Download 14.2 Temperature Change and Heat Capacity

OpenStax College Physics
Instructor Solutions Manual
Chapter 14
CHAPTER 14: HEAT AND HEAT TRANSFER
METHODS
14.2 TEMPERATURE CHANGE AND HEAT CAPACITY
1.
Solution
On a hot day, the temperature of an 80,000-L swimming pool increases by 1.50C .
What is the net heat transfer during this heating? Ignore any complications, such as
loss of water by evaporation.
1 m3
m  V  (1.00  10 kg/m )(80,000 L) 
 8.00  10 4 kg .
1000 L
3
3
Therefore, Q  mcT  (8.00  104 kg)(4186 J/kg  C)(1.50C)  5.02  108 J
2.
Show that 1 cal/g  C  1 kcal/kg  C .
Solution
1 cal
1 kcal 1000 g


 1 kcal/kg  C
g  C 1000 cal 1 kg
3.
To sterilize a 50.0-g glass baby bottle, we must raise its temperature from 22.0C to
95.0C . How much heat transfer is required?
Solution
Q  mcT  (50.0  10 -3 kg)( 840 J/kg  C)(73.0C)  3066 J  3.07  10 3 J
4.
The same heat transfer into identical masses of different substances produces
different temperature changes. Calculate the final temperature when 1.00 kcal of
heat transfers into 1.00 kg of the following, originally at 20.0C : (a) water; (b)
concrete; (c) steel; and (d) mercury.
Solution
Q  mcT  T 
(a) T 
Q
mc
1.00 kcal
 1.00 C  T  20.0 C  1.00 C  21.0 C
(1.00 kg)(1.00 kcal/kg  C)
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Chapter 14
(b) T 
1.00 kcal
 5.0C  T  20.0C  5.0C  25.0C
(1.00 kg)(0.20 kcal/kg  C)
(c) T 
1.00 kcal
 9.26C  T  20.0C  9.26C  29.3C
(1.00 kg)(0.108 kcal/kg  C)
(d) T 
1.00 kcal
 30.0C  T  20.0C  30.0C  50.0C
(1.00 kg)(0.0333 kcal/kg  C)
Rubbing your hands together warms them by converting work into thermal energy. If
a woman rubs her hands back and forth for a total of 20 rubs, at a distance of 7.50 cm
per rub, and with an average frictional force of 40.0 N, what is the temperature
increase? The mass of tissues warmed is only 0.100 kg, mostly in the palms and
fingers.
Solution Let N be the number of hand rubs and F be the average frictional force of a hand
NFd 20(40.0 N)(7.50  10 2 m)
rub: Q  NFd  mcT  T 

 0.171C
mc
(0.100 kg)(3500 J/kg  C)
6.
A 0.250-kg block of a pure material is heated from 20.0C to 65.0C by the addition
of 4.35 kJ of energy. Calculate its specific heat and identify the substance of which it is
most likely composed.
Solution
Q  mcT  c 
Q
1.04 kcal

 0.0924 kcal/kg  C
mT (0.250 kg)( 45.0C)
It is copper.
7.
Suppose identical amounts of heat transfer into different masses of copper and water,
causing identical changes in temperature. What is the ratio of the mass of copper to
water?
Solution
mw cw T  Q  mc cc T
mc cw
1 kcal/kg  C


 10.8
mw cc 0.0924 kcal/kg  C
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8.
Instructor Solutions Manual
Chapter 14
(a) The number of kilocalories in food is determined by calorimetry techniques in
which the food is burned and the amount of heat transfer is measured. How many
kilocalories per gram are there in a 5.00-g peanut if the energy from burning it is
transferred to 0.500 kg of water held in a 0.100-kg aluminum cup, causing a 54.9C
temperature increase? (b) Compare your answer to labeling information found on a
package of peanuts and comment on whether the values are consistent.
Solution (a) Q  mw cw T  mAlcAl T  mw cw  mAlcAl T
(0.500 kg)(1.00 kcal/kg  C)  
Q
 (54.9C)  28.63 kcal
(0.100 kg)( 0.215 kcal/kg  C) 
Q 28.63 kcal

 5.73 kcal/g
mp
5.00 g
(b) A label for unsalted dry roasted peanuts says that 33 g contains 200 calories (kcal),
Q 200 kcal
which is

 6 kcal/g , which is consistent with our results to part (a),
mp
33 g
to one significant figure.
9.
Solution
Following vigorous exercise, the body temperature of an 80.0-kg person is 40.0C . At
what rate in watts must the person transfer thermal energy to reduce the body
temperature to 37.0C in 30.0 min, assuming the body continues to produce energy
at the rate of 150 W? 1 watt = 1 joule/seco nd or 1 W = 1 J/s 
Q  mchuman bodyT  (80.0 kg)(3500 J/kg  C)(40C - 37C)  8.40  10 5 J
Pcooling 
Q
8.40  10 5 J

 4.67  10 2 W
t (30 min)(60 s/1 min)
Thus, Prequired  Pcooling  Pbody  467 W  150 W  617 W .
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10.
Even when shut down after a period of normal use, a large commercial nuclear
reactor transfers thermal energy at the rate of 150 MW by the radioactive decay of
fission products. This heat transfer causes a rapid increase in temperature if the
cooling system fails (1 watt = 1 joule/seco nd or 1 W = 1 J/s and 1 MW = 1 megawatt) .
(a) Calculate the rate of temperature increase in degrees Celsius per second (C/s ) if
the mass of the reactor core is 1.60  105 kg and it has an average specific heat of
0.3349 kJ/kg  C . (b) How long would it take to obtain a temperature increase of
2000C , which could cause some metals holding the radioactive materials to melt?
(The initial rate of temperature increase would be greater than that calculated here
because the heat transfer is concentrated in a smaller mass. Later, however, the
temperature increase would slow down because the 5  105 - kg steel containment
vessel would also begin to heat up.)
Solution
(a) Q  mcT  T 
Q
mc
Recall that 1 W = 1 J/s. Thus T for 1 s is given by
T 
(b) t 
(150  10 6 J)(1 kcal/4186J )
 2.80C  Rate  2.80C/s
(1.60  10 5 kg)(0.0800 kcal/kg  C)
2000C
 714 s  11.9 min
2.80C/s
14.3 PHASE CHANGE AND LATENT HEAT
11.
How much heat transfer (in kilocalories) is required to thaw a 0.450-kg package of
frozen vegetables originally at 0C if their heat of fusion is the same as that of
water?
Solution
Q  mLf  (0.450 kg)(79.8 kcal/kg)  35.9 kcal
12.
A bag containing 0C ice is much more effective in absorbing energy than one
containing the same amount of 0C water. (a) How much heat transfer is necessary
to raise the temperature of 0.800 kg of water from 0C to 30.0C ? (b) How much
heat transfer is required to first melt 0.800 kg of 0C ice and then raise its
temperature? (c) Explain how your answer supports the contention that the ice is
more effective.
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Solution
Instructor Solutions Manual
Chapter 14
(a) Q  mcT  (0.800 kg)(4186 J/kg  C)(30.0C)  1.00  10 5 J
(b) Q  mLf  mcT  (0.800 kg)(334 103 J/kg)  1.005 105 J  3.68 105 J
(c) The ice is much more effective in absorbing heat because it first must be melted,
which requires a lot of energy, then it gains the same amount of heat as the bag
that started with water. The first 2.67  10 5 J of heat is used to melt the ice, then
it absorbs the 1.00  10 5 J of heat as water.
13.
(a) How much heat transfer is required to raise the temperature of a 0.750-kg
aluminum pot containing 2.50 kg of water from 30.0C to the boiling point and then
boil away 0.750 kg of water? (b) How long does this take if the rate of heat transfer is
500 W [ 1 watt = 1 joule/seco nd (1 W = 1 J/s) ]?
Solution
'
(a) Q  mw cw T  mAlcAl T  mw Lv
Q  (2.50 kg)(1.00 kcal/kg  C)(70.0C)  (0.750 kg)(0.215 kcal/kg  C)(70.0C)
 (0.750 kg)(539 kcal/kg)
Q  590.5 kcal  591 kcal
Q
, where P  power and t  time .
P
 4186 J/kcal 
3
t  (590.5 kcal) 
  4.94  10 s
500
W


(b) Q  Pt  t 
14.
Solution
The formation of condensation on a glass of ice water causes the ice to melt faster
than it would otherwise. If 8.00 g of condensation forms on a glass containing both
water and 200 g of ice, how many grams of the ice will melt as a result? Assume no
other heat transfer occurs.
mw Lv  mice Lf  mice  mw
Lv
 580 kcal/kg 
  58.1 g
 (8.00 g) 
Lf
 79.8 kcal/kg 
(Note that Lv for water at 37C is used here as a better approximation than Lv for
100C water.)
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Chapter 14
15.
On a trip, you notice that a 3.50-kg bag of ice lasts an average of one day in your
cooler. What is the average power in watts entering the ice if it starts at 0C and
completely melts to 0C water in exactly one day [
1 watt = 1 joule/seco nd (1 W = 1 J/s) ]?
Solution
P
16.
On a certain dry sunny day, a swimming pool’s temperature would rise by 1.50C if
not for evaporation. What fraction of the water must evaporate to carry away
precisely enough energy to keep the temperature constant?
Solution
Let M be the mass of pool water and m be the mass of pool water that evaporates.
(3.50 kg)(334 kJ/kg  C)
Q mLf


 13.53 W  13.5 W
t
t
86400 s
McT  mLv(37C) 
m
cT
(1.00 kcal/kg  C)(1.50C)


 2.59 10 3
M Lv(37C)
580 kcal/kg
(Note that Lv for water at 37C is used here as a better approximation than Lv for
100C water.)
17.
(a) How much heat transfer is necessary to raise the temperature of a 0.200-kg piece
of ice from  20.0C to 130C , including the energy needed for phase changes? (b)
How much time is required for each stage, assuming a constant 20.0 kJ/s rate of heat
transfer? (c) Make a graph of temperature versus time for this process.
Solution
(a) (i) Heat needed to warm ice to 0C
Q1  mi ci T  (0.200 kg)(2.090 kJ/kg  C)(20C)  8.36 kJ
(ii) Heat needed to melt ice at 0C
Q2  mi Lf  (0.200 kg)(334 kJ/kg)  66.8 kJ
(iii) Heat required to warm 0C water to 100C
Q3  mi cw T  (0.200 kg)(4.186 kJ/kg o C)(100 o C)  83.73 kJ  83.7 kJ
(iv) Heat required to vaporize water at 100C
Q4  mi Lv  (0.200 kg)(2256 kJ/kg)  451.2 kJ  451 kJ
(v) Heat required to warm 100C vapor to 130C
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Q5  mi c v T  (0.200 kg)(1.520 kJ/kg  C)(30C)  9.12 kJ
Total heat required Q  Q1  Q2  Q3  Q4  Q5  619.0 kJ  147.9 kcal  148 kcal .
(b) P 
Q
Q
t 
t
P
(i) t1 
Q1 8.36 kJ

 0.418 s
P 20 kJ/s
(ii) t 2 
Q2 66.8 kJ

 3.34 s
P
20 kJ/s
(iii) t 3 
Q3 83.7 kJ

 4.185 s  4.19 s
P 20 kJ/s
(iv) t 4 
Q4
451 kJ

 22.6 s
P
20 kJ/s
(v) t5 
Q5 9.12 kJ

 0.456 s
P 20 kJ/s
Total time t  t1  t 2  t 3  t 4  t 5  31.00 s  31 s
(c)
18.
In 1986, a gargantuan iceberg broke away from the Ross Ice Shelf in Antarctica. It
was approximately a rectangle 160 km long, 40.0 km wide, and 250 m thick. (a) What
is the mass of this iceberg, given that the density of ice is 917 kg/m 3 ? (b) How much
heat transfer (in joules) is needed to melt it? (c) How many years would it take
sunlight alone to melt ice this thick, if the ice absorbs an average of 100 W/m 2 , 12.00
h per day?
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Solution
Instructor Solutions Manual
Chapter 14
3
3
3
(a) m  V   lwh   (917 kg/m )(160  10 m)(40.0  10 m)(250 m)
 1.467  1015 kg  1.47  1015 kg
(b) Q  mLf  (1.467  1015 kg)(79.8 kcal/kg)(4 186 J/kcal)  4.90  1020 J
3600 s 
16
(c) Qday  (100 W/m 2 )(160  103 m)(40.0  103 m)(12 h) 
  2.765  10 J
 1h 
17
Q
(1.171  10 kcal )(4186 J/kcal)
1y
n

 1.773  10 4 d 
 48.5 y
16
Qday
2.765  10 J
365.25 d
19.
How many grams of coffee must evaporate from 350 g of coffee in a 100-g glass cup to
cool the coffee from 95.0C to 45.0C ? You may assume the coffee has the same
thermal properties as water and that the average heat of vaporization is 2340 kJ/kg (560
cal/g). (You may neglect the change in mass of the coffee as it cools, which will give you an
answer that is slightly larger than correct.)
Solution
The heat gained in evaporating the coffee equals the heat leaving the coffee and
glass to lower its temperature, so that MLv  mc cc ΔT  mg cg ΔT , where M is the
mass of coffee that evaporates. Solving for the evaporated coffee gives:
M 

T (mc cc  mg cg )
Lv
(95.0C  45.0C)
 (350 g)(1.00 cal/g  C)  (100 g)(0.20 cal/g  C)  33.0 g
560 cal/g
20.
(a) It is difficult to extinguish a fire on a crude oil tanker, because each liter of crude oil
releases 2.80  10 7 J of energy when burned. To illustrate this difficulty, calculate the
number of liters of water that must be expended to absorb the energy released by burning
1.00 L of crude oil, if the water has its temperature raised from 20.0C to 100C , it boils,
and the resulting steam is raised to 300 C . (b) Discuss additional complications caused by
the fact that crude oil has a smaller density than water.
Solution
(a) Q  mcw Tw  mLv  mcs Ts
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m

Instructor Solutions Manual
Chapter 14
Q
c w Tw  Lv  cs Ts
2.80  10 7 J
 9.67 kg
(4186 J/kg  C)(80.0C)  2256  10 3 J/kg  (1520 J/kg  C)(200C)
 V  9.67 kg 
1 m3
1L
 3 3  9.67 L
3
1.00  10 kg 10 m
(b) Crude oil is less dense than water, so it floats on top of the water, thereby
exposing it to the oxygen in the air, which it uses to burn. Also, if the water is
under the oil, it is less able to absorb the heat generated by the oil.
21.
The energy released from condensation in thunderstorms can be very large. Calculate the
energy released into the atmosphere for a small storm of radius 1 km, assuming that 1.0
cm of rain is precipitated uniformly over this area.
Solution
We have a phase change Q  mLv . We need to find mass of rain in a cloud of radius
1 km. m  V  (1000 kg/m 3 )(0.01 m)(   106 m 2 )    107 kg . With Q  mLv and
Lv  2256 kJ/kg , we find Q  7  1013 J – about the energy released in the first
atomic bomb explosion.
22.
To help prevent frost damage, 4.00 kg of 0C water is sprayed onto a fruit tree. (a) How
much heat transfer occurs as the water freezes? (b) How much would the temperature of
the 200-kg tree decrease if this amount of heat transferred from the tree? Take the
specific heat to be 3.35 kJ/kg  C , and assume that no phase change occurs.
Solution
(a) Q  mLf  (4.00 kg)(79.8 kcal/kg  C)  319.2 kcal  319 kcal
(b) Q  mcT  T 
Q
319.2 kcal

 2.00C
mc (200 kg)(0.800 kcal/kg  C)
23.
A 0.250-kg aluminum bowl holding 0.800 kg of soup at 25.0C is placed in a freezer.
What is the final temperature if 377 kJ of energy is transferred from the bowl and
soup, assuming the soup’s thermal properties are the same as that of water? Explicitly
show how you follow the steps in the Problem-Solving Strategies for the Effects of Heat
Transfer.
Solution
To bring the system to 0C requires heat, Q , of:
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Q  mAl cAl T  ms cs T
 (0.250 kg)(0.215 kcal/kg  C)  (0.800 kg)(1.00 kcal/kg  C)(25.0C)  21.34 kcal
This leaves (90.0  21.34) kcal  68.66 kcal to freeze all the soup, leaving
Q' '  (68.66  63.84) kcal  4.82 kcal to be removed. So, we can now determine the
final temperature of the frozen soup:
Q"  (mAlcAl  ms cs )T  (mAlcAl  ms cs )( 0C-Tf ).
Tf 

 Q"
mAlcAl  ms cs
 4.82 kcal
  10.6C
(0.250 kg)(0.215 kcal/kg  C)  (0.800 kg)(0.500 kcal/kg  C)
24.
A 0.0500-kg ice cube at  30.0C is placed in 0.400 kg of 35.0C water in a very
well-insulated container. What is the final temperature?
Solution
First bring the ice up to 0C and melt it with heat Q1 :
Q1  mcT1  mLf  (0.0500 kg) 0.500 kcal/kg  C(30.0C)  (79.8 kcal/kg) 
 4.74 kcal
This lowers the temperature of water by T2 :
Q1
4.74 kcal

 11.85C
mc (0.400 kg)(1.00 kcal/kg  C)
New Tw  35.0C  11.85C  23.15C
Q1  mcT2  T2 
Now, the heat lost by the hot water equals that gained by the cold water ( Tf is the
final temperature):
mc c w (Tf  Tc )  mh c w (Th  Tf )
Tf 
mh Th  mcTc (0.400 kg)(296.3 K)  (0.0500 kg)(273.15 K)

 293.7 K  20.6C
mc  m h
0.450 kg
25.
If you pour 0.0100 kg of 20.0C water onto a 1.20-kg block of ice (which is initially at
15.0C ), what is the final temperature? You may assume that the water cools so
rapidly that effects of the surroundings are negligible.
Solution
First, we need to calculate how much heat would be required to raise the
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temperature of the ice to 0C :
Qice  mcT  (1.20 kg)(2090 J/kg  C)(15C)  3.762 10 4 J
Now, we need to calculate how much heat is given off to lower the water to 0C :
Q1  mcT1  (0.0100 kg)(4186 J/kg  C)(20.0C)  837.2 J
Since this is less than the heat required to heat the ice, we need to calculate how
much heat is given off to convert the water to ice:
Q2  mLf  (0.0100 kg)(334 103 J/kg)  3.340 103 J
Thus, the total amount of heat given off to turn the water to ice at 0C :
Qwater  4.177 10 3 J .
Since Qice  Qwater , we have determined that the final state of the water/ice is ice at
some temperature below 0C . Now, we need to calculate the final temperature. We
set the heat lost from the water equal to the heat gained by the ice, where we now
know that the final state is ice at Tf  0C :
Qlost by water  Qgained by ice, or mwater c water T200  mwater Lf  mwater cice T0?  mice cice T15?
Substituting for the change in temperatures (being careful that T is always
positive) and simplifying gives:
mwater [cwater (20C)  Lf  (cice )(0  Tf )]  micecice[Tf  (15C)].
Solving for the final temperature gives Tf 
mwater [cwater (20C)  Lf ]  mice cice (15C)
(mwater  mice )cice
and so finally,
(0.0100 kg)[(4186 J/kg  C)(20C)  334  103 J/kg]
Tf 
(0.0100 kg  1.20 kg)(2090 J/kg  C)
(1.20 kg)(2090 J/kg  C)(15C)

(0.0100 kg  1.20 kg)(2090 J/kg  C)
  13.2C
26.
Indigenous people sometimes cook in watertight baskets by placing hot rocks into
water to bring it to a boil. What mass of 500C rock must be placed in 4.00 kg of
15.0C water to bring its temperature to 100C , if 0.0250 kg of water escapes as
vapor from the initial sizzle? You may neglect the effects of the surroundings and take
the average specific heat of the rocks to be that of granite.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
Let the subscripts r, e, v, and w represent rock, equilibrium, vapor, and water,
respectively.
mr cr (T1  Te )  mv Lv  mw cw (Te  T2 )
mr 
mv Lv  mw cw (Te  T2 )
cr (T1  Te )
(0.0250 kg)(2256 10 3 J/kg)  (3.975 kg)(4186 J/kg  C)(100C  15C)
(840 J/kg  C)(500C  100C)
 4.38 kg

27.
What would be the final temperature of the pan and water in Calculating the Final
Temperature When Heat Is Transferred Between Two Bodies: Pouring Cold Water in a
Hot Pan if 0.260 kg of water was placed in the pan and 0.0100 kg of the water
evaporated immediately, leaving the remainder to come to a common temperature
with the pan?
Solution
Let the subscripts Al, e, v, and w represent aluminum pan, equilibrium, vapor, and
water, respectively.
mAlcAl (T1  Te )  mv Lv  mw cw (Te  T2 ) 
Te 
mAl c AlT1  m w c w T2  m v Lv
m w c w  mAl c Al
 (0.500 kg)(900 J/kg  C)(150C)  (0.250 kg)(4186 J/kg  C)(20.0C) 


 (0.0100 kg)(2256  10 3 J/kg)


Te 


(0.250 kg)(4186 J/kg  C)  (0.500 kg)(900 J/kg  C)




 44.0C
28.
In some countries, liquid nitrogen is used on dairy trucks instead of mechanical
refrigerators. A 3.00-hour delivery trip requires 200 L of liquid nitrogen, which has a
density of 808 kg/m 3 . (a) Calculate the heat transfer necessary to evaporate this
amount of liquid nitrogen and raise its temperature to 3.00C . (Use cp and assume
it is constant over the temperature range.) This value is the amount of cooling the
liquid nitrogen supplies. (b) What is this heat transfer rate in kilowatt-hours? (c)
Compare the amount of cooling obtained from melting an identical mass of 0C ice
with that from evaporating the liquid nitrogen.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
(a) Q  mLv  mcp T  mLv  c p T 
48.0 kcal/kg  (0.248 kcal/kg  C) 
Q  (200  10 3 m 3 )(808 kg/m 3 )

3.00C  (195.8C)

4
 1.57  10 kcal
 1 kW  h
(b) (1.572  10 4 kcal)(4186 J/kcal) 
6
 3.60  10

  18.28 kW  h  18.3 kW  h
J
(c) Qice  mLf  (161.6 kg)(79.8 kcal/kg)  12,895 kcal  1.29  10 4 kcal
29.
Solution
Some gun fanciers make their own bullets, which involves melting and casting the
lead slugs. How much heat transfer is needed to raise the temperature and melt
0.500 kg of lead, starting from 25.0C ?
Q  mcT  mLf  mcT  Lf 
Q  (0.500 kg)  0.0305 kcal/kg  C 327C  25.0C   5.85 kcal/kg   7.53 kcal
14.5 CONDUCTION
30.
(a) Calculate the rate of heat conduction through house walls that are 13.0 cm thick
and that have an average thermal conductivity twice that of glass wool. Assume there
are no windows or doors. The surface area of the walls is 120 m 2 and their inside
surface is at 18.0C , while their outside surface is at 5.00C . (b) How many 1-kW
room heaters would be needed to balance the heat transfer due to conduction?
Solution (a) Q kAT2  T1  2(0.042 J/s  m  C)(120 m 2 )(18.0 C  5.00C)


t
d
0.130 m
3
3
 1.008  10 W  1.01  10 W
(b) 1 one-kilowatt room heater is needed.
31.
The rate of heat conduction out of a window on a winter day is rapid enough to chill
the air next to it. To see just how rapidly the windows transfer heat by conduction,
2
calculate the rate of conduction in watts through a 3.00 - m window that is
0.635 cm thick (1/4 in) if the temperatures of the inner and outer surfaces are
5.00C and 10.0C , respectively. This rapid rate will not be maintained—the inner
surface will cool, and even result in frost formation.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
Q kAT2  T1 

t
d
(0.84 J/s  m  C)(3.00 m 2 )5.00C  (10.0C)

 5953 W  6.0  103 W
-2
0.635  10 m
32.
Calculate the rate of heat conduction out of the human body, assuming that the core
internal temperature is 37.0C , the skin temperature is 34.0C , the thickness of the
tissues between averages 1.00 cm , and the surface area is 1.40 m 2 .
Solution
Q kAT2  T1  (0.2 J/s  m  C)(1.40 m 2 )(37.0C  34.0C)


 84.0 W
t
d
0.0100 m
33.
Suppose you stand with one foot on ceramic flooring and one foot on a wool carpet,
making contact over an area of 80.0 cm 2 with each foot. Both the ceramic and the
carpet are 2.00 cm thick and are 10.0C on their bottom sides. At what rate must
heat transfer occur from each foot to keep the top of the ceramic and carpet at
33.0C ?
Solution
Q kAT2  T1 

t
d
For the wool carpet:
Qw (0.04 J/s  m  C)(80.0  10 -4 m 2 )(33.0C - 10.0C)

 0.368 W
t
0.0200 m
For the ceramic tile:
Qc (0.84 J/s  m  C)(80.0  10 -4 m 2 )( 23C)

 7.73 W
t
0.0200 m
34.
A man consumes 3000 kcal of food in one day, converting most of it to maintain body
temperature. If he loses half this energy by evaporating water (through breathing and
sweating), how many kilograms of water evaporate?
Solution
Q  mLv(37C)  m 
35.
Q
Lv(37C)

1500 kcal
 2.59 kg
580 kcal/kg
(a) A firewalker runs across a bed of hot coals without sustaining burns. Calculate the
OpenStax College Physics
Instructor Solutions Manual
Chapter 14
heat transferred by conduction into the sole of one foot of a firewalker given that the
bottom of the foot is a 3.00-mm-thick callus with a conductivity at the low end of the
range for wood and its density is 300 kg/m 3 . The area of contact is 25.0 cm 2 , the
temperature of the coals is 700C , and the time in contact is 1.00 s. (b) What
temperature increase is produced in the 25.0 cm3 of tissue affected? (c) What effect
do you think this will have on the tissue, keeping in mind that a callus is made of dead
cells?
Solution (a)
kAT2  T1 t
Q
d
(0.0800 J/s  m  C)(25.0  10 -4 m 2 )(700C - 37.0C)(1.00s)

 44.2 J
0.00300 m
(b) Taking the density of the callus to be   300 kg/m 3 , the change in temperature
can be found from:
Q  mcT  VcT 
T 
Q
41.7 J

 1.68C
3
Vc (300 kg/m )(25.0  10 6 m 3 )(3500 J/kg  C)
(c) At a temperature change of  2C , the heat probably won’t do much damage,
since a callus is made of dead cells.
36.
Solution
(a) What is the rate of heat conduction through the 3.00-cm-thick fur of a large
2
animal having a 1.40 - m surface area? Assume that the animal’s skin temperature is
32.0C , that the air temperature is  5.00C , and that fur has the same thermal
conductivity as air. (b) What food intake will the animal need in one day to replace
this heat transfer?
(a)
Q kAT2  T1  (0.023 J/s  m  C)(1.40 m 2 )(37.0C)


 39.71 W  39.7 W
t
d
0.0300 m
 1 kcal 
(b) W  Pt  (39.71 J/s)(8.64  10 4 s)
  819.7 kcal  820 kcal
 4186 J 
37.
A walrus transfers energy by conduction through its blubber at the rate of 150 W
when immersed in 1.00C water. The walrus’s internal core temperature is 37.0C ,
and it has a surface area of 2.00 m 2 . What is the average thickness of its blubber,
which has the conductivity of fatty tissues without blood?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
Q kAT2  T1 


t
d
kAT2  T1  (0.2 J/s  m  C)(2.00 m 2 )(38.0 C)
d

 0.101 m  10.1 cm
Q/t
150 W
38.
Compare the rate of heat conduction through a 13.0-cm-thick wall that has an area of
10.0 m 2 and a thermal conductivity twice that of glass wool with the rate of heat
conduction through a window that is 0.750 cm thick and that has an area of 2.00 m 2 ,
assuming the same temperature difference across each.
Solution
Q kA(T2  T1 )

t
d
, so that
(Q / t ) wall
k A d
(2  0.042 J/s  m  C)(10.0 m 2 )(0.750  10 2 m)
 wall wall window 
(Q / t ) window k window Awindowd wall
(0.84 J/s  m  C)(2.00 m 2 )(13.0  10 2 m)
 0.0288 wall : window, or 35 :1 window : wall
39.
Suppose a person is covered head to foot by wool clothing with average thickness of
2.00 cm and is transferring energy by conduction through the clothing at the rate of
50.0 W. What is the temperature difference across the clothing, given the surface
area is 1.40 m 2 ?
Solution
Q kA(T2  T1 ) kAT



t
d
d
d (Q / t )
(2.00  10 -2 m)(50.0 W)
T 

 17.86C  17.9C
kA
(0.04 J/s  m  C)(1.40 m 2 )
40.
Some stove tops are smooth ceramic for easy cleaning. If the ceramic is 0.600 cm thick
and heat conduction occurs through the same area and at the same rate as computed
in Example 14.6, what is the temperature difference across it? Ceramic has the same
thermal conductivity as glass and brick.
Solution
Q kA(T2  T1 ) kAT



t
d
d
d (Q / t )
(6.00  10 -3 m)(2256 W)
T 

 1046 C  1.05  10 3 K
-2
2
kA
(0.84 J/s  m  C)(1.54  10 m )
OpenStax College Physics
41.
Solution
Instructor Solutions Manual
Chapter 14
One easy way to reduce heating (and cooling) costs is to add extra insulation in the
attic of a house. Suppose the house already had 15 cm of fiberglass insulation in the
attic and in all the exterior surfaces. If you added an extra 8.0 cm of fiberglass to the
attic, then by what percentage would the heating cost of the house drop? Take the
single story house to be of dimensions 10 m by 15 m by 3.0 m. Ignore air infiltration
and heat loss through windows and doors.
Q kAT
. We need to consider all

t
d
6 sides that contribute to the heat loss. We will put the loss through the attic in a
separate part.
The original heat loss by conduction is given by
Q (0.042 J/s/m/ o C)(10 m  3 m  2)  (15 m  3 m  2)  (15 m  10 m) T

t
0.15 m
o
0.042 J/s/m/ C(10 m  15 m) T

0.15 m
 (84 J/s/ C  42 J/s/ o C)T  (126 J/s/ o C)T
If we add 8 cm to the attic, the new addition is
[84 J/s/ o C  (0.042 J/s/m/ o C 150 m 2 )/0.23 m] T  (84 J/s/ o C  27 J/s/ o C)T


 111 J/s/ o C T
So the percentage of savings in heat transfer = (126  111) / 126  12% .
42.
(a) Calculate the rate of heat conduction through a double-paned window that has a
1.50 - m 2 area and is made of two panes of 0.800-cm-thick glass separated by a 1.00cm air gap. The inside surface temperature is 15.0C , while that on the outside is
10.0C . (Hint: There are identical temperature drops across the two glass panes.
First find these and then the temperature drop across the air gap. This problem
ignores the increased heat transfer in the air gap due to convection.) (b) Calculate the
rate of heat conduction through a 1.60-cm-thick window of the same area and with
the same temperatures. Compare your answer with that for part (a).
OpenStax College Physics
Solution
Q
Q1
Instructor Solutions Manual
Q3
Q2
TL T1
d1
T2 TR
d2
d1
In equilibrium, the heat flows across each “slab” are equal.
Q1 Q2 Q3


and T1  TL  TR  T2
t
t
t
(a) a1  K1 , a 2  K 2 , a3  K 3  a1
d1
d2
d3
Q1  Q2 per unit time  a1 A(T1  TL )  a 2 A(T2  T1 )
Q2  Q3 per unit time  a 2 A(T2  T1 )  a3 A(TR  T2 )  a1 A(TR  T2 )
(since a1  a3 )
a1T1  a1TL  a 2T2  a 2T1
a 2T2  a 2T1  a1TR  a1T2
Adding two equations, we obtain:
 a1

(TR  TL )
a1 (TR  TL )  a1 (T1  T2 )  2a 2 (T2  T1 )  T2  T1  
2
a

a
2
1


Q2
a1 a 2 A(TR  TL )
 a 2 A(T2  T1 ) 
t
2a 2  a1
Now, we have
K 1 0.84 J/s  m  C

 105 J/s  m 2  C
d1
0.800  10 -2 m
K
0.023 J/s  m  C
a2  2 
 2.3 J/s  m 2  C
-2
d2
1.00  10 m
TR  15.0C; TL  10.0C . Thus
a1 
Chapter 14
OpenStax College Physics
Instructor Solutions Manual


Chapter 14

Q2 a1a2 A(TR  TL ) 105 J/s  m 2  C 2.3 J/s  m 2  C (1.50 m 2 )25.0C


t
2a2  a1
2(2.3 J/s  m 2  C)  105 J/s  m 2  C
 82.6 W
Since
(b)
Q2 Q1 Q3 Q
Q


   82.6 W  83 W
t
t
t
t
t
Q kA(TR  TL ) (0.84 J/s  m  C)(1.50 m 2 )(25.0C)


 1969 W  1.97  10 3 W
-2
t
d
1.60  10 m
The single-pane window has a rate of heat conduction equal to 1969/83, or 24
times that of a double pane window.
43.
Solution
Many decisions are made on the basis of the payback period: the time it will take
through savings to equal the capital cost of an investment. Acceptable payback times
depend upon the business or philosophy one has. (For some industries, a payback
period is as small as two years.) Suppose you wish to install the extra insulation in
Problem 14.41. If energy cost $1.00 per million joules and the insulation was $4.00
per square meter, then calculate the simple payback time. Take the average T for
the 120 day heating season to be 15.0C .
Q
 126T J/s  C as baseline energy use. So the
t
total heat loss during this period is
Q  (126J/s  C)(15.0C)(120 days) (86.4  103 s/day)  1960  106 J .
We found in Problem 14.41 that
At the cost of $1/MJ, the cost is $1960. From Problem 14.41, the savings is 12% or
$235 / y . We need 150 m 2 of insulation in the attic. At $4 / m 2 this is a $600 cost. So
the payback period is $600 / $235 / year  2.6 years (excluding labor cost) .
44.
For the human body, what is the rate of heat transfer by conduction through the
body’s tissue with the following conditions: the tissue thickness is 3.00 cm, the change
in temperature is 2.00C , and the skin area is 1.50 m 2 . How does this compare with
the average heat transfer rate to the body resulting from an energy intake of about
2400 kcal per day? (No exercise is included.)
Solution
The rate of heat transfer by conduction is
OpenStax College Physics
Instructor Solutions Manual

Chapter 14

Q kAT (0.2 J/s  m o C) 1.50 m 2 2.00 C


 20.0 W.
t
d
3.00  10 2 m
On a daily basis, this is 1,728 kJ/day.
Daily food intake is 2400 kcal/d  4186 J/kcal  10,050 kJ/day.
So only 17.2% of energy intake goes as heat transfer by conduction to the
environment at this T .
14.6 CONVECTION
45.
At what wind speed does 10C air cause the same chill factor as still air at  29C ?
Solution 10 m/s (from Table 14.4)
46.
At what temperature does still air cause the same chill factor as  5C air moving at
15 m/s?
Solution
 26C (from Table 14.4)
47.
The “steam” above a freshly made cup of instant coffee is really water vapor droplets
condensing after evaporating from the hot coffee. What is the final temperature of
250 g of hot coffee initially at 90.0°C if 2.00 g evaporates from it? The coffee is in a
Styrofoam cup, so other methods of heat transfer can be neglected.
Solution Let M be the mass of coffee that is left after evaporation and m be the mass of
coffee that evaporates.
mLv
(2.00 g)(539 kcal/kg)

 4.35C
Mc (248 g)(1.00 kcal/kg  C)
Tf  Ti  T  90.0C  4.35C  85.7C
McT  mLv  T 
48.
(a) How many kilograms of water must evaporate from a 60.0-kg woman to lower her
body temperature by 0.750°C ? (b) Is this a reasonable amount of water to evaporate
in the form of perspiration, assuming the relative humidity of the surrounding air is
low?
OpenStax College Physics
Instructor Solutions Manual
Chapter 14
Solution (a) M is the mass of the woman and m is the mass of water that evaporates:
McT  mLv(37C) 
m
McT (60.0 kg)(0.83 kcal/kg  C)(0.750C)

 6.44  10 2 kg
Lv(37C)
580 kcal/kg
(b) Yes, 64.4 g of water is reasonable. If the air is very dry, the sweat may evaporate
without even being noticed.
49.
Solution
On a hot dry day, evaporation from a lake has just enough heat transfer to balance
the 1.00 kW/m 2 of incoming heat from the Sun. What mass of water evaporates in
1.00 h from each square meter? Explicitly show how you follow the steps in the
Problem-Solving Strategies for the Effects of Heat Transfer.
A  1 m 2 ,  P  1.00 kW  1.00  103 W , so
Pt  Q  mLv(37C)  m 
Pt
Lv(37C)
(1.00  103 W)(3600 s)

 1.48 kg
2430  103 J/kg
(Note that we can use the Lv value at 37C as a closer approximation of the
temperature on a hot day than 100C .)
50.
Solution
51.
One winter day, the climate control system of a large university classroom building
malfunctions. As a result, 500 m 3 of excess cold air is brought in each minute. At what
rate in kilowatts must heat transfer occur to warm this air by 10.0°C (that is, to bring
the air to room temperature)?



Q mcT VcT 1.29 kg/m 3 500 m 3 721 J/kg  C 10.0C 



t
t
t
60.0 s
4
 7.75 10 W  77.5 kW.
P
The Kilauea volcano in Hawaii is the world’s most active, disgorging about 5 105 m3
of 1200°C lava per day. What is the rate of heat transfer out of Earth by convection if
this lava has a density of 2700 kg/m 3 and eventually cools to 30°C ? Assume that the
specific heat of lava is the same as that of granite.
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
Q mcT VcT


t
t
t
For c, we use the specific heat of granite, which is formerly molten rock.



Q
2700 kg/m 3 5  10 5 m 3 840 J/kg  C 1170C 

t
8.64  10 4 s
 1.54  1010 W  2  1010 W  2  10 4 MW
52.
During heavy exercise, the body pumps 2.00 L of blood per minute to the surface,
where it is cooled by 2.00°C . What is the rate of heat transfer from this forced
convection alone, assuming blood has the same specific heat as water and its density
is 1050 kg/m 3 ?
Solution
Q mcT VcT


t
t
t
3
1050 kg/m 2.00  10 -3 m 3 4186 J/kg  C2.00C

 293 W
60 s

53.


A person inhales and exhales 2.00 L of 37.0°C air, evaporating 4.00  102 g of water
from the lungs and breathing passages with each breath. (a) How much heat transfer
occurs due to evaporation in each breath? (b) What is the rate of heat transfer in
watts if the person is breathing at a moderate rate of 18.0 breaths per minute? (c) If
the inhaled air had a temperature of 20.0C , what is the rate of heat transfer for
warming the air? (d) Discuss the total rate of heat transfer as it relates to typical
metabolic rates. Will this breathing be a major form of heat transfer for this person?
Solution (a) Q  mLv(37C)  (4.00 10-5 kg)(2430 103 J/kg)  97.2 J
(b) P 
NQ 18(97.2 J)

 29.2 W
t
60.0 s
(c) Q  mcT  VcT
 (1.29 kg/m 3 )(2.00  10 3 m 3 )(721 J/kg  C)(37.0C  20.0C)
 31.6 J/breath , so the rate of heat loss is :
P
NQ 18(31.6 J)

 9.49 W
t
60.0 s
OpenStax College Physics
Instructor Solutions Manual
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(d) The total rate of heat loss would be 29.2 W + 9.49 W = 38.7 W. While sleeping, our
body consumes 83 W of power, while sitting it ranges 120-210 W. Therefore, the
total rate of heat loss from breathing will not be a major form of heat loss for this
person.
54.
A glass coffee pot has a circular bottom with a 9.00-cm diameter in contact with a
heating element that keeps the coffee warm with a continuous heat transfer rate of
50.0 W. (a) What is the temperature of the bottom of the pot, if it is 3.00 mm thick
and the inside temperature is 60.0°C ? (b) If the temperature of the coffee remains
constant and all of the heat transfer is removed by evaporation, how many grams per
minute evaporate? Take the heat of vaporization to be 2340 kJ/kg.
Solution (a) Q kAT
(Q / t )d
(50.0 W)(3.00  10 3 m)

 T 

 28.07C
t
d
kr 2
(0.84 J/s  m  C) (4.50  10 2 m) 2
T  Ti  T  T  88.07C  88C
(b) Pt  mLv  m 
Pt (50.0 W)(60.0 s)

 1.28 g
Lv
2340J/g
14.7 RADIATION
55.
At what net rate does heat radiate from a 275 - m 2 black roof on a night when the
roof’s temperature is 30.0C and the surrounding temperature is 15.0C ? The
emissivity of the roof is 0.900.
Solution
Q
4
4
 eA(T24  T14 )  (5.67  10 8 J/s  m 2  K 4 )(0.900)(2 75 m 2 ) 288 K   303 K 
t
  21.7 kW


Note that the negative answer implies heat loss to the surroundings.
56.
(a) Cherry-red embers in a fireplace are at 850C and have an exposed area of
0.200 m 2 and an emissivity of 0.980. The surrounding room has a temperature of
18.0C . If 50% of the radiant energy enters the room, what is the net rate of radiant
heat transfer in kilowatts? (b) Does your answer support the contention that most of
the heat transfer into a room by a fireplace comes from infrared radiation?
OpenStax College Physics
Instructor Solutions Manual
Chapter 14
Solution (a) Q 1
 eA(T24  T14 )
t 2
1
4
4
 (5.67  10 8 J/s  m 2  K 4 )(0.980)(0 .200 m 2 ) 291 K   1123 K 
2
  8.80 kW


Note that the negative answer implies heat loss to the surroundings.
(b) This answer is quite large, so it does indeed suggest that the heat put into a room
by a fireplace comes mainly from infrared radiation (which is hotter than red
embers).
57.
Radiation makes it impossible to stand close to a hot lava flow. Calculate the rate of
heat transfer by radiation from 1.00 m 2 of 1200C fresh lava into 30.0C
surroundings, assuming lava’s emissivity is 1.00.
Solution
Q
4
4
 eA(T24  T14 )  (5.67  10 8 J/s  m 2  K 4 )(1.00)(1. 00 m 2 ) 303 K   1473 K 
t
  266 kW
58.
(a) Calculate the rate of heat transfer by radiation from a car radiator at 110°C into a
2
50.0°C environment, if the radiator has an emissivity of 0.750 and a 1.20 - m
surface area. (b) Is this a significant fraction of the heat transfer by an automobile
engine? To answer this, assume a horsepower of 200 hp (1.5 kW) and the efficiency of
automobile engines as 25% .
Solution

(a)

Q
 eA(T24  T14 )
t
4
4
 (5.67  10 8 J/s  m 2  K 4 )(0.750)(1 .20 m 2 ) 323 K   383 K    543 W


(b) Assuming an automobile engine is 200 horsepower and the efficiency of a gasoline
200 horsepower
 800 horsepower
engine is 25%, the engine consumes
25%
Therefore, 600 horsepower is lost due to heating. The radiator transfers
1 hp
543 W 
 0.728 hp from radiation, which is not a significant fraction
746 W
because the heat is primarily transferred from the radiator by other means.
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Chapter 14
59.
Find the net rate of heat transfer by radiation from a skier standing in the shade,
given the following. She is completely clothed in white (head to foot, including a ski
mask), the clothes have an emissivity of 0.200 and a surface temperature of 10.0C ,
the surroundings are at 15.0°C , and her surface area is 1.60 m 2 .
Solution
Q
 eA(T24  T14 )
t
4
4
 (5.67  10 8 J/s  m 2  K 4 )(0.200)(1 .60 m 2 ) 258 K   283 K    36.0 W

60.
Solution

Suppose you walk into a sauna that has an ambient temperature of 50.0°C . (a)
Calculate the rate of heat transfer to you by radiation given your skin temperature is
37.0°C , the emissivity of skin is 0.98, and the surface area of your body is 1.50 m 2 .
(b) If all other forms of heat transfer are balanced (the net heat transfer is zero), at
what rate will your body temperature increase if your mass is 75.0 kg?
(a)
Q
 eA(T24  T14 )
t
4
4
 (5.67  10 8 J/s  m 2  K 4 )(0.98)(1. 50 m 2 ) 323 K   310 K   137 W

(b)

Q  mcT 
T Q 1
(137 W)


 5.24  10 4 C/s  0.0314 C/min
t
t mc (75.0 kg)(3500 J/kg  C)
61.
Thermography is a technique for measuring radiant heat and detecting variations in
surface temperatures that may be medically, environmentally, or militarily
meaningful.(a) What is the percent increase in the rate of heat transfer by radiation
from a given area at a temperature of 34.0°C compared with that at 33.0°C , such
as on a person’s skin? (b) What is the percent increase in the rate of heat transfer by
radiation from a given area at a temperature of 34.0°C compared with that at
20.0°C , such as for warm and cool automobile hoods?
Solution
 307 K  4 
(a) 
  1  100%  1.31%

 306 K 
 307 K  4 
(b) 
  1  100%  20.5%

 293 K 
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62.
The Sun radiates like a perfect black body with an emissivity of exactly 1. (a) Calculate
the surface temperature of the Sun, given that it is a sphere with a 7.00  108 - m
radius that radiates 3.80  10 26 W into 3-K space. (b) How much power does the Sun
radiate per square meter of its surface? (c) How much power in watts per square
meter is that value at the distance of Earth, 1.50  1011 m away? (This number is called
the solar constant.)
Solution
 Q/t
Q
 eAT 4  T  
(a)
2
t
 e( 4r


)
1/ 4
(the surface area of a sphere is 4r 2 )


3.80  10 26 W
T 
8
4
8
2 
 (5.67  10 J/s  m  K )(4 )(7.00  10 m) 
1/4
 5.74  10 3 K
P
P
3.80  10 26 W


 6.17  10 7 W/m 2
(b)
2
8
2
A 4r
4 (7.00  10 m)
(c) Let r be the radius of a sphere with the sun at the center and the earth at a point
on the surface of the sphere.
P
3.80  10 26 W

 1.34  10 3 W/m 2
2
11
2
4r
4 (1.50  10 m)
63.
Solution
A large body of lava from a volcano has stopped flowing and is slowly cooling. The
interior of the lava is at 1200°C , its surface is at 450°C , and the surroundings are at
27.0°C . (a) Calculate the rate at which energy is transferred by radiation from
1.00 m 2 of surface lava into the surroundings, assuming the emissivity is 1.00. (b)
Suppose heat conduction to the surface occurs at the same rate. What is the thickness
of the lava between the 450°C surface and the 1200°C interior, assuming that the
lava’s conductivity is the same as that of brick?
(a)
Q
 eA(T24  T14 )
t
Q
4
4
 (5.67  10 8 J/s  m 2  K 4 )(1.00)(1. 00 m 2 ) 300 K   723 K 
t
 1.503  10 4 W  1.50  10 4 W   15.0 kW


Note the negative answer implies heat lost to the surroundings.
OpenStax College Physics
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Chapter 14
(b) Q kAT


t
d
kAT (0.84 J/s  m  C)(1.00 m 2 )(750 C)
d

 4.19  10  2 m  4.2 cm
4
Q/t
1.503  10 W
64.
Calculate the temperature the entire sky would have to be in order to transfer energy
by radiation at 1000 W/m 2 —about the rate at which the Sun radiates when it is
directly overhead on a clear day. This value is the effective temperature of the sky, a
kind of average that takes account of the fact that the Sun occupies only a small part
of the sky but is much hotter than the rest. Assume that the body receiving the energy
has a temperature of 27.0°C .
Solution
Q
Q/t 

 eA(T24  T14 )  T1   T24 

t
eA 

Q/t
 1000 W/m 2
A
1/ 4


 1000 W/m 2
4
 T1  (300 K) 

8
2
4
(5.67  10 J/s  m  K )(1.00) 

1/4
 401 K
65.
(a) A shirtless rider under a circus tent feels the heat radiating from the sunlit portion
of the tent. Calculate the temperature of the tent canvas based on the following
information: The shirtless rider’s skin temperature is 34.0°C and has an emissivity of
0.970. The exposed area of skin is 0.400 m 2 . He receives radiation at the rate of 20.0
W—half what you would calculate if the entire region behind him was hot. The rest of
the surroundings are at 34.0°C . (b) Discuss how this situation would change if the
sunlit side of the tent was nearly pure white and if the rider was covered by a white
tunic.
Solution
Q
 A 4
4
(a) t  e 2 (T2  T1 ) ,
 Q/t

T2  
 T14 
e( A / 2)

1/ 4
 2(Q / t )


 T14 
 eA

1/ 4

2(20.0 W)
4


(307
K)

8
2
4
2
 (5.67  10 J/s  m  K )(0.970)(0 .400 m )

 321.63 K  48.5C
1/4
OpenStax College Physics
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Chapter 14
(b) A pure white object reflects more of the radiant energy that hits it, so the white
tent would prevent more of the sunlight from heating up the inside of the tent,
and the white tunic would prevent that radiant energy inside the tent from
heating the rider. Therefore, with a white tent, the temperature would be lower
than 48.5C , and the rate of radiant heat transferred to the rider would be less
than 20.0 W.
66.
Integrated Concepts One 30.0°C day the relative humidity is 75.0% , and that
evening the temperature drops to 20.0°C , well below the dew point. (a) How many
grams of water condense from each cubic meter of air? (b) How much heat transfer
occurs by this condensation? (c) What temperature increase could this cause in dry
air?
Solution (a) Let x be the vapor density during the day. Percent relative humidity is equal to
the vapor density divided by the saturation vapor density. Using the values for
relative humidity and saturation vapor density, we have
x
 0.750  x  22.8 g/m 3
30.4 g/m 3
22.8 g/m 3  17.2 g/m 3  5.60 g/m 3
(b) Q  mLv  (5.60  10 3 kg)(539 kcal/kg)  3.02 kcal
(c) Q  mcT  T 
Q
3.02 kcal

 13.6C
mc (1.29 kg)(0.172 kcal/kg  C)
67.
Integrated Concepts Large meteors sometimes strike the Earth, converting most of
their kinetic energy into thermal energy. (a) What is the kinetic energy of a 10 9 kg
meteor moving at 25.0 km/s? (b) If this meteor lands in a deep ocean and 80% of its
kinetic energy goes into heating water, how many kilograms of water could it raise by
5.0°C? (c) Discuss how the energy of the meteor is more likely to be deposited in the
ocean and the likely effects of that energy.
Solution
(a) KE 
1
mv 2  0.5(10 9 kg)(25.0  10 3 m/s) 2  3.125  1017 J  3  1017 J
2
(b) Q  mcT  m 
Q
(0.80)(3.1 25  1017 J)

 1.19  1013 kg  1 1013 kg
cT (4186 J/kg  C)(5.0C)
(c) When a large meteor hits the ocean, it causes great tidal waves, dissipating a large
OpenStax College Physics
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amount of its energy in the form of kinetic energy of the water.
68.
Integrated Concepts Frozen waste from airplane toilets has sometimes been
accidentally ejected at high altitude. Ordinarily it breaks up and disperses over a large
area, but sometimes it holds together and strikes the ground. Calculate the mass of
0°C ice that can be melted by the conversion of kinetic and gravitational potential
energy when a 20.0 kg piece of frozen waste is released at 12.0 km altitude while
moving at 250 m/s and strikes the ground at 100 m/s (since less than 20.0 kg melts, a
significant mess results).
Solution Let M be the mass of the ice block and m be the mass that melts before hitting the
ground.
KEi  PE  Q  KEf
1
1
Mv 2i  Mgh  mLf  Mv 2f
2
2

M gh  0.5(v 2i v 2f )
m
Lf



(20.0 kg) (9.80 m/s 2 )(12.0  10 3 m)  0.5(250 m/s) 2  0.5(100 m/s) 2

 8.61 kg
334  10 3 J/kg
69.
Solution
Integrated Concepts (a) A large electrical power facility produces 1600 MW of “waste
heat,” which is dissipated to the environment in cooling towers by warming air
flowing through the towers by 5.00°C . What is the necessary flow rate of air in m 3 /s ?
(b) Is your result consistent with the large cooling towers used by many large electrical
power plants?
(a) Q  mcT  m 
V
m


Q
1600  10 6 J/s

 4.438  10 5 kg/s
cT (721 J/kg  C)(5.00C)
4.438  10 5 kg/s
V
 3.44  10 5 m 3 /s  F   3.44  10 5 m 3 /s
3
t
1.29 kg/m
(b) This is equivalent to 12 million cubic feet of air per second. That is tremendous.
This is too large to be dissipated by heating the air by only 5C . Many of these
cooling towers use the circulation of cooler air over warmer water to increase the
rate of evaporation. This would allow there to be much smaller amounts of air
necessary to remove such a large amount of heat, because evaporation removes
larger quantities of heat than was considered in part (a).
OpenStax College Physics
70.
Instructor Solutions Manual
Chapter 14
Integrated Concepts (a) Suppose you start a workout on a Stairmaster, producing
power at the same rate as climbing 116 stairs per minute. Assuming your mass is 76.0
kg and your efficiency is 20.0% , how long will it take for your body temperature to
rise 1.00º C if all other forms of heat transfer in and out of your body are balanced?
(b) Is this consistent with your experience in getting warm while exercising?
Solution (a) You produce power at a rate of 685 W, and since you are 20% efficient, you must
P
685 W
have generated: Pgenerated  produced 
 3425 W .
efficiency
0.20
If only 685 W of power was useful, the power available to heat the body is
Pwasted  3425 W  685 W  2.740  103 W .
Q mcT

, so that
t
t
mcT (76.0 kg)(3500 J/kg  C)(1.00C)
t

 97.1 s
Pwasted
2.74  10 3 W
Now, Pwasted 
(b) This says that it takes about a minute and a half to generate enough heat to raise
the temperature of your body by 1.00C , which seems quite reasonable.
Generally, within five minutes of working out on a Stairmaster, you definitely feel
warm and probably are sweating to keep your body from overheating.
71.
Integrated Concepts A 76.0-kg person suffering from hypothermia comes indoors and
shivers vigorously. How long does it take the heat transfer to increase the person’s
body temperature by 2.00º C if all other forms of heat transfer are balanced?
Solution
Q  Pt  mcT 
t
72.
mcT (76.0 kg)(3500 J/kg  C)(2.00C)

 1.25  10 3 s  20.9 min
P
425 W
Integrated Concepts In certain large geographic regions, the underlying rock is hot.
Wells can be drilled and water circulated through the rock for heat transfer for the
generation of electricity. (a) Calculate the heat transfer that can be extracted by
cooling 1.00 km3 of granite by 100°C . (b) How long will it take for heat transfer at
the rate of 300 MW, assuming no heat transfers back into the 1.00 km3 of rock by its
surroundings?
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Chapter 14
Solution (a) Q  mcT  VcT
 (2700 kg/m 3 )(1.00  10 3 m) 3 (840 J/kg  C)(100C)  2.27  1017 J
(b) Pt  Q  t 
73.
Q 2.27  1017 J

 7.57  10 8 s  24.0 y
6
P 300  10 W
Integrated Concepts Heat transfers from your lungs and breathing passages by
evaporating water. (a) Calculate the maximum number of grams of water that can be
evaporated when you inhale 1.50 L of 37.0°C air with an original relative humidity of
40.0%. (Assume that body temperature is also 37.0°C .) (b) How many joules of
energy are required to evaporate this amount? (c) What is the rate of heat transfer in
watts from this method, if you breathe at a normal resting rate of 10.0 breaths per
minute?
Solution (a) To solve this, we calculate the mass of water initially in the breath and subtract
this value from the mass of the water in an exhaled breath at 100% humidity.
Using the saturation vapor density of water at 37C ,

 1.50  10

(44.0 g/m
min  1.50  10 3 m 3 (44.0 g/m 3 )(0.400)  2.64  10 2 g
mex
3
m

3
3
)  6.60  10 2 g
 

m  mex  min  6.60  10 2 g  2.64  10 2 g  3.96  10 2 g
(b) Q  mLv(37C)  (3.96 102 g)(2430 J/g)  96.23 J  96.2 J
(Note that Lv for water at 37C is used here as a better approximation than Lv for
water at 100C. )
(c) P 
74.
NQ 10(96.23 J)

 16.0 W
t
60.0 s
Integrated Concepts (a) What is the temperature increase of water falling 55.0 m
over Niagara Falls? (b) What fraction must evaporate to keep the temperature
constant?
OpenStax College Physics
Solution
Instructor Solutions Manual
Chapter 14
gh (9.80 m/s 2 )(50.0 m)
(a) mgh  mcT  T 

 0.117C
c
4186 J/kg  C
(b) Let M be the mass of water that evaporates.
M gh (9.80 m/s 2 )(50.0 m)
mgh  MLv 


 2.17  10 4
3
m Lv 2430  10 J/kg  C
(Note that Lv for water at 37C is used here as a better approximation than Lv
for water at 100C. )
75.
Integrated Concepts Hot air rises because it has expanded. It then displaces a greater
volume of cold air, which increases the buoyant force on it. (a) Calculate the ratio of
the buoyant force to the weight of 50.0°C air surrounded by 20.0°C air. (b) What
energy is needed to cause 1.00 m3 of air to go from 20.0°C to 50.0°C? (c) What
gravitational potential energy is gained by this volume of air if it rises 1.00 m? Will
this cause a significant cooling of the air?
Solution
(a) PV  nRT  The density of a given volume of air will be proportional to
h ~
1
Th
c ~
1
.
T
1
Tc
The buoyant force is equal to the weight of the displaced cold air (Archimedes’
principle.) Thus,
FB  wc   cVg and wh   hVg
FB  cVg Th 323 K



 1.102
wh  hVg Tc 293 K
(b) Q  mcp T  Vcp T
 (1.29 kg/m 3 )(1.00 m 3 )(721 J/kg  C)(30.0C)  2.79  10 4 J
(c) PE  mgh  (1.29 kg/m 3 )(1.00 m 3 )(9.80 m/s 2 )(1.00 m)  12.6 J
This will not cause a significant cooling of the air because it is much less than the
energy found in part (b), which is the energy required to warm the air from
20.0C to 50.0C .
OpenStax College Physics
Instructor Solutions Manual
Chapter 14
76.
Unreasonable Results (a) What is the temperature increase of an 80.0 kg person who
consumes 2500 kcal of food in one day with 95.0% of the energy transferred as heat
to the body? (b) What is unreasonable about this result? (c) Which premise or
assumption is responsible?
Solution
(a) Q  mcT , so that T 
Q
(0.950)(25 00 kcal)

 36C.
mc (80.0 kg)(0.83 kcal/kg  C)
This says that the temperature of the person is 37C  36C  73C !
(b) Any temperature increase greater than about 3C would be unreasonably large.
In this case the final temperature of the person would rise to 73C 163F .
(c) The assumption that the person retains 95% of the energy as body heat is
unreasonable. Most of the food consumed on a day is converted to body heat,
losing energy by sweating and breathing, etc.
77.
Unreasonable Results A slightly deranged Arctic inventor surrounded by ice thinks it
would be much less mechanically complex to cool a car engine by melting ice on it
than by having a water-cooled system with a radiator, water pump, antifreeze, and so
on. (a) If 80.0% of the energy in 1.00 gal of gasoline is converted into “waste heat” in
a car engine, how many kilograms of 0°C ice could it melt? (b) Is this a reasonable
amount of ice to carry around to cool the engine for 1.00 gal of gasoline
consumption? (c) What premises or assumptions are unreasonable?
Solution
Qgas
(a) Qgas  MLf  M  L
f
Qgas  0.800 1.3  10 8 J. Thus,
M 


0.800 1.3  10 8 J
 311.4 kg  3.1  10 2 kg
3
334  10 J/kg
(b) No, the mass of ice is greater than 1/4 of a ton.
(c) Not all waste heat goes into the engine.
78.
Unreasonable Results (a) Calculate the rate of heat transfer by conduction through a
window with an area of 1.00 m 2 that is 0.750 cm thick, if its inner surface is at
22.0°C and its outer surface is at 35.0°C . (b) What is unreasonable about this result?
(c) Which premise or assumption is responsible?
OpenStax College Physics
Solution
(a)
Instructor Solutions Manual
Chapter 14
Q kAT2  T1  (0.84 J/s  m  C)(1.00 m 2 )(35.0C - 22.0C)


t
d
0.750  10 -2 m
 1456 W  1.46  10 3 W  1.46 kW
(b) This is very high power loss through a window. An electric heater of this power
can keep an entire room warm.
(c) The surface temperatures of the window do not differ by as great an amount as
assumed. The inner surface will be warmer, and the outer surface will be cooler.
79.
Solution
Unreasonable Results A meteorite 1.20 cm in diameter is so hot immediately after
penetrating the atmosphere that it radiates 20.0 kW of power. (a) What is its
temperature, if the surroundings are at 20.0°C and it has an emissivity of 0.800? (b)
What is unreasonable about this result? (c) Which premise or assumption is
responsible?
Q
 20.0 kW  20.0  10 3 W and R  0.600 cm. (Note that the negative
t
sign indicates that the meteorite radiates heat to the surroundings.)
(a) Given
Q/t 

T1   T24 

eA 

1/ 4


 20.0  10 3 W
 (293 K) 4 
8
2
4
2
2 
(5.67  10 J/s  m  K )(0.800)4 (0.600  10 m) 

 5.59  10 3 K
(b) The meteorite has too high a temperature. It would completely melt.
(c) The rate of radiation is probably too high.
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