rang巳 2 to 6 , depending on the temperatures involved. The coefficients of p巳r­
formance for these real devices 缸巳 less than those for ideal , or Carnot, refrigerators and
air conditioners.
In a sense, refrigerators and air conditioners operate like pllmps. They pump heat "uphill"
from a lower temperatme to a higher temperature, just as a water pump forces water uphill
from a lower elevation to a higher elevation. It would be appropriate to call them heat pllmps.
However, the nam巳 "heat pump" is reserved for the device illustrat巳d in Figure 15.15 , which
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The physics of
heat pumps.
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is a home heating applianc巳 The heat pump uses work Iwl to make heat IQ cI from th巳 win. c outdoors (the cold reservoir) flow up the temperature "hill" into a w缸.m house (th巳 hot
try
c u -tr a c k
reservoir). According to 由 e conservation of en巳rgy， the h巳 at pump deposits inside the house
an amount of heat IQHI = Iwl + IQ cI. The air conditioner and the h巳at pump do closely related jobs. The air condition巳r ref吨erates the inside of the house and heats 叩 the outdoors ,
while the heat pump refl吨巳rates the outdoors and heats up the inside. These jobs ar巳 so
closely relat巳d that most heat pump systems serv巳 in a dual capacity, b巳in g 巳q uipped with a
switch that conv巳rts them from heaters in the winter into air conditioners in the summer.
Heat pumps ar巳 popular for home heating in today 's energy-conscious world , and it is
easy to understand why. Suppose that 1000 J of energy is available for home heating.
Figure 15.16 shows that a conventional electric h巳ating system uses this 1000 J to heat a
coil of w iJ.e , just as in a toaste r. A fan blows air across the hot coil , and forc巳d conv巳ction
carries the 1000 J of heat into the house. In contrast, the heat pump in Figur巳 15.15 does not
use the 1000 J directly as h巳 at. Instead , it uses the 1000 J to do the work (magnitude =
Iwl) of pumping heat (magnitud巳= IQ cI) from th巳 cool巳r outdoors into th巳 warmer house
and , in so doing , delivers an amount of en巳皂y IQHI = Iwl + IQcI. With Iwl = 1000 J , this
b巳comes IQHI = 1000 J + IQ cI, so that the heat pump deliv巳rs mor巳 than 1000 J of heat ,
whereas the conventional 巳lectric heating system d巳liv巳rs only 1000 1. The n巳xt 巳xample
shows how the basic r巳lations IQHI = Iwl + IQ cI and IQ cI /IQH I = Tc/TH ar巳山巳d with
heat pumps.
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15.10 REFRIGERATORS , AIR CONDITIONERS , AND HEAT PUMPS
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1000 J
Figure 15.16 This conventional electric
heating system is d巳Iivering 1000 J of
h巳 at to the Ii ving room.
卢寸盯百古有ττ古节古τ百百百[E百百百了 ，
Example 唱 o
A Heat Pump
An ideal , or Carnot, heat pump is used to h巳 at a house to a t巳 mperature of 294 K (21 o C). How much work must th巳 pump do
to deliv巳r 3350 J of heat into the house on a day when the outdoor temperature is 273 K (0 o C) and on another day wh巳n the
outdoor temperature is 252 K (-21 o C)?
Reasoning The conservation of 巳nergy dictates that the h巳 at d巳 livered into the hous巳 (the hot reservoir) equals the energy
from th巳 work don巳 by the heat pump plus th巳 energy in the form of heat tak巳n from th 巳 cold outdoors (the cold res巳rvoir). Th巳
h巳at deliv巳 redinto the house is giv巳 n ， so that w巳 can u s巳 e nergy cons巳rvation to determine the work , provided that we can obtain a valu巳 for the heat taken from the outdoors. Sinc巳 we are dealing with an ideal heat pump , w巳 can obtain this valu 巳 by using Equation 15.14 , which relat巳s th巳 ratio of the magnitudes of the heats for the cold and hot reservoirs to the ratio of the
r巳S巳rvoir temperatures (in kelvins).
Knowns and Unknowns The following data are available:
294K
273 K or 252 K
3350 J
内M
Hn
Unknown Variable
Magnitude of work done by pump
Value
11
Temperature of hot reservoir (interior of house)
Temp巳rature of cold reservoir (outdoors)
Magnitud巳 ofh巳at deliver巳d into house
Symbol
l门UL
T刮 T川
Description
W
Comment
Temperature in kelvins must be used .
Temperature in kelvins must be used.
7
Modeling the Problem
E跚 T川…阳 ofEI陀阳叫叫叩
r咆g白咱叨即y 川皂叨盼
川y沪归川川
ψ
钊创
-c
川
ωOαnserv唰
c∞
刚
vaa创tion prin
川
川叫
inci
肌n1比cα
呻唰
阳叫叩
叫叫
1甲pμ
叫
刷le
( 11
，
‘EEB
，
、、
t ha创t IQH川1=lwl 十 IQcI (Equation 15.12), wh巳re IQHI , Iwl , and IQcI旧， respectively, the
由
magnitudes of the heat deliver巳d into the house (th巳 hot res巳 rvoir) ， th巳 work don巳 by the
heat pump , and the heat taken from th巳 cold outdoors (the cold res巳rvoir) . Solving for Iwl
gives Equation 1 at 出巳 right. In this result, w巳 have a value for IQH I but not for IQ cI. Step
2 deals with this rnissing information.
&
Continued
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，1
'
Iwl=IQHI 一 IQd
given about the ，temp巳臼
也
r创
at山.巳创S TH for 由
t h巳 hotr巳s巳rvoi让r and Tc for 由巳 cold
= TcI凡， which can be solved for IQd:
、
m
时
uormation
rst 、
use 由
th巳
.d o
-i
r陀巳s优巳r凹vo
∞
ir. According to Equation 15.14, IQd/IQHI
、
At
th巳 right ，
we substitute this
Solution Combining the
r巳 sult
r，巳sults
、、11II/
L
、‘、
，，
S，EE-.
，，
QH
王
「M
一一
Q
into Equation 1.
of each step algebraically, w巳白 nd that
?!一 IQ甲|一 IQHI阳
It follow s 由
t ha
创t the
magn
川lÏ tωud巳 oft白
h巳
3mm
m2叫……
纺
e例m
刚叫t阳
u旷r陀肌e
work for
白
th巳
two
giv巳n
outdoor 臼t 巳mp巳ratur巳 s is
恻
W刑仲川…|←川斗=斗节咄|恒陆
Q岛Hi(1 去剖扑非)
= (刃
川叫叫
3350
川川叫
50O川刊叫
Jη叫)(1卜一 击击剖
引) = 124011
E
气咆惚切
仇
i?
0咋
2γγt
印…
e
OT~;)~
飞/飞/
Problem-solvin 日 insight
When applying Equation 15.14 CIQ cI /IQHI =
Tc/T H) to heat pumps , refrigerators , or air
conditioners , be sure the temperatures Tc
and TH are expressed in kelvins ; degrees
Celsius or degrees Fahrenheit will not do.
More work must be done wh巳n the outdoor temperature is lower, because the heat is
pumped up a greater temperature "hill."
Related Homework: Problem 88
It is also possible to sp巳cify a coeffìcient of performanc巳 for heat pumps. However,
unlike refrig巳 rators and air conditioners , the job of a h巳 at pump is to heat , not to cool. As
a result, the coeffìcient of p巳rformance of a heat pump is the ratio of th巳 magnitude of th巳
heat IQHI delivered into th巳 house to the magnitud巳 of the work Iwl required to deliver it:
Coefficient of
performance
Heatpllmp
IQH I
Iw I
(1 5. l7)
The coeffìcient of perform anc巳 d巳pends on the indoor and outdoor temperatures. Commercial units have co巳fficients of about 3 to 4 under favorable conditions.
,j" CHECK
V 。
ωUR U 酌m 巨 RSTANDII
川N
川‘唱4
伊
g iven at the end of the book.)
16. Each drawing represents a hypothetical heat engine or a hypothetical heat pump and
shows the corresponding heats and work. Only one of these hypothetical situations is
allowed in nature. Which is it?
(The answers are
(α)
(b )
(c)
(d)
(e)
17. A refrigerator is kept in a garage that is not heated in the cold winter or air-conditioned
in the hot summer. Does it cost more for this refrigerator to make a kilogram of ice cubes
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:: 11
ENTROPV
Iri 1m
A Carnot engine has the mηax川1m
盯m川1 possible 巳ffì白kωci巳
within it are r巳 V巳rsi
臼削
ib
ble
巳. Irreversible processes , such as
friction , caus巳 real 巳 ngines to operate at less than maximum 巳ffìciency， for th巳y reduce our
ability to use h巳 at to perform work. As an 巳xtr巳 m巳巳xample ， imagine that a hot object is
placed in thermal contact with a cold object, so heat ftow s spontaneously, and hence
盯eversibly， 仕.om hot to cold. Eventually both objects reach the same t巳 mp巳rature ， and
Tc = TH • A Carnot engin巳 using these two objects as heat reservoirs is unable ωdo work ,
because the efficiency of the 巳ngin巳 i s zero [eCamol 二 1 - (TcITH ) = 0]. In general , iπb
versible process巳 s cause us to lose some , but not necessarily all , of the ability to perform
work. This partialloss can be expressed in terms of a conc巳pt called entropy.
To introduce th巳 idea of entropy we recall the relation IQc! / IQHI = TcfTH (Equation
15 .1 4) that applies ωa Carnot engine. It is possible to reaITa吨巳 this 巳quation as IQc! ITc =
IQHIITH, which focuses attention on the heat Q divid巳d by the Ke\vin temp巳刚ure T. The
quantity QIT is called the change in the entropy ð. S:
.... 11
b巳cause 由
t h巳 proc
臼巳s邱s巳臼s oc
∞
C Ul时
甘tn
T
吨
1鸣
g
ð. S
=
(手)R
(1 5.18)
In this expression the temperature T must be in kelvins , and the subscript R refers to the
word "reversible." It can b巳 shown that Equation 15.18 app\ies to any process in which
heat 巳nters (Q is positive) or \eaves (Q is negative) a system reversib\y at a constant temp巳 rature. Such is the case for the heat that ftows into and out of the r巳 servoirs of a Carnot
engine. Equation 15.18 indicat巳 s that the SI unit for 巳 ntropy is a jou\e per kelvin (J /K) .
Entropy, Iik巳 internal energy, is a function of the stat巳 or condition of the system. Only
the state of a system determines the 巳 ntropy S that a system has . Therefore , the change in
entropy ð. S is equal to th巳 entropy of the fìnal state of the system mjnus the entropy of the
initial state.
We can now describe what happens to the entropy of a Carnot engine. As the
engine operates , the 巳ntropy of the hot reservoir decreases , since heat of magnitude IQHI
departs rev巳rsibly at a Kelvin t巳 mp巳rature TH. The cOITesponding change in the entropy is
ð. SH = - IQHIITH, where the llÚ nus sign is needed to indicate a decrease , since the symbol
IQHI denot巳s only th巳 magn山de of the hea t. In contrast, the 巳ntropy of the cold r巳s巳rvoir
increases by an amount ð. Sc = + IQc! ITc , for the 叫巳ct巳d heat reversibly ent巳rs the cold
reservoir at a Kelvin t巳mperature Tc . The total change in entropy is
ð. Sr + ð. Su
becaus巳 IQcI!Tc
IQc l
IQHI
+~- 一一一!... =o
Tc
TH
= IQHIITH according to Equation 15 .14
that th巳 total change in entropy is zero for a Carnot 巳 ngtn巳 is a specifìc illustration of a general r巳sult. It can be proved that when any reversible process occurs , the
chang巳 in th巳巳ntropy of the universe is zero; ð. Suniverse = 0 J/K for a r巳V巳rs ible process. The
word "universe" means that ð. Suniverse takes into account the 巳ntropy changes of a l\ parts of
Th巳 fact
lic
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m
C
w
o
o
m
in the winter or in the summer? (a) In the summer (b) In the winter (c) It costs the
same in both seasons.
18. The coefficient of performance of a heat pump that is removing heat from the cold
outdoors (a) must always be less than one , (b) can be either less than or greater than
one , (c) must always be greater than one.
19. A kitchen air conditioner and a refrigerator both remove heat from a cold reservoir and
deposit it in a hot reservoir. However, the air conditioner 一一一一一一 the kitchen , while the
refrigerator 一一一一一 the kitchen.
(a) coo ls , cools (b) cools , warms (c) warms , warms
(d) warms , cools
20. On a summer day a window air conditioner cycles on and off, according to how the
temperature within the room changes. When are you more likely to be able to fry an egg
on the outside part of the unit? (a) When the unit is on (b) When the unit is 0忏
(c) It does not matter whether the unit is on or off.
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15.11 ENTROPYF 457
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5
Hypothetical reversible process
for irreversible
ð. S for hypothetical
reversible process
process
Figure 15.17 Although the relation
S = (Q/T)R applies to reversible
processes , it can be used as part of an
indirect procedure to find the entropy
change fo r an irreversible process. This
drawing illustrates the procedure
di sc u ss巳d in the tex t.
ð. S
/:::,.
Example 11
The Entropy of the Universe Increases
Fig ur 巳 1 5.18
shows 1200 J of heat flowin g spontaneollsly through a copper rod fro m a hot
reservoir at 650 K to a cold reservoir at 350 K. Det巳rmine th巳 amollnt by which thi s irreversible
proc巳 ss changes 由e entropy of th 巳lI niverse ， assllming that no oth巳 r changes OCCll r.
Hot reservoir
TH = 650 K
./
or
cl
nvnvnu
mdo
rJ
Reasoning The hot-to-cold h巳at flow is irreversible , so th巳 r巳lati on D. S = (Q/ T) R is applied to
a hyp oth巳 tic al pro c巳ss w h 巳 reby the 1200 J of h巳at is taken rev巳 rs ibl y from th 巳 hot reservoir and
added reversibly to the cold reservoir.
叫ζ
Solution The total
fo r each r巳5巳rV01r :
巳 ntropy
change of the lI niverse is the algebraic sum of th巴巴ntropy
1200 J
1200 J
!1 S………
=一一-一一一+
~ "" "、
6 50 K
350 K
U II'
Cold reservoir
TC = 350 K
Figure 15.18 Heat flows spontaneously
fro ll1 a hot reservoir to a cold reservoir.
m
sm巳， th 巳巳 ntropy
The
i rr巳vers ible
、--v-'
、--、--
Entropy lost
by hot res巳rv oi r
Entropy gained
by cold reservoi,
process causes the entropy of the uni ve rse to
=
chang巳s
1+1. 6 J/K I
in cr巳ase
by 1. 6 J/K
Example 11 is a specific illustratio n of a g巳 n era l res ul t: Any irreversible process increases the entropy of the universe. In other words , !1 Sunivcrsc > 0 J/K for an irreve rsible
process. Reversible processes do not alter the e ntropy of the unive rs巳， whereas irr巳versible
processes caus巳 th巳 entropy to increase. Th er巳fore ， th巳 e ntropy of the univ巳rs巳 co ntinu ally
incr巳 a s es ， lik巳 tim巳 it se lf， and entropy is sometimes ca ll 巳d " time's arrow." It can b巳 s h ow n
that this b巳 h avior of the 巳 ntropy of the univ巳rse provides a co mplet巳 ly g巳 n eral s tatem巳 nt
of the s巳c ond law of thermodynamics , whi ch applies not o nly to heat flow but also to all
kinds of other processes.
THE SECOND LAW OF THERMODYNAMICS STATED IN TERMS OF ENTROPY
The total entropy of
th巳 umv巳 rse
does not
chang巳 w h巳 n
a reversible process occurs
(!1Sun 阳
When a n irr巳V巳 r s ibl e process occurs and the e ntropy of th 巳 univer se in c r巳 ases ， the
avai lable fo r doing work d ecr巳 as町， as th 巳 n ex t exam ple illustrates.
e n 巳 rgy
υ 巨臼
xamp
附'恒
e 12
En叫
1
5u
叩
1ψppo s巳 t仙
ha
川t 1200 J of heat is lI sed as inpllt foαr an 巳n g in 巳 lI nd巳创r two dωif配
毛祀
er陀巳n
川
lt cοnditi ons . In
Figllre 15 .19a the heat is supplied by a hot reservoir whose temperature is 650 K. In part b of
th巳 drawin g ， the heat flows irreversibly through a copper rod into a second reservoir whose temp eratllr巳 i s 350 K and then enters th 巳 e n g in e . In e i th 巳 r case , a 150-K reservoir is used as the
o
w
of o n 巳 part of the universe may change
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a r巳versib l e process , but if so , the entropy of a nothe r part changes in the opposite way by the same amount.
What happens to th巳巳ntropy of the univ巳rse w h 巳 n an irreversible process occurs is
more complex , b巳cau se th巳巳xpr巳 ss i on !1 S = (Q/ T) R do巳s not apply directly. However, if
a system changes irreversibly from an initial state to a fin al state , this expression can be
used to calculate !1 S indirectly, as F igur巳 15 .17 indi cates. We imag ine a hypothetical reversible process that causes th巳 sys te m to ch ang巳 betwee n the sαme initial and final states
a nd th巳 n find !1 S for this reversible process. The value o btained for !1 S also applies to th巳
1Iτ巳vers ible proc巳 ss that actually occurs , si nce o nly th 巳 nature of the illÍ tial and final
states , and not th巳 path b巳tw巳巳 n the m , d巳termines !1 S. Example 11 illustrates this indirect
me thod and shows that spontaneo us ( ineversible) proc巳 ss es increase the entropy of the
umverse.
entropy of the universe. To be
己与
Reversible processes do not alter the total
k
th巳巳 nvironm巳 n t.
lic
the system and all parts of
to
bu
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C
lic
k
Irreversible
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CHAPTER 15 THERMO DY NAMICS
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15.11 ENTROPY 459
w
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k
obtain 巳d
lic
o
.d o
m
m
w
o
the maximum amount of work that can be
C
d巳t巳rnúne
to
bu
y
N
y
bu
to
k
lic
C
cold reservoir. For each case,
.c
from
the 1200 J of hea t.
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Reasoning According to Equation 15.11 , th巳 w。此 (magnitude = IWI) obtained from th巳巳n
gine is the product of its efficiency e and the input h巳at (magnitude = IQHI) , or Iwl = elQιFor
a given input heat , the maximum amount of work is obtained when 由巳 efficiency is a maximum
that is , wh巳n 由e engine is a Carnot engine. The efficiency of a Carnot 巳ngine is given by Eq uation
15 .l 5 as eCamot = 1 - Tc/TH • Therefor巳， the efficiency may be deternúned from th巳 Kelvin t巳m­
peratures of th巳 hot and cold reservoirs.
Solution
B矿ore
irreversible
heatflow
Tr
TH
-~
er.....".... ,
150 K
650 K
一一一一 =
0.77
|川
川1=(比
W
E陀
eCa
ωω
化‘a盯1fr口川t1叫
Tr
eCarn O l
。
After
irreversible
( )
150 K
350 K
一 一ι= 1 一 一一-一一 =
'H
0.57
heatflow
|阳川
川1=(问
W
勺
气
E陀
ω
C，a叫
1‘
呐
w巾
h巳 n 由
t h巳
1200 J of inpu川th巳at 比
1s
tak巳 n
the 巳伍 cα1巳 ncy oft由
h巳 Carnot 巳 ng
♂ine
extracted from the input
from 由
t h巳 350-K r巳s巳rvoi让r in s 忧t 巳 ad oft由
h巳 650ι-K re
巳 s巳臼rvoir，
is smalle r. As a r巳 s ult ， less work (680 J versus 920 1) can be
h巳at.
。
Example 12 shows that 240 J less work (920 J - 680 J) can be p巳rformed wh巳n the inis obtained from the hot reservoir with the lower temperature. In other words ，由巳让
reversible process of heat flow through the copp巳r rod causes en巳rgy to becom巳 unavailable
for doing work in the amount of Wunav，巾ble = 240 1. Example 11 shows tha创t this 让i .rr巳ver邱siblee
proc
臼巳s岱s 剖
s imul让tan巳ousl忖
y cωau
时s巳臼s 巾
th巳 巳ntropy 仗
0f 白
th巳 川
umi忖
V巳町rs优
e tωo incαr巳
eas优巳 by an amount
puth巳at
ßS山
n1
u川
tipl忖
y ßS
，乱u山n阳llve陌
川 rse by 150 K , which is the lowest Kelvin temperature in Example 12 , you obtain
Wu 川 ülable
= (150 K) X (1. 6 JIK) = 240 1. This illustrates th巳 following general resul t:
Wunavailable = TOßSuniverse
wh巳r巳 To
(1 5.19)
is th巳 Kelvin t巳 mp巳rature of the cold巳 st heat res巳rvoir. Since irr 巳versible
processes cause the entropy of the universe to increase , they cause 巳n 巳rgy to be degraded ,
in the sense that part of the energy becomes unavailable for th巳 p巳rformance of work.
In contrast , ther巳 is no penalty when reversible processes occur, b巳cause for them
ßSuniverse = 0 J/K , and there is no loss of work.
Entropy can also be interpreted in terms of order and disord巳r. As an example, consider
a block of ice (Figure 15 .20) with each of its H 20 molecules fìxed rigidly in place in a highly
structured and ordered arrangem巳 n t. In comparison , th巳 puddle of water into which the ice
melts is disordered and unorganized , becaus巳 the molecules in a liquid are free to move from
place to place. Heat is required to melt the ice and produce the disorder. Moreover, heat flow
into a system increas巳 s the entropy of the system , accOI由 ng to ßS 二 (Q/T)R. We associate
an increase in 巳 ntropy， th巳n ， with an incr巳as巳 in disorder. Conv巳rs巳ly， we associat巳 adecr巳ase
in entropy with a decrease in disorder or a greater d巳gre巳 of ord巳r. Example 13 illustrates an
order-to-disorder change and th巳 increas巳 of 巳ntropy that accompanies it.
Block 口f
(b)
Figure 15.19 Heat in the amount of
IQI-II = 1200] is used as input for an
engine under two di:ff，巳r巳nt conditions
in parts a and b.
ice
‘ -一气、 PU削e 川ter
Figure 15.20 A block of ice is an
example of an ordered syst巳 m relative
to a puddle of water_
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TH E RMODYNAMICS
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15
N
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CHAPTER
o
c u -tr
.
ack
c
.d o
F ind the change in entropy th at results w hen a 2.3-kg block of ic巳 melts slowly (reversibly) at
273 K (0 o c).
Reasoning Since the phase change occurs reversibly at a constant temp巳ra阳陀 ， the change in entropy can be found by using Equation 15 .18, I:l. S = (Q/T) R' where Q i s 由 e heat absorbed by the
m巳 I ting i c巳. Thi s heat can be determined by using th巳 rel ati o n Q = mL r (Equation 12.5) , where m
is the mass and Lr = 3.35 x 10 5 J/kg is the latent heat of fusion of wat巳r (se巳 Table 12.3).
Solution Using Equation 15 .18 and Equation 12.5 , w e fìnd that thechange in entropy IS
I:l. S
( Q\ _
= 1 一一 |
飞 T
/R
mLr 一 (2.3 kg)(3.35
一一一一 一
T
x
10 5 J/kg
273 K
飞
|
= 1+2.8 X 10 3 J/K 1
a result th at is positive, since th 巳 ice absorb s heat as it m 巳 Its
F ig ure 15.21 shows a n oth 巳 r order-to-d i sorder change that can be described in term s of
entropy.
~ CHECK V 。
ωUR U 酌NDERSTANDI川川'酌
N、唱a
(The answers are given at the end of th e book.)
21. Two equal amounts of water are
mixed togethe r in an insu lated container,
and no work is done in the process. The
initial temperatures of the water are different , but the mi xtu re reaches a uniform
temperature. 0 0 the interna l energy and
entropy of the water increase , decrease ,
or rema in constant as a resu lt of the mi xing process?
Figure 15.21 With the ai d of
巳xp los i ves ， demoLition experts caused
th巳Ki ngdome in S巳attle ， Washington ,
to go from the o rder巳d s tat巳(l ower
entropy) shown in the top ph otograph
to the di so rd er巳d state (higher entropy)
, shown in the bottom photograph
(Anthony Bolante/Reuters/Landov LL C)
Interna l Energy
of the Water
Entropy
of the Water
(a)
Increases
Increases
(b)
Oecreases
Oecreases
(c)
Remains co nstant
Oecreases
(d)
Remains constant
Increases
(e)
Remains constant
Remains constant
22. An event happens somewhere in the universe and , as a result , the entropy of an object changes by - 5 J/K. Consistent with the second law of thermodynamics , which one
(or more) of the following is a possible value for the entropy change for the rest of the
universe? (a) - 5 J/K (b) 0 J/K (c) + 5 J/K (d) + 10 J/K
23. In each of the fo ll owing ca ses , whi ch has the greater entropy, a handful of popcorn
kerne ls or the popcorn that results from t hem; a salad before or after it has been tossed;
and a messy apartment or a neat apartment?
24. A glass of water contains a teaspoon of disso lved suga r. After a while , the water evaporates , leaving behind sugar crystals. The entropy of the sugar crysta ls is less t han the entropy of the d isso lved sugar because the sugar crystals are in a more ordered state. Why
doesn't t his process violate the second law of t hermodynamics? (a) Because , consid ering
what happens to the water, the tota l entropy of the universe also decreases. (b) Because,
considering wh at happens to th e water, the total entropy of the universe increases.
(c) Because the seco nd law does not apply to th is situation.
25. A builder uses lumber to construct a building , which is unfortunately destroyed in a
fire. Thus , the lumber existed at one time or another in three different states: (A) as unused
building material , (B) as a bui lding , and (C) as a burned-out shell of a building. Rank these
three states in order of decreasing entropy (I argest first). (a) C ，日， A (b) A , B, C
(c) C, A , B (d) A , C, B (e) B, A. C
~ II
一
THE THIRD lAW OF THERMODYNAMICS
的
F
To thezeI Otl1，
自 rst，
and
s巳cond
laws of thermodynamics we add
th巳 third
(and last)
law. The third law of thermodynamics indicates that i t is impossible to reach a t巳 mp巳 ra­
ture of absolute z巳1'0.
THE THIRD LAW OF THERMODYNAMICS
It i s not possibl e to l ower th巳 t巳 mp巳ra阳 re of an y sys t巳 m to absolute zero (T = 0 K) in
a 白 nite number of s t巳p s.
m
o
w
w
w
.d o
C
lic
k
Order to D isorder
m
C
lic
k
Example 13
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CALCU LATIONS 461
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CONCEPTS & CAlCUlATIONS
Th巳自 rst
law of thermodynamics is basically a r巳 statem巳 nt of th巳 conservation­
plinciple in terms of heat and work. Example 14 emphasizes this important fact
by showing that th巳 conservation principle and th巳 fìrst law provide th巳 sam巳 approach to
a problem. 1n addition , the example reviews th巳 concept of latent h巳 at of sublimation (s巳e
Section 12.8) and th巳 ideal gas law (se巳 Section 14.2).
of-巳 nergy
Concepts & Calculalïons Example 14
0
The Sublimation of Zinc
The sllblimation of zinc (mass per mole = 0.0654 kg/mol) takes plac巳 at a t巳 mp巳 rature of
6.00 X 102 K , and th巳 latent heat of sllblimation is 1.99 x 106 J/kg. The pressllre remains constant during the sllblimation. Assume that th 巳 zinc vapor can be tr巳ated as a monatomic id巳 al
gas and that the volllm巳 of solid zinc is n巳gligible compar叫 to the corresponding vapor. What
is the change in th巳 internal energy of th巳 z inc when 1.50 kg of zinc sllblimat巳 s ?
Concept Questions and
sllblimation?
Ans而vers
What is sllblimation and what is the latent heat of
Sublimation is the process wh巳reby a sobd phas巳 changes directly into a gas
phase in respons巳 to th巳 inpllt of heat. Th 巳 h巳 at per kilogram n 巳eded to caus巳 th巳 phas巳
change is call巳d th巳 latent heat of sublimation Ls. The heat Q n 巳巳d巳 d to bring abollt the
sllblimation of a mass m of solid material is giv巳 n by Eqllation 12.5 as Q = mLs.
Answer
When a solid phase changes to a gas phase , does the volllme of
and by how mllch?
th巳 material increas巳 OI
d巳crease ，
Answer For a giv巳n mass of material , gases g巳 n巳rally have greater volllmes than solids
do , so the volume of the material incl 巳 as巳 s. Th巳1I1CαI 巳as巳 in volllm巳 i臼s~V= 飞酣一飞oliωd
Sinc巳 t由
h巳 volllm巳 of 由巳 sobd
Vsolid is n巳gbgibly sm
曰1all in comηp缸'1βson tωo 由
th巳 volum巳 of 由巳 gas
we have ~ V =飞回 . Usi吨 the ideal gas law as given in Eqllation 14.1 , it follows that
飞as = nR刃P， so that ~ V = nRT/P. 1n 由is reslllt, n is the n Ul曲目 of moles of mat巳rial ， R 自由e
lI niversal gas constant , and T is th巳 Kelvin temp巳rature .
飞，，， ，
As
th巳 mat巳rial chang巳s
巳 nvironm巳 nt
from a solid to a gas , does it do work on
do work on it? How mllch work is involv巳d?
th巳巳 nvironment
or
do巳 s
the
卜
1
To mak巳 room for itself, 由
t h巳巳 xpanding 盯
r a
m
创t巳创rial m
引1U 创
S t pllsh aga
创I川n
川s创t 仙
t h巳巳 nvi让ron
忏
and , in so doing , does work on the 巳 nvironmen t. Since the pr巳 ssure remalI1 S constant ,
work don巳 by the material is given by Eqllation 15.2 as W = P ~V. Sinc巳 ~V = nRT/P ,
work b巳comes W = P(nRT/ P) = nR T.
Answer
mη1巳 旧
nlt
th巳
th巳
1n this problem we begin with heat Q and realize that it is lI sed for two pllrpos巳 s: First , it mak巳s
the solid change into a gas , which entails a change ~U in th巳 internal energy of th巳 material ，
~U = Ugas - Usolid' Second , it allows th巳巳xpanding material to do work W on th巳巳 nvironmen t.
According to the cons巳rvation-of-巳 nergy principle , how is Q relat巳d to ~U and W?
According to the cons巳rvation-of-巳nergy plinciple , energy can neither b巳 created
nor destroyed, bllt can only be convelted from one form to ano由巳r (see Section 6.8). Ther巳fore，
part of the heat Q is used for ~ U and p也.t for W, with the resnlt that Q = ~U + W
Answer
According to the fìrst law of th巳rmodynamics ， how is Q related to
~U
and W?
As indicat巳d in Equation 15.1 , th巳 fìrst law of thermodynamics is ~U = Q - w.
Rearranging this eqllation gives Q = ~U + W , which is identical to th巳 I 巳 slllt obtain巳d from
the conservation-of-energy principl巳.
Answer
Solution Using the facts that Q =
Q
=
~U
~U
+
+
W
W , Q = mL" and W = nRT, we
or
mL s
=
~U
+
nRT
hav巳 that
k
lic
.d o
o
w
m
C
m
o
~ II
to
bu
y
bu
to
k
lic
C
This law, lik巳 th巳 S巳cond law, can b巳巳Xpl 巳ss巳d in a numb巳r of ways , but a discussion of
th巳m
is b巳yond th巳 scop巳 of this t巳X t. The third law is need巳d to 巳 xplain a nllmber of exper.c
c u -tr a c k
imental observations that cannot b巳 explained by the other laws of thermodynamics.
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15.13 CONCEPTS
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to
bu
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bu
to
k
lic
c
/::"U = mL s - nRT
.d o
ln this result , n is the numb巳r of moles of th巳 id巳 al gas. According to th巳 discussion in S 巳:ction
14.1, the number of moles of gas巳 ous zinc is the mass m of the sample divid巳d by the mass per
mole of zinc or n = m/(0.0654 kg/mol). Th巳 refore ， we find
/::"U
=
mLs - nRT
{ , ^^ " '^^ J \
{ 5 0 kg
\f n~ ，
J
\
= (1. 50kg) l 1. 99 X 10 6 一一 ) - l ^:~~-:.~ , 11 8.31-一一-:::- )(6.00 X 10 2 K)
\ k g ) \ 0 . 0 6 5 4 kg/mol )\mol.K)
^
=12.87 X 10 6
JI
Heat engines can b巳 us巳d to P巳rform work , as we have s巳en in this chapte r. The concept of work , how巳ver， was 白 rst introdllced in Chapter 6 , along with th巳 idea of kinetic energy and the work-e n巳rgy theorem . The next example r 巳views some of the main featur巳s
of heat engines , as well as kinetic energy and th 巳 work-en 巳rgy theor巳 m
Concepts &
Ca
刮Icu
川lat
创a阳。 ns Exampl悟
e15
0 T阳 W呐一 Ene
附叫
1旧
er咱
r即
gy
yT
刊刊
The
怡
1冶
eo陀
Temperature = TH
Each of two Carnot engines uses th巳 same cold reservoir at a 忧t 巳 mp巳ratωllr陀巳 of 275 K for its ex♂111
g
阳
le
巳 re
町
c巳创lV
四
es 1450 J of input hea t. The work from either of these engines is
hallst 忱h巳a创t. Each 巳en咆耶
used to driv巳 a pulley arrang巳 ment that us巳 s a rop巳 to accelerate a 125-kg crate from rest along
a horizontal frictionless sllrface , as Figure 15.22 suggests. With engine 1 the crate attains a
speed of 2.00 m/s , while with 巳 ngin巳 2 it attains a speed of 3.00 m/s . Find the t巳 mperature of
th巳 hot reservoir for each engin 巳
Concept Questions and Answers With which engine is the change in the crate's kinetic
energy greater?
Answer The cha咿 lS grea阳 with engine 2. Ki netic energy is 阻 = tmv 2 , according
to Equation 6.2 , wh巳 re m is the mass of the crate and v is its speed. The change in th巳 ki­
netic energy is the final minus the initial valu 巳， or KE f - KEo. Since the crate starts from
rest , it has zero initial kinetic energy. Thus , the change is equal to th 巳 final kinetic 巳nergy
Since engine 2 gives th巳 crat巳 th 巳 greater final speed , it callses th 巳 greater chang巳 111
kinetic en巳rgy.
Temperature = Tc
Which engine does more work?
Answer The work-e nergy theorem, as stated in Equation 6.3, indicates that the net work
don巳 on an object equals the change in the object's kin巳tic 巳 nergy， or W = KE f - KEo. The
net work is the work done by th巳 net force. ln Figure 15 .22 the sllrface is horizontal , and the
crate does not leave it. Therefore , th巳 llpward normal force that th巳 surface applies to the crate
must balance th巳 downward weight of the crat巳 . Fll此hermore ， the surface is frictionless , so
ther古 is no friction force. The net force acting on the crate, then , consists of th巳 single force
due to the tension in the rope , which arises from the action of the engine. Thus , the work don巳
by the engin巳 is ， in fact , the net work done on 由巳 crate . But w巳 know that engine 2 causes
the crate's kinetic 巳nergy to change by the greater amount, so that engin巳 must do more work.
Figure 15.22 With the aid of pulleys
and a rope , a Carnot engine provides
th巳 work that is us巳d to acce \e rate the
crate from rest along a horizontal
frictionless surface. See Example 15.
For
which 巳ngine
is the
t巳 mperature
of the hot reservoir greater?
Answer Th巳 temperatur巳 ofth巳 hot reservoir for engine 2 is greater. We know that 巳 ngine
2 does more work , but 巳ach 巳ngin 巳 rec 巳ives the same 1450 J of input hea t. Th巳refor毡 ， engine 2 d巳rives more work from th巳 input hea t. ln other words , it is more efficien t. But the
efficiency of a Carnot engin巳 depends only on the Kelvin t巳 mperatures of its hot and cold
reservoirs. Since both 巳 ngine s use the same cold reservoir whose temperature i
Solution According to Equation 15 .11 , the efficiency e of a heat engine is th巳 magnitude of
th巳 work Iwl divid巳d by the magnitude of the input heat IQHI , or e = Iwl/ IQHI. Accordi 吨 to
m
o
m
w
o
c u -tr
.
ack
C
k
lic
C
Solving for /::"U gives
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THERMODYNAMICS
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to
c u -tr a c k
王~
IQHI
TH
Iwl
.d o
m
w
o
m
l
C
lic
k
Equation 15.15 , the efficiency of a Carnot engine is e C arnot = I - Tc/TH , where T c and T H are ,
resp巳ctively，
th巳 K巳lvin t巳 mp巳ratures of the cold and hot res巳 rvoirs. Combining thes巳 two 巳qua­
.c
c u -tr a c k
tions , we have
o
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w
w
w
w
w
C
lic
k
to
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CONCEPT SUMMARY
h a n g e Vi
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But
is the magnitude of th巳 net work done on th巳 crate， and it equals the change in the
2
crate's kinetic 巳阳gy， or
= KE f - KEo =
, according to Equations 6.2 and 6.3.
With this substitution , the effìciency expression becomes
imv
Iwl
l
Tc=imd
一一一
IQHI
TH
Solving for the
t巳mperature
TH , we fìnd
TH
1t
=
1As
exp巳cted ，
。
mV"
21QHI
the value of TH for engine 2 is greater:
Engine 1
TH
=
275 K
l 一(125 kg)(2.00 m/s)吗
[332 K[
2(1450 J)
Engine 2
TH
=
l
一
275 K
(1 25 kg)(3.00 m/s)
句三
= [449 KI
2 (1 450 J)
rrON:CEPT SUMMARV
lf you need more help with a concept , use the Learning Aids noted next to the discussion or equation. Examples (Ex.) are in the text
of this chapter. Go to www.wiley.com/college/cutnell for the following Learning Aids:
Interactive LearningWare (l LW) Concept Simulations (CS) Interactive Solutions (IS) -
Additional examples solved in a fìve-step interactive forma t.
Animated text fìgures or animations of important concepts.
Models for certain types of problems in the chapter homework. The caIc ulations are carried out interactively.
Di scussion
Topic
Learning Aids
15.1 THERMODYNAMIC SYSTEMS AND THEIR SURROUNDINGS A thermodynamic system is the
collection of objects on which attention is being focused , and the surroundings are everything else
in the environmen t. The state of a system is the physical condition of the system , as described by
values for physical parameters , often pressure , volume , and temperature.
Thermal equilibrium
『
『
a巳
咱a
mnur t "u
臼巳
γt
m"
Zerolh law
01 Ihermodynamics
Firsllaw
Ihermodynamics
15.2 THE ZEROTH LAW OF THERMODYNAMICS Two systems are in thermal equilibrium if there
is no net f1 0w of heat between them when they are brought into thermal contac t.
Temperature is th巳 indicator of thermal equiLibrium in the sense that there is no net f1 0w of heat
between two systems in thermal contact that have the same temperatur巳.
The zeroth law of thermodynamics states that two systems individually in thermal equilibrium
with a third system are in thermal equilibrium with each other
15.3 THE FIRST LAW OF THERMODYNAMICS The fìrst law of thermodynarnics states that due to
heat Q and work W, the internal energy of a system changes from its initial value of Uj to a final
value of U f according to
t:.. U
。1
Sign convenlion lor 0 and W
=
Uf
-
Uj
= Q-
Q is positive wh巳 n the system gains heat and
is done by the system and
negativ巳 when
The fìrst law of thermodynamics is the
and the change in the internal 巳 nergy.
(15 .1) Ex. 1, 14
W
negativ巳 when it loses hea t. W is positive
work is done on the system.
cons巳rvation-of-energy
wh巳n
work
principle applied to heat , work ,
.c
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The int巳mal en巳rgy is called a function of state because it depends only on
and not on th巳 method by which the system cam巳 to b巳 in a given state.
Quasi-static process
15.4 THERMAL PROCESSES A thermal process is quasi-static when it occurs slowly enough
that a uniform pressure and temperature exist throughout th巳 system at all times.
nu旷
』H"
snu ar nu o au s
『
pb ed
『
ρb
『
nu旷
LnH
ρU
Isobaric work
l snuE E
ou acseo
Isothermal process
th巳 state
An isobaric process is one that occurs at constant pressur巳 The work W don巳
chang巳s at a constant pressure P from an initial volum巳 Vj to a fìna l volume Vr is
of the system Ex. 2
when
a system
(1 5.2) EX.3
W= P ilV= P(Vr - V;)
An isochoric process is one that takes place at constant volume, and no work is done in such a
process
An isothermal process is
on巳 that
takes place at constant temperature.
plac巳 without
the transfer of hea t.
Adiabalic process
An adiabatic process is one that takes
Work done as the area under
a pressure-volume graph
The work done in any kind of quasi-static process is
giv巳 n
by
th巳 area
under the corresponding Ex. 4
pressure-versus-volum巳 graph.
、、
，，
飞飞
，，，
E，
E‘E
，
RT n
一­
W
vi门v
15.5 THERMAL PROCESSES USING AN IDEAL GAS Wh巳 n n moles of an id巳al gas change quasistatically from an initial volume Vj to a fìnal vo lum巳 Vf at a constant Kelvin temperature T , the work
done is
Work done during an
isolhermal process
\
Ill--
EX.5
(15.3)
/
iiw 15.1
When n moles of a monatomic id巳 al gas change quasi-statically and adiabatically from an initial
temperature T j to a fìnal temperature 鸟， the work done is
Work done during an
adiabalic process
W
= ~ nR(贝一刊
(1 5 .4)
During an adiabatic process , and in addition to the ideal gas law, an
Adiabalic change in pressure
and volume
id巳a l
gas
ob巳ys th巳 relation
PjVj =PrV/
γ
wh巳r巳 γ = cp/cv
h巳 at
is the ratio of the specifìc
(1 5.5) IS 15.27
capacities at constant
press Ul巳 and
constant
volume.
15.6 SPECIFIC HEAT CAPACITIES Th巳 molar specifìc heat capacity C of a substance determin巳s
how much heat Q is added or r巳moved when the temper创ure of n moles of the substance changes
by an amou nt ilT:
Q = CnilT
For a monatomic id巳al gas ,
volum巳 are， respectively ,
th巳 molar
Specilic heal capacities 01 a
monatomic ideal gas
(15.6) IS 15.97
specifìc heat capacities at constant
pr巳ssure
and constant
Cp = %R
(15.7)
Cv = ~ R
( 15.8)
wh巳re R is the ideal gas constan t. For a diatomic ideal gas at moderat巳 temp巳ratur巳 s that do not allow vibration to occ叽 these valu巳s ar 巳 Cp = ~ R and C v = ~ R. For any type of ideal gas , the diff巳rence between Cp and C v is
Cp
The second law
。1 Ihermodynamics (heat
flow slalemenl)
-
Cv = R
(15 .10)
15.7 THE SECOND LAW OF THERMODYNAMICS The second law of thermodynamics can be
stated in a number of equivalent forms. In terms of heat fiow , th巳 s巳cond law declares that h巳at
fiows spontaneollsly from a substance at a higher temperatur巳 to a substance at a lower temp巳 ra­
ture and does not fiow spontaneously in the reverse direction.
15.8
HEAT ENGINES
A heat engine produc巳s work (magnitud巳= 1W 1) from input h巳at (magn i-
tl由= IQHI) that is extracted from a heat res巳rvoir at a relatively 1吨h temperatur巳 . The 巳ngine rejects heat (magnitud巳= IQ cI) into a reservoir at a relatively low temperatur巳.
The 巳ffìci 巳 ncy
Efficiency 01 a heat engine
Conservation 01 energy
lor a heal engine
e of a heat engine is
e
Work done
Inp川 heat
1
1
wl
The cons巳rvation of en巳rgy requires that IQHI must be equal to
IQHI =
Iwl + IQcI
(15.11)
QH 1
Iwl plus IQcI:
( 15.12) Ex.6
k
lic
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m
w
o
c
Function 01 state
c u -tr
C
m
o
.
ack
to
bu
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bu
to
k
lic
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Learning Aids
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Topic
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15
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.c
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C h a n g e Vie
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bu
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k
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.c
By combining Equation 15.12 with Equation 15 .11 ,
wntten as
e=
w
，、"
『
""
lB
"u旷
"u
ρLb
'ψ
、。
『
a ""Bt,
nb
o
tnunu nur nuES
『
--
nnnuuwnur
th e 巳 fficiency
of a heat engine can also
IQcl
.d o
(1 5.13)
I -~工
15.9 CARNOT'S PRINCIPLE AND THE CARNOT ENGINE A r巳V巳 rs ibl e process is on巳 in which both
the system and its environment can be retllrned to exactly the states they were in before the process
occllrred
Carnot's principle is an alternative statement of the second law of thermodynamics. It states that
no irreversible engin巳 operating between two reSerVOiIS at constant te mpera仙 res can have a
greater efiìciency than a reversible engine operating between the same t巳 mperatll res. FlIlthermore ,
all reversib le engines operating between the same temperatures have the same efficiency
,
A Carnot engine is a rev巳rsible engine in which all inpllt heat (magn itude = IQ d) orψnates from
a hot reservoir at a single Kelvin temperatllre TH and all rejected heat (magnitllde = IQ c!) goes
into a cold reservoir at a single Kelvin temperatur巳 Tc . For a Carnot engine
ACarnot engine
IQc l
TC
IQI-II
T1_1
The efficiency eCarnot of a Carnot engine is the maximllm
tween two fixed t巳 mperatures can have:
Elficiency 01 a Carnot engine
(1 5.14)
effic i 巳 ncy
that an engine operating be
Tc
eCamo(
(1 5.15)
'H
Ex. 7. 8. 15
CS
牛j
15.10 REFRIGERATORS, AIR CONDITIONERS, AND HEAT PUMPS Refrigerators, air conditioners , and heat pllmps ar巳 devices that lI tilize work (magnitllde = IWI) to make heat (magnitllde = IQ c!)
f1 0w from a lower Kelvin temperature Tc to a higher Kelv in temperature TH. ln the process (th巳 EX.9
refrigeration process) they deposit heat (magn itude = IQHI) at the hi gher temperature. Th巳 princi- ILW 15.2
ple of the cons巳川tion of en巳rgy 叫 lI ires that IQHI = Iw l + IQ c!
If the refrigeration process is ideal , in the sense that it occllrs revers ibl y, the devices are called EX.10
Carnot devices and the r巳lation IQcl/ IQHI =几 /TH (Eqllation 15.14) holds.
The coefficient of performance of a refrigerator or an air conditioner is
Coelficient 01
pe时日 rmance
(relrigerator 日 r
air conditioner)
Coefficient of
pe巾rmance
IQc I
I wl
( 15. 16)
The coefficient of performance of a heat pllmp is
Coelficient 01 performance
(heat pump)
Co 巳ffi口
k比
ωωi
ci巳创n川
1t of
I QH川|
( 15.17)
严E
p
町
阳，巾
rf
foα111
盯阳
T
15.11 ENTROPY The change in entropy /:,. S for a proc巳ss in which heat Q ent巳rs or leaves a system reversibly at a constant Kelvin temperature T is
Change in entropy
/:,.
S=
(手t
(15. 18) Ex. 11 , 13
where the sllbscript R stands for "reversible."
The second law
01 thermodynamics
(entropy statement)
The second law of thermodynamics can be stated in a nllmber of eqll ivalent forms. ln terms of en- IS 15.79
tropy , the second law states that the total entropy of the lI nivers巳 doe s not change when a reversible proc巳s s occurs (/:"Sunivcrse = 0 J/K) and increases when an irreversible process occurs
(/:"Sunivcrsc > 0 J/K)
Irreversible processes callse energy to be degraded in the sense that part of the energy becomes Ex. 12
unavailable for the p巳rforman ce of work. The en巳 rgy W'IIl川 ilablc that is lI navailab le for doing work
becallse of an irreversible process is
Unavailable work
(15. 19)
Wunav仙
where /:"Sun附rsc is the total entropy change of the lI ni verse and To is the Ke lvin temperature of the
coldest res巳 rvoir into which heat can be rejected
Entropy and disorder
lncreased entropy is associated with a greater degree of disorder and
lesser degree of disorder (more order).
The third law
01 \hermodynamics
15.12 THE THIRD LAW OFTHERMODYNAMICS The third law ofthermodynamics states that it is not
possible to lower the temperature of any system to absolute zero (T = 0 K) in a finite number of steps
decr巳ased
entropy with a
m
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available online. HOWeVel; all of /h e qllω/io川 are avαilable for assig l1l nen/ via an online homework managemel1 t program such as Wi leyPLUS 01 陆bAssigll
Section 15.3
The First Law of Thermodynamics
Section 15.8
1. The 自 rst law of thermodynamics states that the change /1 U in the
Q- W, where Q is
internal energy of a system is given by /1 U
the heat and W is 由巳 work. Both Q and W can be positive or negative
numbers. Q is a positive number if 一一一一一， and W is a positive
number if 一一一一一. (a) the syst巳 m loses heat; work is don巳 by
the system (b) the system loses heat; work is done 011 the system
(c) the system g ains heat; work is done by th巳 sys tem (d) the system gains heat; work is done 011 the system
Section 15.4
Thermal Processes
,
I飞
A
一~
飞
A
B
B
A
C
Volume
ω』2的的ω』且
10. A monatomic ideaJ gas is thennaIJy insulated , so no heat can f1 0w
between it and its surroundings. 1s it possible for the temperature of the
gas ωrise? (a) Yes. The temperature can rise if work is don巳 by the gas.
(b) No. The only way 由 at the temperature can rise is if heat is added to
th巳 gas. (c) Yes. The tempera阳re can rise if work is done 011 由e gas.
/0
15. The three C缸ηot engines shown in the drawing operate with hot
and cold reservoirs whose temperature di.ffeJ 巳nces are J00 K. Rank the
efficiencies of the engines , largest to s mall巳st. (a) All engines have th巳
same efficiency. (b) A, B , C (c) B , A, C (d) C , B , A (e) C, A, B
Volume
8. An ideal monatomic gas expands isothermall y from A to B , as the
graph shows. What can be said about this process? (a) The gas do巳S
no work. (b) No heat enters or leaves
A
the gas. (c) The first law of thermodynamics does not apply to an isothermal process. (d) The ideal gas law is
B
not valid during an isothermal process.
(e) Ther 巳 is no change in the interna l
Volume
energy of th巳 gas
No/e
Carnot's Principle and the Carnot Engine
/
Thermal Processes Using an Ideal Gas
Section 15.5
Heat Engines
ωL2山的ω』且
6. The pressure-volume graph
shows three pa由 s in whlch a gas
expands from an inüiaJ state A to
a fina l state B. The change
/1 UA • B in internaJ energy is the
same for each of the paths. Rank
the paths accord ing to the heat
Q added to the gas , larg巳st to
smallest. (a) 1, 2, 3 (b) 1, 3 , 2
(c) 2, 1, 3 (d) 3, 1, 2 (e) 3, 2 , l
C
c u -tr a c k
13. A heat e ngin巳 takes heat QH from a hot reservoir and uses part of
this energy to perform work W. Assuming that QH cannot be changed,
how can the 巳伍 ciency of the engine be improved? (a) 1ncrease the
work W; the heat Qc rejected to the cold reservoir increases as a resul t. (b) 1ncreas巳 the work W; the heat Qc rejected to 由e cold reservoir remains unchanged. (c) 1ncrease the work W; the heat Qc
r句 ected to the cold reservoir decreas巳s as a resul t.
(d) Decrease 由e
work W; the heat Qc rejected to the cold res巳rvoir remains unchanged
(e) Decrease the work W; the heat Qc rejected to the cold reservoir
decreases as a res ul t.
Section 15.9
ω」3的的由ι
」
4. The drawing shows th巳 ex­
pansion of three ideal gases.
Rank th巳 gases according
to the work they do , largest
to smalles t. (a) A , B , C
(b) A and B (a tie) , C (c) B
and C (a tie) , A (d) B , C , A
(e) C, A, B
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lnstruc/ors: Mos/ 旷 /h e hOlllework problems in /his chap/er 0/在
Section 15.10 Refrigerators , Air Conditioners ,
and Heat Pumps
17. A refrigerator operates for a certain time , and the work done by
the electricaJ energy during thls time is W = 1000 J. What can be said
about the heat d eliver巳d to the room contai ning the refrigerator?
(a) The heat d巳 liv巳red to the room is less than 1000 J. (b) The heat
de li vered to the room is equal to 1000 J. (c) The heat delivered to
the room is greater than 1000 J.
Section 15.11
Entropy
19. Heat is transf，巳rred from the sun to the earth via e l巳ctromagnetic
waves (see Chapter 24). Because of this transfer, the entropy of
the sun 一一一一一一一， the entropy of the earth _一一一一一一， and the
entropy of 由e sun-e arth system 一一一一一 (a) increases , decreases ,
decreases (b) decreases , increases , increas巳s (c) increases , increases , increases (d) increases , decreases , increases (e) decreases ,
increases , decreases
available
fo r assignmen/ via an online homework
manageme
凹
e l1/ plV
η g ra
削
川
11n 川
s IIch
ht
αωS
川州
川il均
w
l ey
e 凡
P 川
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U
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o ， 陆b
协
'As
叫
s
川
tυ
ll1era
曰lC/I.川川
vJ汀I~σ
y.
See Preface for additional details.
ssm Solution is in the Student Solutions 岛1anua l.
www Solution is available online at www.wiley.comJcollege/cutnell
Section 15.3
The First Law of Thermodynamics
1. 1n moving out of a dormitory at the end of th巳 semester， a
studen t does J. 6 X 104 J of work. 1n th 巳 proces s ， his internal
This icon represents a biomedical application.
energy d 巳 creases by 4.2 X 10 4 J. Determine each of th 巳 fol­
lowing quantities (including the algebraic sign) : (a) W (b) /1 U
(c) Q
.c
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川
F
日 ow s 川
i ntωot由
he sys 忧t 巳 m
n a l 巳nerg y
dllring 由
t h巳
of the system .
pro ce s s.
Find the chang巳 in the inter-
5. ssm When on巳 gallon of gaso line is bllrned in a car engine ,
1.19 X 108 J of internal energy is released. S lI ppose that 1.00 x 108 J
of this energy ftows directly into the sllrrollndings (engine block and
exhallst system) in the form of heat. If 6.0 x 105 J of work is reqll ired
to make the car go one mi le , how many mi les can the car travel on one
gallon of gas?
6. Three moles of an ideal monatomic gas are at a temperatur 巳 of
345 K. Then , 2438 J of heat is added to the gas , and 962 J of work is
done o n it. What is the fin al temperatllre of the gas?
*7. 电馨，. In exercising , a weight lifter loses 0.150 kg of wat巳r throllgh
巳va poration ， the heat reqll 让ed to eva porat巳 the water coming
from the weight lifter's body. The work done in lifting w巳 ights is
l. 40 X 105 J. (a) Assllming that the latent heat of vaporization of
perspiration is 2 .42 X 106 J/kg , find the chan g巳 in the internal energy
of the weight lifte r. (b) Determin巳 the minimllm nllmber of nlltrition al Calories of food (l nlltritional Calorie = 4186 J) that mllst be
consllmed to replac巳 the loss of internal energy.
Section 15.4
Thermal Processes
8. A gas lI ndergoes isochoric heating , dllrin g which it gains 5470 J of
heat and attains a pressllre of 3 .45 X 105 Pa. Following this , it experiences an isobaric compression that is also adiabatic , in which its vol
3
3 Find the total change in the
lI me decreases by 6. 84 X 10- m
internal energy of the gas fo r th is two-step process. Be sure to inclllde
the algebraic sign (+ or -) of the tota l change in the internal energy
9. ssm When a .22-calib巳r rifte is fired , the 巳xpanding gas from the
bllrning gllnpowder creates a pressllre behind the bllllet. This pressllre
callses the force that pllshes the bllllet throllgh the barrel. The barr巳 l
has a length of 0.61 m and an opening whos巳 radills is 2.8 X 10- 3 m
A bllllet (mass = 2.6 X 10- 3 kg) has a speed of 370 mls after passing throllgh thi s barre l. Ignore 白 i ctio n and determine the average
pressur巳 of the expanding gas .
10. (
A system gains 2780 J of heat at a constant pressure of
1. 26 X 105 Pa , and its int巳rnal energy increases by 3990 J. What is the
change in the volllme of the system , and is it an increase or a decrease?
11. A gas , whi le expanding lI nd巳 r isobaric conditions , does 480 J of
work. The pressllre of the gas is 1.6 X 10 5 Pa , and its initial vo lllme
is 1. 5 X 10 m3 What is the fina l volllme of the gas?
•
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ν，
vl
<lJ
.t 3.0 x
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w
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7.0 x 10 5
C
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3. ssm A system does 164 J of work on its 巳 nvironm巳 nt and gains
77 J of heat in the process. Find the change in the internal energy of
(a) th巳 system and (b) the environme nt
13. The pressllre and vo llI me of a gas are changed
alo ng the path ABCA.
Using the data shown in the
graph , determine the work
done (inclllding the a lgebraic sig n) in each segment
of the path: (a) A to B ,
(b) B to C , and (c) CtoA
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2. 嗖替，. In a game offootball olltdoors on a cold day, a play巳rwill b巳gin to feel ex hallsted after lI sing approximately 8.0 X 105 J
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of internal energy. (a) One play巳 r， dressed too lightly for the
weather, has to leave the ga me after losin g 6.8 x 100 J of heat. How
mllch work has he done? (b) Another play巳 r， wearing clothes that
offer better protection against heat loss , is ab le to remain in th巳 g a m巳
long enollgh to do 2.1 x 100 J of work. What is the magnitllde of the
heat that he has lost?
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阵卡
14. R ef巳 r to M lI ltipl eConcept Exampl 巳 3 to se巳
2.0 x 10- 3 5.0 x 10- 3
how the concepts pertinent
to thi s problem are lI sed.
Volume , m3
The pressllre of a gas reProblem 13
mains constant while th巳
temperatllre , volllme , and internal energy of the gas increase by
53.0 C O , 1 .40 X 10- 3 m 3, and 939 J , respectively. The mass of the gas
is 24.0 g , and its specifi c heat capacity is 1080 J/(kg. C O ). Determine
th巳 pl 巳 ss ure.
15. ssm A system gains 1500 J of heat, while the internal energy of
the sys tem increases by 4500 J and the volllme decreases by 0.010 m3
Assllme that the pressllre is co nstant and find its valll巳.
*1 6. A piece of alllminllm has a volllm巳 of l. 4
X 10- 3 m 3 • The coefficient of volllme expansion for a llllllinllm is ß = 69 X 10- 6 (C O ) 斗 ，
The temperature of this object is raised from 20 to 320 o
How
mllch work is done by th巳 expanding alllminllm if the air pressllre is
1. 01 X 10 0 Pa?
c.
。' 17 .
ssm www A mon atomi c ideal gas 巳x pands isobarically. Using the
first law of thermodynamics , prove that the heat Q is positive , so that
it is impossible for heat to ftow Ollt of th巳 ga s
"18. Refer to the drawing that accompanies Problem 9 l. When a system changes from A to B along the path shown on th巳 pressure- versll s­
volllm巳 graph ， it gains 2700 J of heat. What is the change in the
internal energy of the system ?
!"" 19. Water is heated in an open pan where the air pressllre is one atmo sp h巳 re. The water remain s a liqllid , which 巳xpands by a small
amollnt as it is heated. D etermin e the ratio of the work done by th巳
wat巳 r to the heat absorb巳d by the wate r.
Section 15.5
Thermal Processes Using an Ideal Gas
20. Five moles of a monatomic ideal gas expand adiabatically, and its
temperatllre decreases from 370 to 290 K. Determine (a) the work
done (i nclllding the algebraic sign) by the gas , and (b) the change
in its internal energy.
2 1. ssm Three moles of an ideal gas are compressed from 5.5 X 10- 2
to 2.5 X 10- 2 m 3 . D lI ring the compression , 6 .1 X 103 J of work is
done on the gas , and heat is r巳 mo v巳d to keep the telllperature of th巳
gas constant at all times . Find (a) I1 U, (b) Q, and (c) the temperature of the gas
22. (
Three moles of neo n 巳xpand i sotherm均 to 0.250 from
0.100 m 3 . Into the gas ftow s 4.75 X 103 J of heat. Assllming 由 at neon
is an ideal gas , find its temperatllre.
23. The temperatllre of a monatolllic ideal gas remains constant during a process in which 4700 J of heat ftow s Ollt of the gas. How mllch
work (inclllding the prop巳 r + or - sign) is done?
24. The pressure of a monato mic ideal gas (γ= ~) dOllbles dllring an
adiabatic co mpres sion. What is the ratio of the final volllme to the initi al volllme?
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A岱
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th巳 ratio
of 由
th巳
ml凶
ti 创
a 1 volum
丑1巳 to
the fìn al volume.
创. ② An ideal gas i s 时巳n 山叫h th巳巾ee processes
•
•
•
(A
B, B
C , and C
A)
shown in the draw ing. In genB
巳ral ， for each process the int巳rnal enenrv U of the !!:as can 主
vl
change because heat Q can be
0..
added to or removed fro m the
gas and work W can be done by
the gas or on the gas. For the
three processes shown in the
draw in σfìll in the fìve missin!!:
b entries in the
C
followin σ tabl巳.
Process
!:!.U
Q
W
A 一c> B
(b)
+561 J
(a)
B• C
+4303 J
(c)
十 3740
C 一兮 A
(d)
(e)
+2867 J
!:!. U
A to B
•
B to C
o
J
W
Q
fectly insulated cylinder that is fìtted with a movable piston. The ini
tial pressure of the gas is 1. 50 X 10 5 Pa. The piston is pushed so as to
compress the gas , with the result that the Kelvin temperature doubles.
What is the fìnal pressure of the gas?
m m
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"4 1. Suppose that 3 1.4 J of heat is added to an idea1 gas. The gas expands
at a constant pressure of 1.40 X 10 4 Pa while changing its vo lume 仕um
3.00 X 10- 4 to 8.00 X LO- 4 m3 The gas is not monatomic , so the r在 lati on
Cp = ~ R do巳s not apply. (a) Detennine 出巳 chang巳 in the intemal energy of the gas. (b) Calculate its molar specifìc heat capacity Cp .
•
0.200
0 .4 00
Vo lume , m3
'*43. One mole of neon, a monatomic gas , starts out at co nditions of
standard temperature and pressure. The gas is heated at constant volume until its pressure is tripled , then further heated at constant pressure until its volum巳 is doubled . Assume that neon behaves as an ideal
gas . For the entire process , fìnd the heat added to the gas
m
lic
(a) What
percentage of the heat being suppli ed to the gas is used to increase the
internal energy of the gas? (b) What percentage is used for doing
the work of expansion?
*42. A monatomic ideal gas is heated while at a constant volume of
1. 00 X 10 m 3, usi ng a ten-watt heater. The pressure of the gas increases by 5.0 X 104 Pa. How long was the heater on?
Rd
O
38. (
Under constant-volume con出ions， 3500 J of hωis added
to 1.6 moles of an ideal gas. As a result , the t巳 m peratu re of the gas in
creases by 75 K. How much heat would be required to cause the same
temperature change under co n s tant-press Ul 巳 conditio n s? Do not assume anythin g about whether the gas is monatomic , diatomic , etc.
1'40. A monatomic ideal gas expands at co nstant pressure.
D to A
斗 30. ~ A monatomic ideal gas ( γ= ~) is contained within a per-
*3 1. The pressure and volume
of an ideal monatomic gas
change fro m A to B to C, as
th巳 drawing shows. The
curved line between A and C
is an isotherm. (a) Deter
mine the total heat for the
process and (b) state whether
the flow of heat is into or out
of the gas.
37. ssm Heat is added to two identical samples of a monatomic ideal
gas . In the fìrst sample the heat is added while the volum巳 of the gas
is kept constant , and the h巳at causes the temperature to rise by 75 K.
In the second sample , an identical amount of heat is added while th巳
pr巳 ssure (but not the vo lume) of the gas is kept conSlan t. By how
much does the temperature of this sample increase?
39. Heat Q is added to a monatomic ideal gas at constant pressure.
As a res ult , the gas does work W. Find the ratio QI W.
C ωD
Vo lume
Specific Heat Capacities
。
ω
isotherm
'"V ,..,. ...
- -ñ↑电芝F寸---- 400.0-K isotherm
01
}
\
2OO .O-K isotherm
Section 15.6
36. A monatomic ideal σas m a nσid container is heated from 217 K
to 279 K by adding 8500 J of h巳 at. How many moles of gas are th巳 re
in the container?
ssm The drawing refers to one mole of a monatomic ideal gas
and shows a proc巳 ss that has fo ur steps , two isobaric (A to B , C to D )
and two isochoric (B to C, D to A). Complete the following table by
calculating !:!.U, W, and Q (including th巳 a l geb rai c signs) for each of
the four steps.
吐\
+ \ 800
. 0-K
\f 三\ l r
~~~.~"
525 K on the
and 275 K on the righ t. The partition is then allowed to move slowly
(i .e. , qu as i-staticall y) to the right , unti l
th巳 pressures on each sid巳 of the partition are the same. Find the 自 nal temperatures on the (a) left and (b) right.
35. ssm The temperature of 2.5 mol of a monatomic ideal gas is
350 K. The internal energy of this gas is doubled by the add ition of
hea t. How much heat is needed when it is added at (a) constant volume and (b) constant pressure?
Volume
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34. Three moles of a monatomic ideal !!:as are heated at a constant volume of 1.50 m3 . The amount of heat added is 5.24 X 103 J. (a) What
is the change in the temperature of th巳 gas? (b) Find the c han g巳 m
its internal energy. (c) Determine the change in press ure.
的
.t
**33. ' ssm The drawi ng shows an adiabatically isolated cylinder that is
dωivid ed ini汹
ti 剖
a 11忖
y into two identical part岱s by an adi旧
abatic par川ti让tlωon.
Both 剖
s id巳 s contam o n巳 m o l e of a
o
m
o
"27. Refer to Interactive Solution 15.27 at www.wiley.co mJcollege/
cutnell for help in solving this problem. A dies巳l 巳 ngine does not use
spark plugs to ignite the fuel and air in the cylinders. Instead , the temperature required to ignite the fuel occurs because the pistons comt ha
创t air at an ini凶
ti 创
a1 臼
t 巳 mperaωr陀
e
press the air in the cylinders. Suppose 由
of 21 oC 沁
i s compressed adiab 剖
a tl比
cally tωo a 忧
t emp巳ratur巳 of 688 o
C
to
k
lic
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26. Heat is Ç1 dded isothermally to 2.5 mol of a monatomic ideal gas
Th巳 temperature of the gas is 430 K. How much heat must b巳 ad ded
tomak，巳 the volume of th巳 gas double?
"*32. Beginning with a pressure of 2.20 x 105 Pa and a vo lume of
6.34 X 10 - 3 此 an ideal monatomic gas (γ= ~) undergoes an adiaw
.c
.d o
c
ack
batic expansion such that its fìnal pressure is 8.15 X 104 Pa. An alter- u - t r
native process leading to the same fìnal state begins with an isochoric
cooling to the fìnal pressure , fo llowed by an isobari c expansion to the
fìnal volume. How mu ch more work does the gas do in the ad iabatic
process than in the alternative process?
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25. ssm A monatomic id巳al gas has an initial temperature of 405 K
This
gas
expands and does the same amount of work whether the
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巳xpansion is adiabatic or isotherma l. When the expansion is adiabatic , the fìn al temp巳rature of the gas is 245 K. What is the ratio of
the fìnal to the initial volum巳 when the expansion is isothermal?
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CHAPTER 15 THERMODYNAMICS
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悦 ③ Carnot eng ine A has an e ffi ciency of 0. 60， 川 Carnot en-
to
44.
gine B has an effici ency o f 0.80. Both engines utili ze the sawme
. d o hot k . c
c u -tr a c
rese rvoir, whi ch has a te mperature of 650 K and de livers 1200 J of
h 巳a t to eac h e ng ine. Find th e magni tude of the work produ ced by
eac h eng ine and th e te mperatures of the co ld reservo irs that they
use.
哩害，.
Heat engines take input energy in the form of heat, use
~ so me of th at energy to do work , and ex haust the re m a ind e l二
Similarly, a person can be viewed as a heat e ngine that takes an input
of internal energy, uses so me of it to do work , and gives 0仔 th e rest
as heat. Suppose that a trained athl巳te can fun cti on as a heat eng ine
with an effi ciency of 0. 11 . (a) What is the magnitude of the intern al
巳nergy that the athlete uses in order to do 5 .1 X 104 J of work?
(b) Determine 出巳 mag nitu d巳 of the h巳 at the athl ete gives o1'f.
c u -tr a c
57. Concept Simulation 15.1 at wwwλviley. comJcollege/cutnell illustrates the co n c巳 pts pertinent to thi s problem. A Carnot engine operates between temperatures of 650 and 350 K. To improv巳 th巳
efficiency of the engine , it is d ec id ed 巳 i t h er to raise the temperature
of the hot reservo ir by 40 K or to lower th巳 te mpe ratu re of th巳 co ld
reservo ir by 40 K. Wh ich change gives the greatest improve ment?
Justify your answer by calcul atin g the ef白 c i e n cy in eac h cas巳
45. ssm Multiple-Concept Exa ll1 ple 6 deals with the concepts that
ar巳 i ll1 po rta nt in this proble ll1 . ln doing 16 600 ] of work , an engine
叫 ects 9700 ] of hea t. Wh at is the effi ciency of the engine?
with an effi c i 巳 ncy of 0.22 rejects 9900 ] of
What is the magnitude of the work th at the engine
does in one second ?
46. A
l awnm ow e r 巳n gin e
h eat 巳very s巳co nd .
47. Due to a tune-up , the effi ciency of an automobile eng ine increases by 5.0%. For an input heat of 1300 J , how much more wo rk
does the engine produce after the tune-up th an befo re?
*咱
4咄8. Y ~ 叼
A 5旦2-1
均
E鸣g 川
盯m1ω
肌川
ωou川l川I川
O饥
B
叼
v 巳r忧t 比
i ca
刽1
distance of 730 m. At 由
t h巳 tω
op ， she 比
is
ag
，引
a山
in 挝
a tl陀巳s创t.
In
t h巳 process趴， h巳r body generates 4. 1 X 106 ] o f 巳 ne rgy vla mη巳 tabo li c
ln fac t, her body acts like a heat engine , the effi ciency 0 1'
which is given by Equati on 15.1 I as e = IwI/IQHI , where Iwl is the
magn 山de of the work she does and IQHI is the magnitude 01' the in
put hea t. F ind her efficiency as a heat eng ine
严ro
P
町
c巳
郎ss臼巳 s.
不 49.
ssm www Due to design changes , the efficiency of an engine inar
creases fro m 0.23 to 0 .42. For th巳 s 创me
inc
αreas巳 t由
h巳 work don巳 by the mη1 α0I 巳巳f白 cωie n t 巳 n ♂
g in e and 1 巳du c巳 th e
amount of h巳 at r巳句'J ect优巳d tωo the cold 1 巳s巳r vo ir. Find 巾
t h巳 ratiω
o oft由
h巳 h巳 a t
r陀
巳J庐
ec
αt巳
e d tω
o 由
t h巳 cold 陀
res巳
创rv
刊
0町
1 r fo r the ill1 proved engine to that for the
original engi ne.
"*50. An engine has aI1 efficiency e\ . The engine takes input heat of
magnitude IQHI fro m a hot res巳rvo ir and d巳li vers wo rk of magnitude
Iw\l. The heat 叫 ected by thi s 巳叩 ne is used as input heat fo r a s巳c­
ond engine , whi ch has an effi ciency e2 and deli vers work of mag ni tude Iw2 1. The overall efficiency of this two-e 叩 ne device i s 由e
magnitude of the total work deliver巳d (lw\1 + IW2 1) div id巳d by the
magnitude IQHI of the input hea t. Find an ex pression fo r the overaII
efficiency e in terlllS of e\ and e2'
Section 15.9
Carnot's Principle and the Ca rnot E ngine
5 1. A Carnot engine operates w ith an efficiency of 27 .0% when the
temperature of its cold reservoir is 275 K. Ass ullling that the temperature of the hot reservoir remains the same , what must b巳 the temperature of the cold reservoir in order to increase the e ffi c i巳 n cy to 32.0%?
r.
58. The hot reservo ir fo r a Carnot eng ine has a t巳 mpe ra tu re of 890 K ,
while the cold reservo ir has a te mperature of 670 K. The heat input
fo r this engine is 4800 J. The 670-K reservo ir also serves as the hot
reservoir fo r a second Carnot e n σ in e. This second engine uses the rejected heat of the first engine as input and ex tracts additi onal work
from it. The I吗 ec ted heat from the seco nd engine goes into a r巳 se r­
vo ir that has a te mperature of 420 K. Find th巳 to ta l work d e li v巳red by
the two engines.
。
气 59 .
ssm A power plant taps steam superheated by geothermal energy
to 505 K (the te ll1 p巳 ra ture of the hot reservo ir) and uses the stearn to
do work in turning th巳 turbin e of an electric generator. The stearn is
then converted back into wat巳r in a condenser at 323 K (the temperature of the cold res巳rvo ir) ， after which the water is pumped back dow n
into the earth where it is h巳a ted again . Th巳 o utput power (work per uni t
time) of the plant is 84 000 ki lowatts. De t巳 rllline (a) the max imum
effi ciency at whi ch this plant can operate and (b) the min imum
amount of rejected heat th at ll1 ust be 1 巳 moved from th巳 c onde n se r
every twenty-four hours.
王 60.
Suppose that the gasoline in a car engine burns at 63 1 oC , while
exhaust temperature (th巳 temperature of the cold reservo ir) is 139 OC
and the outdoor temperature is 27 o Ass ume that th巳 e n g in e ca n be
t l 巳 a te d as a Ca rn o t 巳 n g in巳 (a gross oversimpl ifi cati on). ln an attempt
to increase mil eage performance , an inventor builds a seco nd engine
that fun ctions between the ex haust and outdoor te mp e ratur巳 s and
uses the exhaust heat to produce additi ona l work. Ass ume th at the inventor's eng ine can also be treated as a Carnot eng ine. Determine the
rati o of th e total work produ ced by both engines to that produ ced by
the first engine a lone.
由e
c.
位 "' 6 1.
ssm The draw ing (not to scale) shows the way in which th巳 pres
sure and vo lum巳 c h a n ge fo r an ideal gas that is used as the working
substance in a Carnot e n g in巳 Th e gas begins at point a (pressure =
P a, volum
52. Five thousand joules of heat is put into a Carn ot engine whose
hot and cold reservoirs have te ll1 peratures of 500 and 200 K , respecti vely. How much h巳at is co nverted into work?
53. ssm A Carnot engine has an efficiency of 0.700 , and the temperature of its cold reservoir is 378 K. (a) De te rmin巳 th e temperature
of its hot reservoir. (b) If 5230 ] of heat is rejec ted to th巳 co ld reservoir, what amount of heat is put into the eng ine?
ll|
ω』2山的ω』
ι
55. An engine does 18 500 J of work and rejects 6550] of heat into
a cold reservoir whose temperature is 285 K. Wh at would be the
sma Il est possible temperature of the hot res巳rvo ir ?
A.，
54. A Carnot engine operates with a large hot reservoir and a mu ch
sll1 aIler cold reservoir. As a result , the temperature of the hot r巳servo lI
remains constant while the t巳 mp e rature of the cold res巳 rvo ir slowl y
increases. This te ll1 perature ch a n g巳 d ec reases the e ffi ciency 0 1' the engine to 0.70 fro m 0.75. F ind the ratio of th巳 fi n a l te mpe ratur巳 of the
cold reservoir to its initi al temperature.
m
lic
k
Heat E ngines
Isotherm
temperature =
c
Volume 一-
TH
Isoth erm
temperature = Tc
o
c
k.
C
m
o
k
lic
C
Section 15.8
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A nuclear-fu 巳 led e Iectric power plant uti Ii zes a so-called "boiling water reacto r." ln this type of reactor, nuclear energy causes wa ter under pressure to boil at 285 oC (the temp巳rature of the hot
reservoir). After the steam does the work of turning the turbine of an
electric generator, th巳 steam is converted back into water in a cond巳 nser at 40 oC (th巳 temperature of the cold reservoir). To keep the
cond巳 nser at 40 oC , the rejected heat must be carried away by SO Ill巳
Ill eans-for example , by water frolll a rive r. The plant operat巳 s at
three-fourths of its Carnot effici 巳 ncy， and the electrical output poweI
of the plant is 1. 2 x 10 9 watts. A river with a water flow rate of
1. 0 X 10 5 kg/s is avai lable to relllove the rejected heat frolll the
plan t. F ind the nUlllber of Celsius d巳grees by which th 巳 telllpe rature
of the river rises.
喝 72.
63. ssm www The temperatures indoors and outdoors are 299 and
312 K , respectively. A Carnot air cond iti on巳r depo sits 6.12 X 105 J of
h巳 at outdoors. How much heat is removed from the hou se?
64. ③ Thein础。fa Carnot 阶1gen阳山 maintain巳d at a temperature of 277 K , whi le th巳 temperature in the kitchen is 299 K. Using
2500 J of work , how much heat can this refrigerator remove from its
inside compartment?
65. A refrigerator operates betwe巳 n t巳 mp巳ratures of 296 and 275 K.
What wOllld be its maximllm coefficient of performance?
66. (
Two Carnot air c叫 tIOn叽 A and B ，也 e removing heat
from different rooms. The olltside t巳 mperature is the sa me for both
rooms , 309.0 K. Th巳 room serviced by unit A is kept at a temp巳rature
of 294.0 K , while the room serviced by lI nit B is kept at 301.0 K. The
heat remov巳d from either room is 4330 J. For both units , find th巳
magnitllde of th巳 work reqllired and the magnitude of the heat deposited olltside.
67. A Carnot refrigerator is lI sed in a kitchen in which the temperature is kept at 301 K. This refrigerator lI ses 241 J of work to remove
2561 J of heat from the food inside. What is th巳 temperature inside
the refrigerator?
68. A heat pllmp removes 2090 J of h巳at from the outdoors and delivers 3140 J of heat to the inside of a hOlls巳 (a) How much work
does the heat pump need? (b) What is the coefficient of performance of the heat pllmp?
69. A Carnot heat pump operates between an outdoor temperature
of 265 K and an indoor temp巳 rature of 298 K. Find its co巳ffic i ent of
performance
lnteractive LearningWare 15.2 at www.wiley.comlcollege/cutnell
explores one approach to problellls such as this. Two kilograms of liquid water at 0 oC is put into th巳 fre巳zer compartment of a Carnot re
frigerator. Th巳 temperature of the compartment is -15 oC , and the
temperatllre of the kitchen is 27 o If th 巳 cost of electrical energy is
ten cents per kilowatt. hour, how much does it cost to make two kilograms of ice at 0 oC7
c.
*73. ssm www A Carnot refrigerator transfers heat from its inside
(6 .0 OC) to the room air outside (20.0 oc) . (a) Find the coeffici巳 nt
of performance of the refrig巳 rator. (b) D巳termine the magnitude of
th巳 ml I1l mllm work needed to cool 5.00 kg of water from 20.0 to
6.0 oC when it is placed in the refrigerato r.
中" 74.
A Carnot engine lI S 巳 s hot and cold r巳 servoirs that have temperatures of 1684 and 842 K , respectiv巳Iy . The input heat for thi s
engine is IQHI. The work delivered by the engine is used to operate a Carnot heat pump. T he pump removes heat from the 842-K
reservoir and puts it into a hot reservoir at a temperature T'. The
amount of heat removed from the 842-K [巳servoir is also IQHI. Find
th巳 t巳 mperature T'
Section 15.11
Entropy
75. Consider three engines that each lI se 1650 J of heat from a hot
re s巳 rvo ir (temperature = 550 K). These thr巳巳 engines reject heat to a
cold re s巳 rvoir (temperature = 330 K). Engine 1 rej 巳cts 1120 J of heat
Engi ne Il r巳J 巳cts 990 J of hea t. Engine III rejects 660 J of hea t. One
of the 巳 ngin es operates reversibly, and two operate irreversibly.
However, of th巳 two irreversible engines , one violat巳s the second law
of thermodynamics and could not exis t. For each of the 巳 ngine s de
termine the tota l entropy change of the universe , which is the sllm of
the entropy changes of th巳 hot and cold reservoirs. On th巳 basi s of
your calculations , identify which engine op巳rates reversibly , which
operates irreversibly and cOllld exist , and which operates irr巳versibly
and cOllld not 巳XIS t.
76. H巳 at Q ftows spontaneously from a reservoir at 394 K into a
reservoir that has a low巳r temperature T. B 巳cause of the spontan 巳ous
flow , thirty percent of Q is rendered unavailable for work when a
Carnot engine operates betwe巳n the reservoir at temp巳rature T and a
reservoir at 248 K. Find the temperature T.
77. s s m F ind the change in entropy of th 巳 H 2 0 molecules when
(a) thre巳 ki l ograms of ice melts into wat巳 r at 273 K and (b) three
ki lograllls of water changes into steam at 373 K. (c) On the basis
of th巳 answers to parts (a) and (b) , discll ss which change creates
more di sorder in the co ll 巳ction of H 20 molecules
78. On a cold day , 24 500 J of h 巳 at Ie aks out of a house. The inside t巳 mperature is 21 oC , and the outside temp巳 rature is 一 15 C
What is th巳 Increas巳 in th巳 entropy of the universe that thi s heat
loss prodllc巳 s7
0
*70. The wattage of a commerc ial ice Ill aker is 225 W and is the rate
at which it does work. The ice maker operates jllst like a refrigerator or an air conditioner and has a coe仔icient of performance of
3.60. The water going into the unit has a temperature of 15.0 oC , and
the ic巳 maker produces ic巳 cllbes at O.O o Ignoring th 巳 work
needed to keep stored ice from melting , find the maximum alllount
(i n kg) of ice that the lI nit can prodllce in one day of continuous
operatIOn.
c.
*79. Refer to Interactlve Solutlon 15.79 at www.wiley.comlcoHege/
cutnell to review a method by which this probl巳 m can be solved.
(a) After 6.00 kg of water at 85 .0 oC is rnixed in a perf巳ct thermos
with 3.00 kg of ice at 0 .0 oC , the mixture is aIIowed to reach equilibrilllll. When heat is added to or removed from a solid or liqllid of
m
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Section 15.10 Refrigerators , Air Conditioners ,
and Heat Pumps
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!'71. ssm Review Conceptual Example 9 before attempting this probw
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lelll. A window air conditioner has an av巳rage coefficient of performc u -tr a c k
ance of 2.0 . This unit has b巴巴 n placed on the ftoor by th巳 bed ， in a
futile attempt to cool th巳 b巳droom. During this attempt 7 .6 X 104 J of
h巳at is pll Il ed in the 白 ont of the uni t. Th巳 room IS S巳 aIed and contains
3800 11101 of air. Assuming that the molar specific heat capacity of the
1 S C \I = ~R， d巳创t巳rm1l1巳 th巳 r门IS巳 111 忧肌
mp巳削Ir巳 caus巳d by op巳则 n
alr 罔
t由
h巳 al町r co ndition巳 r in this mη1ann巳r
lic
reservoir of the 巳 ngine . Then , from point b to point c (pressur巳 = Pc,
. c = Vç ) , the gas expands adiabatically. Next , the gas is
volume
c u -tr a c k
compressed isothermally at temperature TC from point c to point d
(pressure = Pd , volum巳= Vd ). During this compression , heat of magn山 de IQcI is r句巳ct巳d to the cold reservoir of the engin巳. Fina lI y, the
gas is compress巳 d adiabatica lI y from point d to point a , where the gas
is back in its initial stat巳 Th巳 overaIl proc巳 s s a to b to c to d is ca lI ed
a Carnot cycle. Prove for this cycle that IQcI/ IQHI = Tc/TH
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CHAPTER 15 THERMODYNAMICS
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An irreversible engine operates between temperatures of
"81. (
852 and 314 K. 1t absorbs 1285 J of h巳 at from the hot reservoir
and does 264 J of work. (a) What is the change I:!. Sunjvcrse in the
entropy of the universe associated with the operation of this
engine? (b) If the engine were rev巳 rsible ， what would b 巳 the
magnitude Iwl of the work it would have don巳， assuming that it
op巳rated between th 巳 same temperatures and absorbed the same
heat as the irreversible engine? (c) Using th巳 results of parts (a) and
n叫
d 由
th 巳 d
副if
旺
f巳I 巳 nce b巳创tw巳巳 n 由
t h巳 work 严
p ro
刀
oduc巳 d by 巾
t h巳 陀r啕它
E
(b) , fin阳
V巳rsibl巳 and in 巳V巳 rsibl 巳巳 ngm 巳s
出 80.
Heat flows from a res巳 r­
voir at 373 K to a reservoir at
Problem. 80
Af日 DITIONAL
PROBLEMS
82. One-half mole of a monatomic ideal gas expands adiabatically
and does 610 J of work. By how many kelvins does its temperature
change? Specify whether the change is an increase or a decreas巳.
83. ssm One-h a1 f mole of a monatomic ide a1 gas absorbs 1200 J of
heat whiJe 2500 J of work is done by the gas. (a) What is th巳 temp町，
ature change of the gas? (b) 1s the change an increase or a decrease?
84. Multiple-Concept Examp l巳 6 deals with th巳 same concepts as
this problem do巳s. What is the efficiency of a heat engine that uses an
input heat of 5.6 X 104 J and rejects 1. 8 X 104 J of heat?
85. A gas is contained in a chamber such as that in Figure 15 .4
Suppose that the region outsid巳 the ch创nber is evacuated and the total
mass of 出巳 block and the movable piston is 135 kg. When 2050 J of
heat flows into the gas , the internal 巳 nergy of the gas increases by
1730 J. What is the distance s through which th巳 piston rises?
86. Engin巳 1 has an efficiency of 0.18 and requires 5500 J of input
heat to perform a certain amount of work. Engine 2 has an efficiency
of 0.26 and performs the sam巳 amount of work. How much input heat
does the second engine require?
87. ssm A process occurs in which the entropy of a system increases
by 125 J/K. During the process , the energy that becomes unavailable
for doing work is zero. (a) Is this process reversible or irreversible?
Give your reasoning. (b) Determine the change in the entropy of the
surroundings .
88. See Multiple-Concept Example 10 to review the concepts that 盯巳
important in this problem . The water in a deep underground well is
used as the cold reservoir of a Carnot heat pump that maintains the
temperature of a house at 301 K. To d巳 posit 14200 J of heat in the
house , the heat pump requires 800 J of work. Determine the temp巳r
ature of the well water
89. A Carnot air conditioner maintains the temperature in a house at
297 K on a day when the temperature outside is 311 K. What is the
coefficient of performance of the air conditioner?
90. A Carnot engine has an efficiency of 0 .40. The Kelvin temperature of its hot reservoir is quadrupled , and the Kelv in temperature of
its cold reservoir is doubled. What is the 巳fficiency that results from
these changes?
9 1. ssm (a) Using the data
presented in the accompanying
pressure-versus-volume graph ,
estimate the magnitude of the
work done wh巳 n the syst巳m
changes 台om A to B to C along
the path shown. (b) Determine
whether the work is don巳 by the
system or on the system and ,
henc巳， whether the work is posItrve or negatrve.
*92. Refer to the drawing in Problem 12 , where th巳 curv巳 between A
and B is now an i soth巳 rm. An ideal gas begins at A and is changed
along the horizontalline from A to C and then along the vertical line
from C to B. (a) Find the heat for the process ACB and (b) determine whether it flows into or out of the gas
写 93.
ssm Suppo s巳 a monatomic ideal gas is contained within a vertical cylinder that is fitted with a movable piston . The piston is frictionless and has a negligible mass. The area of the piston is 3.14 X 10- 2 m2,
and the pressure outside the cylinder is 1.01 X lO s Pa. H巳 at (2093 J)
is removed from th巳 gas. Through what d i stanc巳 does the piston
drop?
;'94. An air conditioner ke巳ps the inside of a house at a temp巳rature
of 19.0 oC when the outdoor temp巳rature is 33.0 OC. Heat , leaking
into the hous巳 at the rat巳 of 10 500 joules p巳r second , is removed
by th巳 air conditioner. Assuming that the air conditioner is a Carnot
air conditioner, what is the work p巳 r second that must be done
by the electrical energy in order to keep the inside temperature
constant?
m
o
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k
273 K through a 0.35-m copper rod with a cross-sectional ar巳 a of
9 .4 X 10- 4 m2 (see the drawing). The heat then leaves the w273-K
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c u -tr a c k
reservoir and 巳 nters a Carnot engine , which uses part of this h巳atto
do work and r巳j 巳 cts the remainder to a third reservoir at 173 K.
How much of the heat l eavinσthe
373-K reservoir is r巳 ndered unb
available for doing work in a period of 2.0 min?
o
k
lic
C
mass m and speci 白 c heat capacity
c, the chang巳 in entropy
.c
c u -tr a c k
can be shown to be I:!. S =
mc In(Tf/Tj ) , where T j and T f
are th巳 initial and final Kelvin
t巳 mperatures. Using this expr巳ssion and the change in 巳n­
tropy for melting , find the
change in entropy that occurs
(b) Should the entropy of the
u l1lvers巳 incr巳ase or decreas巳
as a result of the mixing
process? Give your reasoning
and state whether your answer
in part (a) is consistent with
your answer here.
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ADDITIONAL PROBLEMS 471
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98. 吨，. Even at I 町， the human body ge n e刚es hea t. The heat arises
6.00
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2.00
4.00
6.00
8.00
10.0
12.0
Vo lume, m3
"96. Th巳 s un is a sphere wi th a radius of 6.96 x 108 m and an average s 1I rface t巳 mperature of 5800 K. Determine th巳 amollnt by which
the sun 's thermal radiation increases the 巳 ntropy of th巳 entlre 1I 111V巳 rse each second . Ass ume that th巳 slln is a perfect blackbody, and
that the average t巳 mp巳 rature of th巳 rest of the uni vers巳 i s 2.73 K. Do
not consider the therm a l radiation absorbed by the slln from th e
l 巳 s t of the lI niverse.
ij becalls巳 of the body 's m巳tabolism-that 岖 ， the chemical reactions that 缸 e always occ urrin g in the body to generate energy. ln
room s designed for use by large groups , adequate ve ntil ation 0 1' ail
conditioning mllst be provid巳d to remov巳 thi s hea t. Consider a c1 ass
ro0 1l1 containing 200 stlldents. As s llm 巳 that the llletabolic rate of gen
巳ratin g h巳at is 130 W for each stlldent and that the heat accumlllat巳S
dllring a fiftY-lllinut巳 lectllre. ln addition , ass llme that the air has a mo
lar specific heat of C v = ~R and 阳t the rOO Ill (volume = 1200 m3 ,
initial press ure = 1.01 X 105 Pa , and initial telllperature = 21 o c) is
sealed Shll t. lf all the heat g巳 nerat巳d by th巳 stud e nts were absorbed by
the air, by how lllllCh would the 创l' telllp巳ra ture ri se during a lecture?
"*99. ssm Engine A receives thre巳 tlm巳 s lllore inpllt heat , prodllces 自 ve
tillles more work , and rejects two times more heat than engin巳 B. Find
the efl白 ciency of (a) engine A and (b) 巳n g in 巳 B
常 1' 100.
The work don巳 b y one巳盯mo l 巳e of a lllona
川
创tω
a
川
阳
O
in expanding adiabaticaU飞ly 巴
i s 825 J. Th巳 initia
川
\1 忧t 巳 lllp巳 ra阳I 巳 and volullle of 由
t h巳 gas ar陀
巳 393 K and 0.100 m 3 • Obtain (a
创) 白
th巳白 n 创
a 1 忧t 巳 mp巳町r­
创tu
a
川l'巳 缸
an叫
d
(仙
b) 由
t h巳 final volume of the gas
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"97. Interactive 5olution 15.97 at www.wiley.comlcollege/cutnell
offers one approach to thi s problem. A lìfteen-watt h eat巳r is lI sed wto. d o
.c
c u -tr a c k
heat a monatomic ideal gas at a co nstant pl 巳 ss ure of 7.60 x 105 Pa.
During the process , the 1 .40 X 10- 3 m3 vo lume of the gas increases
by 25.0%. How long was the heater on?
w
"95. A monatomic ideal gas expands from point A to point B along th 巳
path shown
in the drawing. (a) Deterrnine the work done by the gas
.c
c u -tr a c k
(b) The temperature of the gas at point A is 185 K. Wh at is its temP巳 rature at point B? (c) How much heat has been added to or removed from the gas during the process?
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