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612 TEMPERATURE AND HEAT
CHAPTER
12 TEMPERATURE AND HEAT
CONCEPTUAL QUESTIONS
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1.
REASONING AND SOLUTION The length L0 of the tape rule changes by an amount ∆L
when its temperature changes by an amount ∆T, where ∆L is given by ∆L = α L0 ∆T and α
is the coefficient of linear expansion. When the length of the tape rule changes, the location
of the graduations changes and makes the rule inaccurate. For year-round outdoor use, the
range of ∆T will be large; therefore, to minimize the thermal expansion of the rule we would
want the rule to be made from a material with a small coefficient of linear expansion. Since
the coefficient of linear expansion of steel is less than that of aluminum, we would choose a
steel tape rule for highest accuracy.
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2.
REASONING AND SOLUTION The distance between the two fine lines engraved near
the ends of the bar is unique only at a specified temperature. When the temperature of the
bar is raised, the length of the bar increases, and the distance between the engraved lines
increases. Since the bar was an international standard for length, it was essential that the
distance between the lines always remained the same; therefore, it was important that the
bar be kept at a constant temperature.
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3.
REASONING AND SOLUTION The plate is made of aluminum; the spherical ball is
made of brass. The coefficient of linear expansion of aluminum is greater than the
coefficient of linear expansion of brass. Therefore, if the plate and the ball are heated, both
will expand; however, the diameter of the hole in the aluminum plate will expand more than
the diameter of the brass ball. In order to prevent the ball from falling through the hole, the
plate and the ball must be cooled. Both the diameter of the hole in the plate and the
diameter of the ball will contract. The diameter of the hole will decrease more than the
diameter of the ball, thereby preventing the ball from falling through the hole.
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4.
REASONING AND SOLUTION Reinforced concrete is concrete that is reinforced with
embedded steel rods. For this arrangement to be practical, it is essential that the coefficient
of linear expansion of concrete is the same as that of steel. When the reinforced concrete is
heated or cooled, both the concrete and the steel must expand or contract by the same
amount. If the coefficients of linear expansion were not the same, the concrete and steel
would expand and contract at different rates when heated or cooled. If the steel expanded or
contracted more than the concrete, it would place internal stresses on the concrete, thereby
causing it to weaken and possibly crack.
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Chapter 12 Conceptual Questions
5.
613
REASONING AND SOLUTION A rod is hung from an aluminum frame such that there is
a small gap between the rod and the floor. The rod and the frame are heated uniformly; both
the frame and the rod will expand. The linear expansion of the vertical portion of the frame
and the linear expansion of the rod can be determined from Equation 12.2: ∆L = α L0 ∆T .
a. If the rod is made of aluminum, the fractional increase in length ∆L / L0 is the same for
the vertical portion of the frame and the rod. Since the rod is initially shorter than the
vertical portion of the frame, the rod will never be as long as the vertical portion of the
frame. Therefore, the rod will never touch the floor.
b. If the rod is made of lead, the heated rod could touch the floor. The coefficient of linear
expansion of lead is greater than that of aluminum. Therefore, the fractional change in the
length of the rod will be greater than the fractional change in the length of the vertical
portion of the frame. If the temperature is raised high enough, the length of the rod could
expand enough to fill the gap and touch the floor.
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6.
REASONING AND SOLUTION A simple pendulum is made using a long thin metal wire.
From Equations 10.6 and 10.16, we know that the period of the pendulum is proportional to
L / g , where L is the length of the wire. When the temperature drops, the length of the
wire decreases; therefore, the period of the pendulum decreases.
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7. REASONING AND SOLUTION The bottom of the pot bows outward, because it is a kind
of bimetallic strip. Table 12.1 shows that the coefficient of linear thermal expansion for
copper is α = 17 × 10-6 (Cº)-1, while for steel it is α = 12 × 10-6 (Cº)-1. Thus, as the pot is
heated, the copper expands more than the steel. Since the copper is on the outside, the
bottom of the pot bows outward.
8.
SSM REASONING AND SOLUTION A hot steel ring fits snugly over a cold brass
cylinder. The temperatures of the ring and the cylinder are, respectively, above and below
room temperature. As the assembly equilibrates to room temperature, the diameter of the
steel ring contracts while the diameter of the brass cylinder expands. The ring and the
cylinder, therefore, fit much more snugly than they did initially. It becomes nearly
impossible to pull the ring off the cylinder once the assembly has equilibrated at room
temperature.
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9.
REASONING AND SOLUTION The coefficient of volume expansion of Pyrex glass is
only about 39 percent that of common glass. Therefore, the fractional change in volume,
∆V / V 0 , of a Pyrex baking dish is significantly less than that of common glass when both
objects are taken from a refrigerator and placed in a hot oven. The Pyrex dish, therefore, is
less likely to crack from thermal stress.
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614 TEMPERATURE AND HEAT
10. REASONING AND SOLUTION An ordinary mercury-in-glass thermometer works on the
following principle. When the thermometer is placed in contact with the object whose
temperature is to be measured, the thermometer comes into thermal equilibrium with the
object. As thermal equilibrium is reached, the glass tube and the enclosed mercury expand
or contract, depending on whether the thermometer heats up or cools down. The coefficient
of volume expansion of mercury is about 7 times greater than that of ordinary glass. Thus,
the expansion or contraction of the mercury column relative to the scale on the glass is
evident when one reads the thermometer. If the coefficients of volume expansion of mercury
and glass were both the same, both would expand or contract by the same amount. The
reading on the thermometer would never change.
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11. REASONING AND SOLUTION According to Archimedes' principle, any fluid applies a
buoyant force to an object that is partially or completely immersed in it; the magnitude of
the buoyant force equals the weight of the fluid that the object displaces. Therefore, the
magnitude of the buoyant force exerted on an object immersed in water is given by
FB = ρ waterVg , where ρwater is the density of water, and V is the volume displaced by the
immersed object.
As shown in Figure 12.20, the density of cold water (above 4 °C) is greater than the
density of warm water (above 4 °C). Therefore, cold water provides a greater buoyant force
than warm water.
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12. REASONING AND SOLUTION When the bulb of a mercury-in-glass thermometer is
inserted into boiling water, the glass is heated first. Therefore, the glass envelope expands.
The cavity that contains the mercury increases in volume, and the mercury level drops
slightly. Then, the mercury begins to be heated, and its volume expands. Since the
coefficient of volume expansion for mercury is much greater than that of the glass, the
mercury expands to a greater extent than the glass cavity, and the mercury level then begins
to rise.
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13. REASONING AND SOLUTION When an amount of heat Q is added to an object of mass
m, its temperature will increase. According to Equation 12.4, the amount by which the
temperature will increase is given by ∆T = Q / cm , where c is the specific heat capacity of
the object.
Two different objects are supplied with equal amounts of heat. The objects could have
the same specific heat capacities, but they could have different masses. The objects could
have the same masses, but they could have different specific heat capacities. Or, the objects
could have both different masses and different specific heat capacities. In any of these cases,
the temperature changes of the objects would not be the same.
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14. REASONING AND SOLUTION Two objects are made from the same material, but have
different masses and temperatures. When the two objects are placed in contact, they will
come into thermal equilibrium and reach a common temperature. The heat gained by the
cooler object must be equal to the heat lost by the warmer object, so that cm1∆T1 = cm2 ∆T2 ,
Chapter 12 Conceptual Questions
615
where one object has mass m1 and the other object has mass m2. Since the objects are made
of the same material, they have the same specific heat capacities so that m1∆T1 = m2 ∆T2 or
∆T2 = ( m1 / m2 ) ∆T1 . Suppose that m1 is greater than m2. Then, ∆ T2 is greater than ∆ T1 , so
the less massive object must undergo the greater temperature change.
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15. SSM REASONING AND SOLUTION Two identical mugs contain hot coffee from the
same pot. One mug is full, while the other is only one-quarter full. According to Equation
12.4, the amount of heat that must be removed from either cup to bring it to room
temperature is Q = cm∆T , where m is the mass of coffee in the cup, c is the specific heat
capacity of the coffee and ∆T is the temperature difference between the initial temperature
of the coffee and room temperature. Since the mug that is full contains a larger mass m of
coffee, more heat Q must be removed from the full mug than from the mug that is only onequarter full. Therefore, the mug that is full remains warmer longer.
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16. REASONING AND SOLUTION As heat flows from the higher-temperature air to the
lower-temperature water bottle, it passes through the wet sock. Most of the heat causes the
water in the sock to evaporate, and only a relatively small amount of heat reaches the water
bottle. Thus, as long as the sock remains wet, the water bottle remains cool.
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17. REASONING AND SOLUTION An alcohol rub has a cooling effect on a sick patient for
the following reason. In order for the alcohol to evaporate, heat must be supplied to it. The
evaporating alcohol removes heat from the skin of the sick patient. The evaporating alcohol,
therefore, helps to lower the body temperature of the patient.
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18. REASONING AND SOLUTION Your head feels colder under an air-conditioning vent
when your hair is wet than when it is dry. When your hair is wet, the water evaporates. In
order to evaporate, heat must be supplied to the water. Therefore, heat is removed from your
head as the water evaporates, thereby causing a cooling effect on your head.
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19. REASONING AND SOLUTION Suppose that the latent heat of vaporization of H2O were
one-tenth of its actual value.
a. Other factors being equal, it would take only one-tenth the amount of heat to "boil away"
a pot of water on the stove. Assuming that the stove supplies heat at a constant rate, it
would, therefore, take a shorter time for a pot of water to boil away.
b. The evaporative cooling mechanism of the human body would be less effective. When a
given amount of water evaporates from the skin, it would now require only one-tenth as
much heat. Therefore, only one-tenth as much heat would be removed from the skin surface
for a given amount of perspiration.
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616 TEMPERATURE AND HEAT
20. REASONING AND SOLUTION Orchard owners sometimes spray a film of water over
fruit blossoms to protect them when a hard freeze is expected. The thin film of water will
freeze before the water in the tissues of the blossoms. When the thin film of water freezes at
0 °C, each kilogram of water releases an amount of heat equal to the latent heat of fusion.
Some of this heat will warm and, therefore, protect the blossoms from reaching the
damaging temperature of – 4 °C.
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21. REASONING AND SOLUTION When a liquid is heated, its equilibrium vapor pressure
increases as the temperature increases, as shown in Figure 12.33 for water. When the vapor
pressure of the liquid is equal to the atmospheric pressure above the liquid, the liquid begins
to boil. Therefore, at a higher altitude, where the atmospheric pressure is less than at lower
altitudes, boiling is more easily achieved. At a higher altitude, the vapor pressure does not
have to be raised as much to reach the boiling point, and the temperature required to boil a
liquid at the higher altitude is less than that at a lower altitude.
A camping stove is used to boil water on a mountain. Since water will boil at a lower
temperature at higher altitudes, it does not necessarily follow that the same stove can boil
water at lower altitudes. If the stove cannot provide sufficient heat Q to raise the
temperature of the water to the temperature at which the vapor pressure of the liquid is equal
to the atmospheric pressure at the lower altitude, the water will not boil.
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22. REASONING AND SOLUTION If a bowl of water is placed in a closed container and
water vapor is pumped away rapidly enough, the liquid water turns to ice. As water vapor is
pumped away, the pressure in the jar is reduced. The liquid attempts to reestablish the
equilibrium vapor pressure, so more water molecules enter the gaseous state. As the water
vaporizes, heat must be supplied in the form of the latent heat of vaporization. Because of
the heat loss, the remaining liquid is cooled. If this heat loss occurs rapidly enough, the
surroundings cannot replenish the heat loss, and, when sufficient heat has been removed, the
remaining liquid turns to ice.
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23. REASONING AND SOLUTION
If medical instruments were sterilized in an open
container of water, the water would be subjected to atmospheric pressure. Therefore, the
highest possible temperature that could be attained is 100 °C. In order to raise the
temperature above 100 °C, we would have to raise the pressure above the liquid, as
suggested by Figure 12.33. An autoclave heats medical instruments in water under high
pressure. We can see from Figure 12.33 that at pressures above atmospheric pressure, the
boiling point of water is greater than 100 °C. Therefore, by sterilizing medical instruments
in an autoclave, it is possible to attain temperatures much greater than 100 °C.
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24. SSM REASONING AND SOLUTION A bottle of carbonated soda is left outside in
subfreezing temperatures. The temperature of the soda is "subfreezing," but the soda
remains in the liquid state. Because the bottle is sealed, the pressure above the soda remains
too high for the soda to freeze. When the soda is brought inside and opened, the pressure
drops to atmospheric pressure. Because the temperature of the soda is below the freezing
point at atmospheric pressure, the soda freezes.
Chapter 12 Conceptual Questions
617
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25. REASONING AND SOLUTION A bowl of water is covered tightly and allowed to sit at a
constant temperature of 23 °C for a long time. Since the bowl is tightly covered, none of
the water vapor can escape, and since the bowl is allowed to sit for a long time at a constant
temperature, the vapor above the liquid attains equilibrium with the liquid. The partial
pressure of the water vapor will equal the equilibrium vapor pressure of water at 23 °C, and
the relative humidity will be 100%.
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26. REASONING AND SOLUTION Dew forms on Tuesday night and not on Monday night,
even though Monday night is the cooler night. Since dew did not form on Monday night, we
can conclude that the temperature on Monday night was above the dew point. The partial
pressure of water in the air must have been greater on Tuesday night; then, the dew point
would occur at a higher temperature. As long as the temperature on Tuesday is below the
dew point, dew will form.
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27. REASONING AND SOLUTION Two rooms in a house have the same temperature. One
of the rooms contains an indoor swimming pool. The partial pressure of water vapor in the
air will be greater in the room that contains the indoor pool. Therefore, the dew point will be
higher in the room with the pool. On a cold day, the windows of one of the rooms are
"steamed up." Water will condense on the window when the temperature of the air next to
the glass falls below the dew point. Since only one of the rooms has "steamed" windows,
and the dew point is higher for the room with the pool, then the outside temperature must be
less than the dew point of the room with the pool, but greater than the dew point for the
room without the pool. Therefore, the windows are "steamed up" in the room with the pool.
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28. REASONING AND SOLUTION A jar is half filled with boiling water. The lid is then
screwed on the jar. The temperature of the jar and the lid will be close to the boiling point of
water. As the jar is cooled to room temperature, the lid and the jar contract. Since the
coefficient of linear expansion of the metal jar lid is greater than the coefficient of linear
expansion of the glass jar, the lid will contract more than the jar. Therefore, the lid will fit
more snugly when the jar has cooled down to room temperature and be difficult to remove.
In addition, as the temperature of the water vapor inside the jar drops, the pressure inside
becomes less than the outside pressure. Thus, there is a net force pushing the lid onto the jar,
making the lid harder to turn.
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618 TEMPERATURE AND HEAT
CHAPTER 12
TEMPERATURE AND HEAT
PROBLEMS
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1.
REASONING AND SOLUTION The temperature of –273.15 °C is 273.15 Celsius
degrees below the ice point of 0 °C. This number of Celsius degrees corresponds to
SSM
⎛ (9/5) F° ⎞
(273.15 C°) ⎝
= 491.67 F°
1 C° ⎠
Subtracting 491.67 F° from the ice point of 32.00 °F on the Fahrenheit scale gives a
Fahrenheit temperature of –459.67 °F
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2.
REASONING
a. According to the discussion in Section 12.1, the size of a Fahrenheit degree is smaller
than that of a Celsius degree by a factor of 95 ; thus, 1 F° = 95 C° . This factor will be used to
find the temperature difference in Fahrenheit degrees.
b. The size of one kelvin is identical to that of one Celsius degree (see Section 12.2),
1 K = 1 C°. Thus, the temperature difference, expressed in kelvins, is the same at that
expressed in Celsius degrees.
SOLUTION
a. The difference in the two temperatures is 34 °C − 3 °C = 31 C°. This difference,
expressed in Fahrenheit degrees, is
⎛ 1 F°
Temperature difference = 31 C° = ( 31 C° ) ⎜ 5
⎝ 9 C°
⎞
⎟ = 56 F°
⎠
b. Since 1 K = 1 C°, the temperature difference, expressed in kelvins, is
⎛ 1K
Temperature difference = 31 C° = ( 31 C° ) ⎜
⎝ 1 C°
⎞
⎟ = 31 K
⎠
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3. REASONING
a. The relationship between the Kelvin temperature T and the Celsius temperature TC is
given by T = TC + 273.15 (Equation 12.1).
b. The relationship between the Kelvin temperature T and the Fahrenheit temperature TF can
be obtained by following the procedure outlined in Examples 1 and 2 in the text. On the
Chapter 12 Problems
619
Kelvin scale the ice point is 273.15 K. Therefore, a Kelvin temperature T is T - 273.15
kelvins above the ice point. The size of the kelvin is larger than the size of a Fahrenheit
degree by a factor of 95 . As a result a temperature that is T - 273.15 kelvins above the ice
point on the Kelvin scale is 95 (T - 273.15) Fº above the ice point on the Fahrenheit scale.
This amount must be added to the ice point of 32.0 ºF. The relationship between the Kelvin
and Fahrenheit temperatures, then, is given by
TF = 95 (T − 273.15 ) + 32.0
SOLUTION
a. Solving Equation 12.1 for TC, we find that
Day
TC = T − 273.15 = 375 − 273.15 = 102 °C
Night
TC = T − 273.15 = 1.00 × 102 − 273.15 = −173 °C
b. Using the equation developed in the reasoning, we find
Day
Night
4.
TF = 95 (T − 273.15 ) + 32.0 = 95 ( 375 − 273.15 ) + 32.0 = 215 °F
(
)
TF = 95 (T − 273.15 ) + 32.0 = 95 1.00 × 102 − 273.15 + 32.0 = −2.80 ×102 °F
REASONING AND SOLUTION
a. The Kelvin temperature and the temperature on the Celsius scale are related by
Equation 12.1: T = Tc + 273.15, where T is the Kelvin temperature and Tc is the Celsius
temperature. Therefore, a temperature of 77 K on the Celsius scale is
Tc = T – 273.15 = 77 K – 273.15 K =
–196 ° C
b. The temperature of –196 °C is 196 Celsius degrees below the ice point of 0 °C. Since
9
1 C° = 5 F° , this number of Celsius degrees corresponds to
⎛ 9 F° ⎞
196 C°⎜ 5
⎟ = 353 F°
⎝ 1 C° ⎠
Subtracting 353 Fahrenheit degrees from the ice point of 32.0 °F on the Fahrenheit scale
gives a Fahrenheit temperature of –321 ° F .
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620 TEMPERATURE AND HEAT
5.
REASONING AND SOLUTION
expressed in Fahrenheit degrees, is
SSM
The difference between these two averages,
98.6 °F − 98.2 °F = 0.4 F°
Since 1 C° is equal to
9
5
F°, we can make the following conversion
⎛ 1 C° ⎞
(0.4 F°) ⎜
= 0.2 C°
⎝ (9/5) F°⎠
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6.
REASONING AND SOLUTION The space invaders temperature of 58 °I is 58 °I – 25 °Ι
or 33 I° above the Space invaders ice point. Additionally, the space invaders degree is
100 °C − 0 °C
= 0.76 C°/I°
156 °I − 25 °I
times smaller than the Celsius degree. Now
58 °I = (33 °I)(0.76 C°/I°) = 25 °C
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7.
SSM
REASONING
The temperature at any pressure can be determined from the
equation of the graph in Figure 12.4. Since the gas pressure is 5.00 × 103 Pa when the
temperature is 0.00 °C , and the pressure is zero when the temperature is –273.15 °C, the
slope of the line is given by
∆P 5.00 × 103 Pa – 0.00 Pa
=
= 18.3 Pa/C°
∆T 0.00 °C – (–273.15 °C)
Therefore, the equation of the line is ∆ P = (18.3 Pa/C°)∆T , or
P – P0 = (18.3 Pa/C°)(T − T0 )
where P0 = 5.00 × 103 Pa when T0 = 0.00 °C .
SOLUTION Solving the equation above for T, we obtain
P − P0
2.00 × 103 Pa − 5.00 × 10 3 Pa
+ T0 =
+ 0.00 °C = –164 °C
18.3 Pa/C°
18.3 Pa/C°
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T=
Chapter 12 Problems
8.
621
REASONING AND SOLUTION The Rankine and Fahrenheit degrees are the same size,
since the difference between the steam point and ice point temperatures is the same for both.
The difference in the ice points of the two scales is 491.67 – 32.00 = 459.67. To get
Rankine from Fahrenheit this amount must be added, so TR = TF + 459.67 .
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9.
REASONING The change in length ∆L of the pipe is proportional to the coefficient of
linear expansion α for steel, the original length L0 of the pipe, and the change in
temperature ∆T . The coefficient of linear expansion for steel can be found in Table 12.1.
SOLUTION The change in length of the pipe is
−1
∆L = α L0 ∆T = ⎡⎣1.2 × 10−5 ( C° ) ⎤⎦ ( 65 m ) ⎣⎡18 °C − ( −45 °C ) ⎦⎤ = 4.9 × 10−2 m
(12.2)
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10. REASONING The change in temperature is the final temperature T minus the initial
temperature T0, or ∆T = T − T0 . Thus, T = T0 + ∆T . When the bat is heated, its length
changes by an amount given by Equation 12.2 as ∆L = α L0 ∆T , where α is the coefficient
of linear expansion, and L0 is the initial length of the bat. Solving this expression for ∆T
and substituting the result into T = T0 + ∆T will allow us to determine the final temperature
of the bat.
SOLUTION Solving ∆L = α L0 ∆T for the change in temperature and substituting the
result into T = T0 + ∆T gives
T = T0 + ∆T = T0 +
∆L
0.000 16 m
= 17 °C +
= 25 °C
α L0
⎡ 23 × 10−6 ( C° )−1 ⎤ ( 0.86 m )
⎣
⎦
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11.
REASONING AND SOLUTION The steel in the bridge expands according
to Equation 12.2, ∆L = αL0 ∆T . Solving for L0 and using the value for the coefficient of
thermal expansion of steel given in Table 12.1, we find that the approximate length of the
Golden Gate bridge is
SSM
WWW
∆L
0.53 m
=
= 1500 m
–6
α ∆ T 12 × 10 (C° ) –1 (32 ° C − 2 ° C)
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L0 =
[
]
622 TEMPERATURE AND HEAT
12. REASONING AND SOLUTION
a. The radius of the hole will be
larger
when the plate is heated, because the hole
expands as if it were made of copper.
b. The expansion of the radius is ∆r = α r0∆T. Using the value for the coefficient of
thermal expansion of copper given in Table 12.1, we find that the fractional change in the
radius is
∆r/r0 = α ∆T = [17 × 10–6 (C°)–1](110 °C − 11 °C) = 0.0017
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13. REASONING AND SOLUTION The fractional change in the length of the beam is given
by ∆L = α L0∆T to be ∆L/L0 = α ∆T. Taking the value for the coefficient of thermal
expansion a for steel from Table 12.1, we find that
⎛ 1 C° ⎞
∆L ⎡
−1
= ⎣12 × 10−6 ( C° ) ⎦⎤ ⎣⎡105 °F − ( −15 °F ) ⎦⎤ ⎜ 9 ⎟ = 8.0 × 10−4
⎜ F° ⎟
L0
⎝5 ⎠
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14. REASONING AND SOLUTION The change in the coin’s diameter is ∆d = α d0∆T,
according to Equation 12.2. Solving for a gives
∆d
2.3 × 10 −5 m
=
= 1.7 ×10 –5 (C°) –1
(12.2)
d 0 ∆ T (1.8 × 10 −2 m)(75 C° )
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α=
15.
REASONING AND SOLUTION Assuming that the rod expands linearly with heat,
we first calculate the quantity ∆L / ∆T using the data given in the problem.
SSM
∆L
8.47 × 10–4 m
–5
=
= 1.13 × 10 m/C°
∆T 100.0 °C – 25.0 °C
Therefore, when the rod is cooled from 25.0 °C, it will shrink by
∆L = (1.13 × 10 –5 m/C °) ∆T
= (1.13 × 10 –5 m/C °) (0.00 °C – 25.0 °C) = –2.82 × 10 –4 m
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16. REASONING Each section of concrete expands as the temperature increases by an amount
DT. The amount of the expansion DL is proportional to the initial length of the section, as
indicated by Equation 12.2. Thus, to find the total expansion of the three sections, we can
Chapter 12 Problems
623
apply this expression to the total length of concrete, which is L0 = 3(2.4 m). Since the two
gaps in the drawing are identical, each must have a minimum width that is one half the total
expansion.
SOLUTION Using Equation 12.2 and taking the value for the coefficient of thermal
expansion for concrete from Table 12.1, we find
−1
∆L = α L0 ∆T = ⎡12 × 10−6 ( C° ) ⎤ ⎡⎣3 ( 2.4 m ) ⎤⎦ ( 32 C° )
⎣
⎦
The minimum necessary gap width is one half this value or
1
2
⎡12 × 10−6 ( C° )−1 ⎤ ⎡3 ( 2.4 m ) ⎤ ( 32 C° ) = 1.4 ×10−3 m
⎦
⎣
⎦⎣
17. REASONING AND SOLUTION ∆L = αL0∆T gives for the expansion of the aluminum
∆LA = αALA∆T
(1)
and for the expansion of the brass
∆LB = αBLB∆T
(2)
Taking the coefficients of thermal expansion for aluminum and brass from Table 10.1,
adding Equations (1) and (2), and solving for ∆T give
∆T =
∆LA − ∆LB
1.3 × 10−3 m
=
= 21 C°
α A LA + α B LB ⎡ 23 × 10−6 ( C° )−1 ⎤ (1.0 m ) + ⎡19 × 10−6 ( C° )−1 ⎤ ( 2.0 m )
⎣
⎦
⎣
⎦
The desired temperature is then
T = 28 °C + 21 C° = 49 °C
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18. REASONING Young’s modulus Y can be obtained from F = Y ( ∆L/L0 ) A (Equation
10.17), where F is the magnitude of the stretching force applied to the ruler, ∆L and L0 are
the change in length and original length, respectively, and A is the cross-sectional area.
Solving for Y gives
F L0
Y=
A ∆L
624 TEMPERATURE AND HEAT
The change in the length of the ruler is given by ∆L = a L0∆T (Equation 12.2), where α is
the coefficient of linear expansion and ∆T is the amount by which the temperature changes.
Substituting this expression for ∆L into the equation above for Y gives the desired result.
SOLUTION Substituting ∆L = a L0∆T into Y = F L0 / ( A ∆L ) and using the fact that
∆T = 39 C°, we find that
Y=
F L0
A ∆L
=
(
F L0
A α L0 ∆T
)
F
1.2 × 103 N
=
= 7.7 × 1010 N/m 2
1
−
5
2
5
−
−
Aα ∆T (1.6 ×10 m ) ⎡⎣ 2.5 ×10 ( C° ) ⎤⎦ ( 39 C° )
______________________________________________________________________________
=
19.
REASONING AND SOLUTION Recall that ω = 2π / T , Equation 10.6,
where ω is the angular frequency of the pendulum and T is the period. Using this fact and
Equation 10.16, we know that the period of the pendulum before the temperature rise is
given by T1 = 2π L0 / g , where L0 is the length of the pendulum. After the temperature
SSM
WWW
has risen, the period becomes (using Equation 12.2), T2 = 2π [L 0 + α L0 ∆ T ]/ g .
Dividing these expressions, solving for T2 , and taking the coefficient of thermal expansion
of brass from Table 12.1, we find that
T2 = T1 1 + α ∆T = (2.0000 s) 1 + (19 × 10 –6 /C° ) (140 C°) = 2.0027 s
______________________________________________________________________________
20. REASONING The length of either heated strip is L0 + DL, where L0 is the initial length and
DL is the amount by which it expands. The expansion DL can be expressed in terms of the
coefficient of linear expansion α, the initial length L0, and the change in temperature DT,
according to Equation 12.2. To find the change in temperature, we will set the length of the
heated steel strip equal to the length of the heated aluminum strip and solve the resulting
equation for DT.
SOLUTION According to Equation 12.2, the expansion is DL = α L0DT.
equation we have
L0, Steel + αSteel L0, Steel ∆T = L0, Aluminum + α Aluminum L0, Aluminum ∆T
∆LSteel
∆LAluminum
L0, Steel (1 + αSteel ∆T ) = L0, Aluminum (1 + α Aluminum ∆T )
Using this
Chapter 12 Problems
625
We know that the steel strip is 0.10 % longer than the aluminum strip, so that
L0, Steel = (1.0010 ) L0, Aluminum . Substituting this result into the equation above, solving for
DT, and taking values for the coefficients of linear expansion for aluminum and steel from
Table 12.1give
(1.0010 ) L0, Aluminum (1 + αSteel ∆T ) = L0, Aluminum (1 + α Aluminum ∆T )
0.0010 + (1.0010 ) αSteel ∆T = α Aluminum ∆T
∆T =
0.0010
0.0010
=
= 91 C°
1
−
α Aluminum − (1.0010 ) αSteel 23 ×10−6 ( C° ) − (1.0010 ) ⎡12 ×10−6 ( C° )−1 ⎤
⎣
⎦
21. REASONING AND SOLUTION The initial diameter of the sphere, ds, is
ds = (5.0 × 10–4)dr + dr
(1)
where dr is the initial diameter of the ring. Applying ∆L = αL0∆T to the diameter of the
sphere gives
∆ds = αsds∆T
(2)
and to the ring gives
∆dr = αrdr∆T
(3)
If the sphere is just to fit inside the ring, we must have
ds + ∆ds = dr + ∆dr
Using Equations (2) and (3) in this expression and solving for ∆T give
∆T =
d r − ds
α s ds − α r d r
Substituting Equation (1) into this result and taking values for the coefficients of thermal
expansion of steel and lead from Table 10.1 yield
∆T =
(
−5.0 × 10−4
)
⎡ 29 × 10−6 ( C° )−1 ⎤ 5.0 × 10−4 + 1 − 12 × 10−6 ( C° )−1
⎢⎣
⎥⎦
= − 29 C°
626 TEMPERATURE AND HEAT
The final temperature is
Tf = 70.0 °C − 29 C° = 41 °C
______________________________________________________________________________
22. REASONING AND SOLUTION
a. As the wheel heats up, it will expand. Its radius, and therefore, its moment of inertia, will
increase. Since no net external torque acts on the wheel, conservation of angular momentum
applies where, according to Equation 9.10, the angular momentum is given by: L = Iω ,
where I is the moment of inertia and ω is the angular velocity. When the moment of inertia
increases at the higher temperature, the angular speed must decrease in order for the angular
momentum to remain the same.
Thus, the angular speed of the wheel decreases as the wheel heats up .
b. According to the principle of conservation of angular momentum,
I 0ω 0 = I f ω f
N
N
Initial angular
momentum, L0
Final angular
momentum, Lf
Solving for ω f , we have, treating the bicycle wheel as a thin-walled hollow hoop ( I = MR2 ,
see Table 9.1)
2
⎛R ⎞
⎛ MR 2 ⎞
⎛ I0 ⎞
0
0
ωf = ω0 ⎜ ⎟ = ω0 ⎜
⎟
2 ⎟ = ω0 ⎜
⎝ If ⎠
⎝ MRf ⎠
⎝ Rf ⎠
According to Equation 12.2, ∆R = α R0 ∆T , and the final radius of the wheel at the higher
temperature is,
Rf = R0 + ∆R = R0 + α R0 ∆T = R0 (1 + α ∆T )
Therefore, taking the coefficient of thermal expansion a for steel from Table 12.1, we find
that the angular speed of the wheel at the higher temperature is
⎡
⎤
R0
ωf = ω0 ⎢
⎥
⎣ R0 (1 + α ∆T ) ⎦
2
1 ⎞2
= ω0 ⎛
⎝ 1+ α∆T ⎠
2
⎧
⎫
1
= (18.00 rad/s) ⎨
⎬ = 17.83 rad/s
–6
–1
⎩1 + [12 × 10 (C°) ][300.0 °C – (–100.0 °C)]⎭
______________________________________________________________________________
Chapter 12 Problems
23.
627
REASONING The change in length of the wire is the sum of the change in length
of each of the two segments: ∆L = ∆Lal + ∆Lst . Using Equation 12.2 to express the changes
in length, we have
SSM
αL0 ∆T = α al L0al ∆T + α st L0st ∆T
Dividing both sides by L0 and algebraically canceling ∆T gives
⎛ L0al ⎞
⎛
⎟ + α st ⎜ L0st ⎞⎟
⎝ L0 ⎠
⎝ L0 ⎠
α = α al ⎜
The length of the steel segment of the wire is given by L0st = L0 − L0al . Making this
substitution leads to
⎛L
α = α al ⎜
0al
⎝ L0
⎞
⎛ L −L ⎞
0
0al
⎟ + α st ⎜
⎟
L
⎠
⎝
⎠
0
⎛L ⎞
⎛L ⎞
⎛L ⎞
= α al ⎜ 0al ⎟ + α st ⎜ 0 ⎟ − α st ⎜ 0al ⎟
⎝ L0 ⎠
⎝ L0 ⎠
⎝ L0 ⎠
This expression can be solved for the desired quantity, L0al / L0 .
SOLUTION Solving for the ratio (L0al / L0 ) and taking values for the coefficients of
thermal expansion for aluminum and steel from Table 12.1 gives
L0al
α − α st
19 × 10–6 (C°) –1 − 12 × 10 –6 (C°) –1
=
=
= 0 .6
L0
α al − α st 23 × 10 –6 (C°) –1 − 12 × 10 –6 (C°)–1
______________________________________________________________________________
24. REASONING AND SOLUTION Let L0 be the length of the wire before heating. After
heating the wire will have stretched an amount ∆Lw = αwL0∆T, while the gap in the concrete
"stretches" an amount ∆Lc = αc L0∆T.
The net change in length of the wire is then ∆Lc − ∆Lw = (αc − αw) L0∆T. The additional
stress needed to produce this change is
Stress =
Y ( ∆Lc − ∆Lw )
L0
= Y (α c − α w ) ∆T
628 TEMPERATURE AND HEAT
Taking values for the coefficients of thermal expansion of concrete and steel from Table
12.1 and the value for Young’s modulus Y of aluminum from Table 10.1, we find that the
additional tension in the wire is
(
)
(
T = ( Stress ) π R 2 = Y (α c − α w )∆T π R 2
(
)
)
(
−1
−1
= 6.9 × 1010 N/m 2 ⎡⎢12 × 10−6 ( C° ) − 23 × 10−6 ( C° ) ⎤⎥ (185 °C − 35 °C ) π 3.0 × 10−4 m
⎣
⎦
)
2
= − 32 N
The new tension in the wire is, then, 50.0 N – 32 N = 18 N .
______________________________________________________________________________
25. REASONING AND SOLUTION The volume V0 of an object changes by an amount ∆ V
when its temperature changes by an amount ∆ T ; the mathematical relationship is given by
Equation 12.3: ∆V = β V0 ∆T . Thus, the volume of the kettle at 24 °C can be found by
solving Equation 12.3 for V0 . According to Table 12.1, the coefficient of volumetric
expansion for copper is 51 × 10–6 (C°)–1 . Solving Equation 12.3 for V0 , we have
∆V
1.2 × 10 –5 m3
=
= 3.1× 10 –3 m3
β ∆T [51×10 –6 (C°) –1 ](100 °C – 24 °C)
______________________________________________________________________________
V0 =
26. REASONING We can identify the liquid by computing the coefficient of volume
expansion β and then comparing the result with the values of β given in Table 12.1. The
relation ∆V = β V0 ∆T (Equation 12.3) can be used to calculate β .
SOLUTION Solving Equation 12.3 for β, we have that β = ∆V / (V0 ∆T ) . The change in the
volume of the liquid is ∆V = 1.500 L –1.383 L = 0.117 L . Therefore, the coefficient of
volume expansion for the unknown liquid is
β=
∆V
0.117 L
=
= 9.50 ×10 –4 (C°) –1 = 950 ×10 –6 (C°) –1
V0 ∆T (1.500 L)(97.1 °C –15.0 °C)
A comparison with the values of β in Table 12.1 indicates that the liquid is gasoline .
____________________________________________________________________________________________
27. REASONING The change DV in the interior volume of the shell is given by Equation 12.3
as DV = βV0DT, where β is the coefficient of volume expansion, V0 is the initial volume, and
DT is the increase in temperature. The interior volume behaves as if it were filled with the
Chapter 12 Problems
surrounding silver. The interior volume is spherical, and the volume of a sphere is
629
4 π r3 ,
3
where r is the radius of the sphere.
SOLUTION In applying Equation 12.3, we note that the initial spherical volume of the
interior space is V0 = 43 π r 3 , so that we have
∆V = β V0 ∆T = β
( 43 π r 3 ) ∆T
(
−1
= ⎡57 × 10−6 ( C° ) ⎤ 43 π 2.0 × 10−2 m
⎣
⎦
)
3
(147 °C − 18 °C ) =
2.5 × 10−7 m3
We have taken the coefficient of volume expansion for silver from Table 12.1.
28. REASONING The density ρ of a substance is defined as its mass m divided by its volume
V0 (Equation 11.1); ρ = m/V0. When the temperature rises, the volume increases by an
amount ∆V = β V0 ∆T (Equation 12.3), where β is the coefficient of volume expansion and
∆T is the change in temperature. Since the mass of the fluid does not change as the
temperature rises, the increase in volume means that the density decreases.
SOLUTION
a. The density of the fluid at 0 °C is
ρ=
m 825 kg
=
= 705 kg/m3
V0 1.17 m3
b. When the temperature rises to 20.0 °C from 0 °C the volume increases to V0 + ∆V, where
∆V = β V0 ∆T . Therefore, the density becomes
ρ=
=
m
m
m
=
=
V0 + ∆V V0 + β V0 ∆T V0 (1 + β ∆T )
825 kg
(1.17 m3 ) {1 + ⎡⎣1.26 ×10−3 ( C° )
−1 ⎤
}
= 688 kg/m3
⎦ ( 20.0 °C − 0 °C )
______________________________________________________________________________
29.
SSM
REASONING The increase DV in volume is given by Equation 12.3 as DV = βV0DT,
where β is the coefficient of volume expansion, V0 is the initial volume, and DT is the
increase in temperature. The lead and quartz objects experience the same change in volume.
630 TEMPERATURE AND HEAT
Therefore, we can use Equation 12.3 to express the two volume changes and set them equal.
We will solve the resulting equation for DTQuartz.
SOLUTION Recognizing that the lead and quartz objects experience the same change in
volume and expressing that change with Equation 12.3, we have
β LeadV0 ∆TLead = β QuartzV0 ∆TQuartz
∆VLead
∆V
Quartz
In this result V0 is the initial volume of each object. Solving for DTQuartz and taking values
for the coefficients of volume expansion for lead and quartz from Table 12.1 gives
−6
⎡
⎤
β Lead ∆TLead ⎣87 ×10 ( C° ) ⎦ ( 4.0 C° )
=
=
= 230 C°
−1
β Quartz
1.5 × 10−6 ( C° )
−1
∆TQuartz
30. REASONING AND SOLUTION Both the coffee and beaker expand as the temperature
increases. For the expansion of the coffee
∆Vc = βwV0 ∆T
and for the expansion of the beaker
∆Vb = βb V0 ∆T
The excess expansion of the coffee, hence the amount which spills, is
∆V = ∆Vc − ∆Vb = (βw − βb) V0 ∆T
Taking the coefficients of volumetric expansion bw and bc for coffee (water) and glass
(Pyrex) from Table 12.1, we find
∆V = [207 ×10 –6 (C°) –1 − 9.9 ×10 –6 (C°) –1 ](0.50 ×10 –3 m3 )(92 °C –18 °C)
= 7.3 × 10 –6 m3
______________________________________________________________________________
Chapter 12 Problems
631
31. SSM REASONING AND SOLUTION Both the gasoline and the tank expand as the
temperature increases. The coefficients of volumetric expansion bg and bs for gasoline and
steel are available in Table 12.1. According to Equation 12.3, the volume expansion of the
gasoline is
−1
∆Vg = β gV0 ∆T = ⎡⎢950 × 10−6 ( C° ) ⎤⎥ (20.0 gal)(18 C°) = 0.34 gal
⎣
⎦
while the volume of the steel tank expands by an amount
−1
∆Vs = βsV0 ∆T = ⎡⎢36 × 10−6 ( C° ) ⎤⎥ (20.0 gal)(18 C°) = 0.013 gal
⎣
⎦
The amount of gasoline which spills out is
∆Vg − ∆Vs = 0.33 gal
______________________________________________________________________________
32. REASONING AND SOLUTION Both the water and pipe expand as the temperature
increases. For the expansion of the water
∆Vw = bw V0 ∆T
and for the expansion of the pipe
∆Vp = bc V0 ∆T
The initial volume of the pipe and water is V0 = p r2L The reservoir then needs a capacity of
∆V = ∆Vw − ∆Vp = (bw − bc) V0 ∆T
Taking the coefficients of volumetric expansion bw and bc for water and copper from Table
12.1, we find
∆V = (207 × 10–6 C°–1 − 51 × 10–6 C°–1) p (9.5 × 10–3 m)2(76 m)(54 C°) = 1.8 × 10 –4 m3
______________________________________________________________________________
33. REASONING The change in volume of the cylindrical mercury column is proportional to
the coefficient of volume expansion β for mercury, its original volume V0, and the change
∆T in temperature. The volume of a cylinder of radius r and height ∆L is ∆V = π r 2 ∆L . The
coefficient of volume expansion for mercury can be found in Table 12.1.
SOLUTION The change in the volume of the mercury is
632 TEMPERATURE AND HEAT
π r ∆L = β V0 ∆T
2
(12.3)
∆V
Solving for ∆L gives
(
)
−4
3
⎡
⎤
β V0 ∆T ⎣1.82 × 10 ( C° ) ⎦ 45 mm (1.0 C° )
∆L =
=
= 9.0 mm
2
π r2
π 1.7 × 10−2 mm
−1
(
)
______________________________________________________________________________
34. REASONING AND SOLUTION
a. The apparent weight of the sphere will be larger
after
it cools. This is because the
sphere shrinks while cooling, displacing less water, and hence decreasing the buoyant force
acting on it.
b. The weight of the submerged sphere before cooling is
W0 = ρ gV0 − ρwgV0
The weight of the submerged sphere after cooling is
W = ρ gV0 − ρwgV
The weight difference is
∆W = W − W0 = −ρwg∆V
where ∆V = V − V0. The volume change is ∆V = β V0∆T. Now, ∆W = −ρwgβ V0∆T. The
sphere’s original volume is
V0 = (4/3)π r 3 = (4/3)(π)(0.50 m)3 = 0.52 m3
The coefficient of volumetric expansion b for aluminum is given in Table 12.1, so we have
∆W = −(1.0 × 103 kg/m3)(9.80 m/s2)[69 × 10–6 (C°)–1](0.52 m3)(−5.0 × 101 C°) = 18 N
______________________________________________________________________________
35.
REASONING In order to keep the water from expanding as its temperature
increases from 15 to 25 °C, the atmospheric pressure must be increased to compress the
water as it tries to expand. The magnitude of the pressure change ∆P needed to compress a
substance by an amount ∆V is, according to Equation 10.20, ∆P = B(∆V / V0 ). The ratio
∆V / V0 is, according to Equation 12.3, ∆V / V0 = β ∆T . Combining these two equations
yields
∆P = B β∆T
SSM
Chapter 12 Problems
633
SOLUTION Taking the value for the coefficient of volumetric expansion b for water from
Table 12.1, we find that the change in atmospheric pressure that is required to keep the
water from expanding is
−1
∆P = (2.2 ×109 N/m 2 ) ⎡ 207 × 10−6 ( C° ) ⎤ (25 °C − 15 °C)
⎣
⎦
1 atm
⎛
⎞
= 4.6 × 106 Pa ⎜
⎟ = 45 atm
5
⎝ 1.01× 10 Pa ⎠
______________________________________________________________________________
(
)
36. REASONING The change DV in volume is given by Equation 12.3 as DV = βV0DT, where β
is the coefficient of volume expansion, V0 is the initial volume, and DT is the increase in
temperature. The increase in volume of the mercury is given directly by this equation, with
V0 being the initial volume of the interior space of the brass shell minus the initial volume of
the steel ball. If the space occupied by the mercury did not change with temperature, the
spillage would simply be the increase in volume of the mercury. However, the space
occupied by the mercury does change with temperature. Both the brass shell and the steel
ball expand. The interior volume of the brass shell expands as if it were solid brass, and this
expansion provides more space for the mercury to occupy, thereby reducing the amount of
spillage. The expansion of the steel ball, in contrast, takes up space that would otherwise be
occupied by mercury, thereby increasing the amount of spillage. The total spillage, therefore,
is DVMercury - DVBrass + DVSteel.
SOLUTION Table 12.1 gives the coefficients of volume expansion for mercury, brass, and
steel. Applying Equation 12.3 to the mercury, the brass cavity, and the steel ball, we have
634 TEMPERATURE AND HEAT
Spillage = ∆VMercury − ∆VBrass + ∆VSteel
= β MercuryV0, Mercury ∆T − β BrassV0, Brass ∆T + βSteelV0, Steel ∆T
(
) (
)
= ⎡182 × 10−6 ( C° ) ⎤ ⎡ 1.60 ×10−3 m3 − 0.70 × 10−3 m3 ⎤ (12 C° )
⎦
⎣
⎦⎣
Mercury
−1
(
)
− ⎡57 ×10−6 ( C° ) ⎤ 1.60 × 10−3 m3 (12 C° )
⎣
⎦
Brass
−1
(
)
+ ⎡36 ×10−6 ( C° ) ⎤ 0.70 × 10−3 m3 (12 C° ) = 1.2 × 10−6 m3
⎣
⎦
Steel
−1
37.
REASONING The cavity that contains the liquid in either Pyrex
thermometer expands according to Equation 12.3, ∆Vg = β gV0 ∆T . On the other hand, the
volume of mercury expands by an amount ∆Vm = β mV0 ∆T , while the volume of alcohol
expands by an amount ∆Va = β a V0 ∆T . Therefore, the net change in volume for the mercury
thermometer is
∆Vm − ∆Vg = (β m − β g )V0 ∆T
SSM
WWW
while the net change in volume for the alcohol thermometer is
∆Va − ∆Vg = ( β a − β g )V0 ∆T
In each case, this volume change is related to a movement of the liquid into a cylindrical
region of the thermometer with volume πr 2 h , where r is the radius of the region and h is
the height of the region. For the mercury thermometer, therefore,
hm =
( β m − β g )V0 ∆T
πr
2
Similarly, for the alcohol thermometer
ha =
(β a − β g )V0∆T
πr
2
These two expressions can be combined to give the ratio of the heights, ha /hm .
Chapter 12 Problems
635
SOLUTION Taking the values for the coefficients of volumetric expansion for methyl
alcohol, Pyrex glass, and mercury from Table 12.1, we divide the two expressions for the
heights of the liquids in the thermometers and find that
ha
hm
=
βa − βg
βm
1200 × 10 –6 (C° ) −1 − 9.9 × 10 –6 (C° ) −1
=
= 6.9
− βg
182 × 10 –6 (C° ) −1 − 9.9 × 10 –6 (C° ) −1
Therefore, the degree marks are 6.9 times further apart on the alcohol thermometer than
on the mercury thermometer.
______________________________________________________________________________
When the temperature is 0.0 °C, P = ρ0gh0, and when
38. REASONING AND SOLUTION
the temperature is 38.0 °C, P = ρgh. Equating and solving for h gives h = (ρ0/ρ)h0. Now
ρ0/ρ = V/V0, since the mass of mercury in the tube remains constant. Then h = (V/V0)h0.
Now,
∆V = V − V0 = βV0∆T
or
V/V0 = 1 + β ∆T
Therefore,
h = (1 + β ∆T)h0 = {1 + [182 × 10−6 (C°)−1](38.0 °C − 0.0 °C)}(0.760 m) = 0.765 m
where we have taken the value for the coefficient of volumetric expansion b for mercury
from Table 12.1.
______________________________________________________________________________
39.
REASONING According to Equation 12.4, the heat required to warm the pool can
be calculated from Q = cm∆T . The specific heat capacity c of water is given in Table 12.2.
In order to use Equation 12.4, we must first determine the mass of the water in the pool.
Equation 11.1 indicates that the mass can be calculated from m = ρ V , where ρ is the
density of water and V is the volume of water in the pool.
SSM
SOLUTION Combining these two expressions, we have Q = cρ V∆T , or
Q = [4186 J/(kg ⋅C°) ](1.00 ×10 3 kg/m 3 )(12.0 m × 9.00 m × 1.5 m) (27 ° C − 15 °C) = 8.14 × 10 9 J
Using the fact that 1 kWh = 3.6 × 106 J , the cost of using electrical energy to heat the water
in the pool at a cost of $0.10 per kWh is
⎛ $0.10 ⎞
= $230.
⎝ 3.6 × 106 J ⎠
______________________________________________________________________________
9
(8.14 × 10 J)
636 TEMPERATURE AND HEAT
40. REASONING AND SOLUTION We require that the heat gained by the cold water equals
the heat lost by the hot water, i.e.,
Qcw = Qhw
Therefore,
ccwmcw∆Tcw = chwmhw∆Thw
Since the specific heat capacity is that of water in each case, ccw = chw, then
mcw(36.0 °C – 13.0 °C) = mhw(49.0 °C – 36.0 °C)
Then
mcw = (0.57)mhw
We also know that mcw + mhw = 191 kg. Substituting for mcw and solving for mhw, we have
191 kg
= 121 kg
1.57
______________________________________________________________________________
mhw =
41. REASONING AND SOLUTION The heat released by the blood is given by Q = cm∆T, in
which the specific heat capacity c of the blood (water) is given in Table 12.2. Then
∆T =
Q
2000 J
=
= 0.8 C°
cm [4186 J/(kg ⋅ C°)](0.6 kg)
Therefore,
Tf = Ti − ∆T = 36.2 °C
______________________________________________________________________________
42. REASONING We assume that no heat is lost through the chest to the outside. Then, energy
conservation dictates that the heat gained by the soda is equal to the heat lost by the
watermelon in reaching the final temperature Tf . Each quantity of heat is given by Equation
12.4, Q = cm∆T , where we write the change in temperature ∆T as the higher temperature
minus the lower temperature.
SOLUTION Starting with the statement of energy conservation, we have
Heat gained by soda = Heat lost by watermelon
(cm∆T )soda = (cm∆T )watermelon
Since the watermelon is being treated like water, we take the specific heat capacity of water
from Table 12.2. Thus, the above equation becomes
[3800 J/(kg ⋅ C°)](12 × 0.35 kg)(Tf – 5.0 °C) = [4186 J/(kg⋅ C°)](6.5 kg)(27 °C – Tf )
Chapter 12 Problems
637
Suppressing units for convenience and algebraically simplifying, we have
1.6 × 104 Tf – 8.0 × 10 4 = 7.3 × 10 5 – 2.7 × 10 4 Tf
Solving for Tf , we obtain
8.1× 10 5
= 19 °C
4.3 × 104
______________________________________________________________________________
Tf =
43. REASONING The metabolic processes occurring in the person’s body produce the heat that
is added to the water. As a result, the temperature of the water increases. The heat Q that
must be supplied to increase the temperature of a substance of mass m by an amount DT is
given by Equation 12.4 as Q = cmDT, where c is the specific heat capacity. The increase DT
in temperature is the final higher temperature Tf minus the initial lower temperature T0.
Hence, we will solve Equation 12.4 for the desired final temperature.
SOLUTION From Equation 12.4, we have
Q = cm∆T = cm (Tf − T0 )
Solving for the final temperature, noting that the heat is Q = (3.0 × 105 J/h)(0.50 h) and
taking the specific heat capacity of water from Table 12.2, we obtain
(
)
(
3.0 × 105 J/h ( 0.50 h )
Q
= 21.00 °C +
= 21.03 °C
Tf = T0 +
cm
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ 1.2 ×103 kg
)
44. REASONING Since the container of the glass and the liquid is being ignored and since we
are assuming negligible heat exchange with the environment, the principle of conservation of
energy applies. In reaching equilibrium the cooler liquid gains heat, and the hotter glass loses
heat. We will apply this principle by equating the heat gained to the heat lost. The heat Q that
must be supplied or removed to change the temperature of a substance of mass m by an
amount DT is given by Equation 12.4 as Q = cmDT, where c is the specific heat capacity. In
using this equation as we apply the energy-conservation principle, we must remember to
express the change in temperature DT as the higher minus the lower temperature.
SOLUTION Applying the energy-conservation principle and using Equation 12.4 give
cLiquid m∆TLiquid = cGlass m∆TGlass
Heat gained by liquid
Heat lost by glass
638 TEMPERATURE AND HEAT
Since it is the same for both the glass and the liquid, the mass m can be eliminated
algebraically from this equation. Solving for cLiquid and taking the specific heat capacity for
glass from Table 12.2, we find
cLiquid =
45.
cGlass ∆TGlass
∆TLiquid
⎡840 J/ ( kg ⋅ C° ) ⎤⎦ ( 83.0 C° − 53.0 C° )
=⎣
= 2500 J/ ( kg ⋅ C° )
( 53.0 C° − 43.0 C° )
REASONING Let the system be comprised only of the metal forging and the oil.
Then, according to the principle of energy conservation, the heat lost by the forging equals
the heat gained by the oil, or Qmetal = Qoil . According to Equation 12.4, the heat lost by the
forging is Qmetal = cmetal mmetal (T0metal − Teq ) , where Teq is the final temperature of the
system at thermal equilibrium. Similarly, the heat gained by the oil is given by
Qoil = coil moil (Teq − T0oil ) .
SSM
SOLUTION
Qmetal = Qoil
c metal mmetal (T0metal − Teq ) = c oil moil (Teq − T0oil )
Solving for T0metal , we have
T0metal =
c oil moil (Teq − T0oil )
c metal mmetal
+ Teq
or
[2700 J/(kg ⋅C°)](710 kg)(47 °C − 32 °C)
+ 47 °C = 940 °C
[430 J/(kg ⋅C°)](75 kg)
______________________________________________________________________________
T0metal =
46. REASONING When heat Q is supplied to the silver bar, its temperature changes by an
amount ∆T. The relation between Q and ∆T is given by Equation 12.4 as Q = c m ∆T , where
c is the specific heat capacity of silver and m is the mass. Solving this equation for m yields
m=
Q
c ∆T
(1)
When the temperature of the bar changes by an amount ∆T, the change ∆L in its length is
given by Equation 12.2 as ∆L = α L0 ∆T . Solving this equation for ∆T gives
∆T =
∆L
α L0
(2)
Chapter 12 Problems
639
where L0 is the initial length and α is the coefficient of linear expansion for silver.
Substituting the expression for ∆T in Equation (2) into Equation (1) gives
m=
α Q L0
Q
Q
=
=
c ∆T
c ∆L
⎛ ∆L ⎞
c⎜
⎟
⎝ α L0 ⎠
SOLUTION Taking the values for α and c from Tables 12.1 and 12.2, respectively, the
mass of the silver bar is
m=
α Q L0
=
⎡⎣19 ×10−6 ( C° )−1 ⎤⎦ ( 4200 J )( 0.15 m )
= 1.2 × 10−2 kg
⎡⎣ 235 J/ ( kg ⋅ C° ) ⎤⎦ ( 4.3 × 10 m )
______________________________________________________________________________
c ∆L
−3
47. REASONING According to Equation 6.10b, the average power is the change in energy
divided by the time. The change in energy in this problem is the heat supplied to the water
and the coffee mug to raise their temperature from 15 to 100 ºC, which is the boiling point of
water. The time is given as three minutes (180 s). The heat Q that must be added to raise the
temperature of a substance of mass m by an amount DT is given by Equation 12.4 as
Q = cmDT, where c is the specific heat capacity. This equation will be used for the water and
the material of which the mug is made.
SOLUTION Using Equation 6.10b, we write the average power P as
P=
Change in energy QWater + QMug
=
Time
Time
The heats QWater and QMug each can be expressed with the aid of Equation 12.4, so that we
obtain
QWater + QMug cWater mWater ∆T + cMug mMug ∆T
P=
=
Time
Time
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ ( 0.25 kg )(100.0 °C − 15 °C )
=
+ ⎡⎣920 J/ ( kg ⋅ C° ) ⎤⎦ ( 0.35 kg )(100.0 °C − 15 °C )
180 s
= 650 W
The specific heat of water has been taken from Table 12.2.
48. REASONING When heat Q is supplied to the block, its temperature changes by an amount
∆T. The relation between Q and ∆T is given by
640 TEMPERATURE AND HEAT
Q = c m ∆T
(12.4)
where c is the specific heat capacity and m is the mass. When the temperature of the block
changes by an amount ∆T, the change ∆V in its volume is given by Equation 12.3 as
∆V = β V0 ∆T , where β is the coefficient of volume expansion and V0 is the initial volume
of the block. Solving for ∆T gives
∆T =
∆V
β V0
Substituting this expression for ∆T into Equation 12.4 gives
⎛ ∆V ⎞
Q = c m ∆T = c m ⎜
⎟
⎝ β V0 ⎠
SOLUTION The heat supplied to the block is
⎛ ∆V ⎞ ⎡⎣ 750 J/ ( kg ⋅ C° ) ⎤⎦ (130 kg ) (1.2 ×10 m )
Q = cm⎜
=
= 4.0 × 105 J
⎟
−1
−
−
5
2
3
⎡⎣6.4 × 10 ( C° ) ⎤⎦ ( 4.6 × 10 m )
⎝ β V0 ⎠
______________________________________________________________________________
−5
49.
3
REASONING AND SOLUTION According to Equation 12.4, the total heat per
kilogram required to raise the temperature of the water is
SSM
Q
= c∆T = [4186 J/(kg⋅ C°)] (32.0 °C) = 1.34 × 10 5 J/kg
m
The mass flow rate, ∆m / ∆t , is given by Equations 11.7 and 11.10 as ∆m / ∆t = ρAv = ρQv ,
where ρ and Qv are the density and volume flow rate, respectively. We have, therefore,
∆m
3
3
–6
3
–3
= (1.000 × 10 kg/m )(5.0 × 10 m /s) = 5.0 × 10 kg/s
∆t
Therefore, the minimum power rating of the heater must be
(1.34 × 10 5 J/kg)(5.0 × 10–3 kg/s) = 6.7 × 10 2 J/s = 6.7 × 10 2 W
______________________________________________________________________________
50. REASONING AND SOLUTION The steel ball and aluminum plate expand when heat is
added to the system. Once the correct amount of heat has been added, the two have the
same radius and the ball fits into the plate. Upon heating, the change in length is given by
∆L = α L0∆T, where ∆L = (Lf − L0).
Chapter 12 Problems
641
We require that the final lengths of the steel and aluminum be equal, i.e., (Lf)st = (Lf)al.
Substituting for ∆L and equating the final lengths yields
(Lf)st = (L0)st(1 + αst∆T) = (Lf)al = (L0)al(1 + αal∆T)
Solving for ∆T and using values for the coefficients of thermal expansion for aluminum and
steel from Table 12.1, we find that
∆T =
0.0010
= 91 C°
α al − (1.0010 ) α st
Taking the values for the specific heat capacities for steel and aluminum from Table 12.2,
we find that the amount of heat that must be added is, therefore,
Q = (cm∆T)st + (cm∆T)al
Q = [452 J/(kg⋅C°)](1.5 kg) (91 C°) + [9.00 × 102 J/(kg⋅C°)](0.85 kg)(91 C°) = 1.3 × 105 J
______________________________________________________________________________
51.
REASONING Heat Q1 must be added to raise the temperature of the aluminum in
its solid phase from 130 °C to its melting point at 660 °C. According to Equation 12.4,
Q1 = cm∆T . The specific heat c of aluminum is given in Table 12.2. Once the solid
aluminum is at its melting point, additional heat Q2 must be supplied to change its phase
from solid to liquid. The additional heat required to melt or liquefy the aluminum is
Q2 = mLf , where Lf is the latent heat of fusion of aluminum. Therefore, the total amount of
heat which must be added to the aluminum in its solid phase to liquefy it is
SSM
Qtotal = Q1 + Q2 = m(c∆T + Lf )
SOLUTION Substituting values, we obtain
{
}
Qtotal = (0.45 kg) [9.00 × 102 J/(kg ⋅ C°)](660 °C –130 °C) + 4.0 × 105 J/kg = 3.9 × 105 J
______________________________________________________________________________
52. REASONING
a. When water changes from the liquid to the ice phase at 0 °C, the amount of heat released
is given by Q = mLf (Equation 12.5), where m is the mass of the water and Lf is its latent
heat of fusion.
b. When heat Q is supplied to the tree, its temperature changes by an amount ∆T. The
relation between Q and ∆T is given by Equation 12.4 as Q = c m ∆T , where c is the specific
642 TEMPERATURE AND HEAT
heat capacity and m is the mass. This equation can be used to find the change in the tree’s
temperature.
SOLUTION
a. Taking the latent heat of fusion for water as Lf = 33.5 × 104 J/kg (see Table 12.3), we
find that the heat released by the water when it freezes is
Q = mLf = ( 7.2 kg ) ( 33.5 ×104 J/kg ) = 2.4 ×106 J
b. Solving Equation 12.4 for the change in temperature, we have
Q
2.4 × 106 J
=
= 5.3 C°
c m ⎡⎣ 2.5 × 103 J/ ( kg ⋅ C° ) ⎤⎦ (180 kg )
______________________________________________________________________________
∆T =
53. REASONING AND SOLUTION
a. The latent heat of vaporization Lv of water is given in Table 12.3. To change water at
100.0 °C to steam we have from Equation 12.5 that
Q = mLv = (2.00 kg)(22.6 × 105 J/kg) = 4.52 × 106 J
b. For liquid water at 0.0 °C we must include the heat needed to raise the temperature to the
boiling point. This heat is depends on the specific heat c of water, which is given in Table
12.2. Using Equations 12.4 and 12.5, we have Q = (cm∆T)water + mLv
Q = [4186 J/(kg⋅C°)](2.00 kg)(100.0 C°) + 4.52 × 106 J= 5.36 × 10 6 J
______________________________________________________________________________
54. REASONING As the body perspires, heat Q must be added to change the water from the
liquid to the gaseous state. The amount of heat depends on the mass m of the water and the
latent heat of vaporization Lv, according to Q = mLv (Equation 12.5).
SOLUTION The mass of water lost to perspiration is
⎛ 4186 J ⎞
⎟
⎝ 1 Calorie ⎠
( 240 Calories ) ⎜
Q
=
= 0.42 kg
6
Lv
2.42 × 10 J/kg
______________________________________________________________________________
m=
55.
REASONING From the principle of conservation of energy, the heat lost by the
coin must be equal to the heat gained by the liquid nitrogen. The heat lost by the silver coin
SSM
Chapter 12 Problems
643
is, from Equation 12.4, Q = ccoin mcoin ∆Tcoin (see Table 12.2 for the specific heat capacity
of silver). If the liquid nitrogen is at its boiling point, –195.8 °C, then the heat gained by the
nitrogen will cause it to change phase from a liquid to a vapor. The heat gained by the
liquid nitrogen is Q = mnitrogen Lv , where mnitrogen is the mass of liquid nitrogen that
vaporizes, and Lv is the latent heat of vaporization for nitrogen (see Table 12.3).
SOLUTION
Qlost by = Qgained by
coin
nitrogen
c coin mcoin ∆Tcoin = mnitrogen Lv
Solving for the mass of the nitrogen that vaporizes
m nitrogen =
=
c coin m coin ∆Tcoin
Lv
[235 J/(kg⋅ C°)](1.5 × 10 –2 kg)[25 °C – (–195.8 °C)]
= 3.9 × 10 −3 kg
2.00 × 10 J/kg
______________________________________________________________________________
5
56. REASONING AND SOLUTION The block of ice must undergo a change in temperature,
followed by a change in phase, and then another change in temperature,
Q = Qice + Qice-to-water + Qwater
Q = (cm∆T)ice + mLf + (cm∆T)water
Table 12.2 gives values for the specific heats of ice and water. Table 12.3 gives the latent
heat of fusion of water. Using these values, we have
4.11 × 106 J = [2.00 × 103 J/(kg⋅C°)](10.0 kg)(10.0 C°) + (10.0 kg)(33.5 × 104 J/kg)
+ [4186 J/(kg⋅C°)](10.0 kg)Tf
Solving for Tf we obtain, Tf = 13 °C .
______________________________________________________________________________
57. REASONING AND SOLUTION The heat required is Q = mLf + cm∆T, where m = rV.
See Table 12.2 for the specific heat c and Table 12.3 for the latent heat Lf. Thus,
Q = rVLf + crV∆T
Q = (917 kg/m3)(4.50 × 10–4 m)(1.25 m2){33.5 × 104 J/kg
644 TEMPERATURE AND HEAT
+ [2.00 × 103 J/(kg⋅C°)](12.0 C°)} = 1.85 × 10 5 J
______________________________________________________________________________
58. REASONING Since there is no heat lost or gained by the system, the heat lost by the
coffee in cooling down must be equal to the heat gained by the ice as it melts plus the heat
gained by the melted water as it subsequently heats up. The heat Q lost or gained by a
substance is given by Equation 12.4 as Q = cm∆T, where c is the specific heat capacity (see
Table 12.2), m is the mass, and ∆T is the change in temperature. The heat that is required to
change ice at 0 °C into liquid water at 0 °C is given by Equation 12.5 as Q = miceLf, where
mice is the mass of ice and Lf is the latent heat of fusion for water (see Table 12.3). Thus,
we have that
mcoffee ccoffee ∆Tcoffee = mice Lf + mice cwater ∆Twater
Heat lost by coffee
Heat gained by ice
and liquid water
The mass of the coffee can be expressed in terms of its density as mcoffee= ρcoffeeVcoffee
(Equation 11.1). The change in temperature of the coffee is ∆Tcoffee= 85 °C – T, where T is
the final temperature of the coffee. The change in temperature of the water is ∆Twater =
T – 0 °C. With these substitutions, the equation above becomes
ρcoffeeVcoffee ccoffee ( 85 °C − T ) = mice Lf + mice cwater (T − 0 °C )
Solving this equation for the final temperature gives
T=
− mice Lf + ρ coffeeVcoffee ccoffee ( 85 °C )
ρcoffeeVcoffee ccoffee + mice cwater
− ( 2 ) (11 × 10−3 kg )( 3.35 × 105 J/kg ) + (1.0 × 103 kg/m3 )(150 × 10−6 m3 ) ⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ ( 85 °C )
=
(1.0 × 103 kg/m3 )(150 × 10−6 m3 ) ⎡⎣4186 J/ ( kg ⋅ C°)⎤⎦ + ( 2 ) (11 × 10−3 kg ) ⎡⎣ 4186 J/ ( kg ⋅ C°)⎤⎦
= 64 °C
______________________________________________________________________________
59.
REASONING According to the statement of the problem, the initial state of the
system is comprised of the ice and the steam. From the principle of energy conservation,
the heat lost by the steam equals the heat gained by the ice, or Qsteam = Qice . When the ice
and the steam are brought together, the steam immediately begins losing heat to the ice. An
amount Q1(lost) is released as the temperature of the steam drops from 130 °C to 100 °C, the
boiling point of water. Then an amount of heat Q2(lost) is released as the steam condenses
SSM
Chapter 12 Problems
645
into liquid water at 100 °C. The remainder of the heat lost by the "steam" Q3(lost) is the heat
that is released as the water at 100 °C cools to the equilibrium temperature of
Teq = 50.0 °C . According to Equation 12.4, Q1(lost) and Q3(lost) are given by
Q1(lost) = csteam msteam (Tsteam − 100.0 °C) and Q3(lost) = cwater msteam (100.0 °C – Teq )
Q2(lost) is given by Q2(lost) = msteam Lv , where Lv is the latent heat of vaporization of water.
The total heat lost by the steam has three effects on the ice. First, a portion of this heat
Q1(gained) is used to raise the temperature of the ice to its melting point at 0.00 °C. Then, an
amount of heat Q2(gained) is used to melt the ice completely (we know this because the
problem states that after thermal equilibrium is reached the liquid phase is present at
50.0 °C). The remainder of the heat Q3(gained) gained by the "ice" is used to raise the
temperature of the resulting liquid at 0.0 °C to the final equilibrium temperature. According
to Equation 12.4, Q1(gained) and Q3(gained) are given by
Q1(gained) = cice mice (0.00 °C – Tice ) and Q3(gained) = cwater mice (Teq – 0.00 °C)
Q2(gained) is given by Q2(gained) = mice Lf , where Lf is the latent heat of fusion of ice.
SOLUTION According to the principle of energy conservation, we have
Qsteam = Qice
Q1(lost) + Q2(lost) + Q3(lost) = Q1(gained) + Q2(gained) + Q3(gained)
or
csteam msteam (Tsteam − 100.0 °C) + msteam Lv + cwater msteam (100.0 °C – Teq )
= cice mice (0.00 °C –Tice ) + mice Lf + cwater mice (Teq – 0.00 °C)
Values for specific heats are given in Table 12.2, and values for the latent heats are given in
Table 12.3. Solving for the ratio of the masses gives
msteam
mice
=
cice (0.00 °C –Tice ) + Lf + cwater (Teq – 0.00 °C)
csteam (Tsteam − 100.0 °C) + Lv + cwater (100.0 °C – Teq )
646 TEMPERATURE AND HEAT
⎡ 2.00 × 103 J/ ( kg ⋅ C° ) ⎤ [ 0.0 °C–(–10.0°C) ] + 33.5 ×104 J/kg + ⎡⎣ 4186 J/ ( kg ⋅ C° ) ) ⎤⎦ (50.0 °C–0.0 °C)
⎣
⎦
=
⎡⎣ 2020 J/ ( kg ⋅ C° ) ⎤⎦ (130 °C − 100.0 °C) + 22.6 × 105 J/kg + ⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ (100.0 °C–50.0 °C)
or
msteam
= 0.223
mice
______________________________________________________________________________
60. REASONING To freeze either liquid, heat must be removed to cool the liquid to its freezing
point. In either case, the heat Q that must be removed to lower the temperature of a
substance of mass m by an amount DT is given by Equation 12.4 as Q = cmDT, where c is the
specific heat capacity. The amount DT by which the temperature is lowered is the initial
temperature T0 minus the freezing point temperature T. Once the liquid has been cooled to
its freezing point, additional heat must be removed to convert the liquid into a solid at the
freezing point. The heat Q that must be removed to freeze a mass m of liquid into a solid is
given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion. The total heat to be
removed, then, is the sum of that specified by Equation 12.4 and that specified by
Equation 12.5, or QTotal = cm (T0 - T) + mLf. Since we know that same amount of heat is
removed from each liquid, we can set QTotal for liquid A equal to QTotal for liquid B and
solve the resulting equation for Lf, A - Lf, B.
SOLUTION Setting QTotal for liquid A equal to QTotal for liquid B gives
cAm (T0 - TA) + mLf, A = cBm (T0 - TB) + mLf, B
Noting that the mass m can be eliminated algebraically from this result and solving for
Lf, A - Lf, B, we find
Lf, A − Lf, B = cB (T0 − TB ) − cA (T0 − TA )
= ⎡⎣ 2670 J/ ( kg ⋅ C° ) ⎤⎦ ⎡⎣ 25.0 °C − ( −96.0 °C ) ⎤⎦
− ⎡⎣1850 J/ ( kg ⋅ C° ) ⎤⎦ ⎡⎣ 25.0 °C − ( −68.0 °C ) ⎤⎦ = 1.51× 105 J/kg
61. REASONING Since all of the heat generated by friction goes into the block of ice, only this
heat provides the heat needed to melt some of the ice. Since the surface on which the block
slides is horizontal, the gravitational potential energy does not change, and energy
conservation dictates that the heat generated by friction equals the amount by which the
kinetic energy decreases or QFriction = 12 Mv02 − 12 Mv 2 , where v0 and v are, respectively, the
initial and final speeds and M is the mass of the block. In reality, the mass of the block
Chapter 12 Problems
647
decreases as the melting proceeds. However, only a very small amount of ice melts, so we
may consider M to be essentially constant at its initial value. The heat Q needed to melt a
mass m of water is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion.
Thus, by equating QFriction to mLf and solving for m, we can determine the mass of ice that
melts.
SOLUTION Equating QFriction to mLf and solving for m gives
QFriction = 12 Mv02 − 12 Mv 2 = mLf
m=
(
M v02 − v 2
2 Lf
) = ( 42 kg ) ⎡⎣( 7.3 m/s )2 − ( 3.5 m/s )2 ⎤⎦ = 2.6 ×10−3 kg
2 ( 33.5 × 104 J/kg )
We have taken the value for the latent heat of fusion for water from Table 12.3.
62. REASONING The gold solidifies at its melting point. Therefore, its temperature does not
change, and the heat that it loses is given by Equation 12.5 as Qg = mg Lf, g , where mg and
Lf, g are, respectively, the mass and latent heat of fusion of gold.
The heat gained by the water is used in two ways:
a. Some of it raises the temperature of the water from 23.0 °C to its boiling point
at 100.0 °C. The amount of heat required is given by Q = cw mw ∆T (Equation
12.4), where cw and mw are, respectively, the specific heat capacity and mass of
water, and ∆T is the change in its temperature.
b. Some of it changes the phase of the water at 100.0 °C from a liquid to a vapor.
The amount of heat required is given by Q = mw Lv, w (Equation 12.5), where
Lv, w is the latent heat of vaporization of water.
From energy conservation, the heat lost by the gold is equal to the heat gained by the water.
Thus, we have
mg Lf, g = cw mw ∆T + mw Lv, w
(1)
Heat lost
by gold
Heat gained by water
This equation can be solved for the mass of the water.
SOLUTION Solving Equation (1) for the mass of the water, we obtain
648 TEMPERATURE AND HEAT
mw =
mg Lf, g
cw ∆T + Lv, w
=
( 0.180 kg ) ( 6.28 ×104
J/kg )
[4186 J/(kg ⋅ C°)](100.0 °C − 23.0 °C) + 22.6 ×10 J/kg
5
= 4.38 × 10−3 kg
_____________________________________________________________________________________________
63.
REASONING The system is comprised of the unknown material, the
glycerin, and the aluminum calorimeter. From the principle of energy conservation, the heat
gained by the unknown material is equal to the heat lost by the glycerin and the calorimeter.
The heat gained by the unknown material is used to melt the material and then raise its
temperature from the initial value of –25.0 °C to the final equilibrium temperature of
Teq = 20.0 °C .
SSM
WWW
SOLUTION
Qgained by = Qlost by
unknown
m u Lf + c u mu ∆Tu
+ Qlost by
glycerine
=
calorimeter
c gl m gl ∆Tgl + c al m al ∆Tal
Taking values for the specific heat capacities of glycerin and aluminum from Table 12.2, we
have
(0.10 kg) Lf + [160 J/(kg ⋅ C ° )](0.10 kg)(45.0 C° ) = [2410 J/(kg ⋅ C °)](0.100 kg)(7.0 C °)
+ [9.0 × 10 2 J/(kg ⋅ C° )](0.150 kg)(7.0 C °)
Solving for Lf yields,
Lf = 1.9 × 10 4 J/kg
______________________________________________________________________________
64. REASONING AND SOLUTION Using the value given for the specific heat c of water
given in Table 12.2, we find that the energy released by the water is
Q = (cm∆T)water + mLf
Q = [4186 J/(kg⋅C°)](840 kg)(10.0 C°) + (840 kg)(3.35 × 105 J/kg) = 3.2 × 10 8 J
The 2.0-kW heater provides 2.0 × 103 J/s, so that the time is
⎛ 3.2 × 10 8 J ⎞ ⎛ 1 hr ⎞
⎟ = 44 hr
⎜
⎟⎜
⎝ 2.0 × 10 3 J/s ⎠ ⎝ 3600 s ⎠
______________________________________________________________________________
65.
REASONING In order to melt, the bullet must first heat up to 327.3 °C (its melting
point) and then undergo a phase change. According to Equation 12.4, the amount of heat
necessary to raise the temperature of the bullet to 327.3 °C is Q = cm(327.3 °C − 30.0 °C) ,
SSM
Chapter 12 Problems
649
where m is the mass of the bullet. The amount of heat required to melt the bullet is given by
Qmelt = mLf , where Lf is the latent heat of fusion of lead.
The lead bullet melts completely when it comes to a sudden halt; all of the kinetic energy of
the bullet is converted into heat; therefore,
KE = Q + Qmelt
1
mv 2
2
= cm(327.3 ° C − 30.0 °C) + mL f
The value for the specific heat c of lead is given in Table 12.2, and the value for the latent
heat of fusion Lf of lead is given in Table 12.3. This expression can be solved for v, the
minimum speed of the bullet for such an event to occur.
SOLUTION Solving for v, we find that the minimum speed of the lead bullet is
v = 2 Lf + 2 c (327.3 °C − 30.0 °C)
v = 2(2.32 × 10 4 J/kg) + 2[128 J/(kg ⋅ C°)] (327.3 °C − 30.0 °C) = 3.50 × 102 m/s
______________________________________________________________________________
66. REASONING AND SOLUTION The steel band must be heated so that it can expand to fit
the wheel. The diameter of the band must increase in length by an amount
∆L = 6.00 × 10 –4 m . Also, ∆L = α L0∆T, where the coefficient of thermal expansion a of
steel is given in Table 12.1, so that
∆T =
∆L
6.00 ×10−4 m
=
= 5.0 × 101 C°
6
1
−
−
α L0 (12 ×10 C° ) (1.00 m )
The heat to expand the steel band comes from the heat released from the steam as it changes
to water. Therefore Qstm = Qsb, or
mstmLv + (cmstm∆T)water = cmsb∆T
Substituting the values for the latent heat of vaporization of water (see Table 12.3), the
specific heat capacity of water (see Table 12.2), and the specific heat capacity of steel (see
Table 12.2) gives
mstm(22.6 × 105 J/kg) + [4186 J/(kg⋅C°)]mstm(100.0 °C − 70.0 °C)
= [452 J/(kg⋅C°)](25.0 kg)(50.0 C°)
650 TEMPERATURE AND HEAT
Solving for the mass of the steam, we obtain mstm = 0.237 kg .
______________________________________________________________________________
67.
SSM
REASONING AND SOLUTION From inspection of the graph that accompanies
this problem, a pressure of 3.5 × 106 Pa corresponds to a temperature of 0 °C. Therefore,
liquid carbon dioxide exists in equilibrium with its vapor phase at 0 °C when the vapor
pressure is 3.5 × 106 Pa .
______________________________________________________________________________
68. REASONING The definition of percent relative humidity is given by Equation 12.6 as
follows:
Partial pressure of water vapor
×100
Percent relative humidity =
Equilibrium vapor pressure of
water at the existing temperature
Using R to denote the percent relative humidity, P to denote the partial pressure of water
vapor, and PV to denote the equilibrium vapor pressure of water at the existing temperature,
we can write Equation 12.6 as
P
R=
× 100
PV
The partial pressure of water vapor P is the same at the two given temperatures. The relative
humidity is not the same at the two temperatures, however, because the equilibrium vapor
pressure PV is different at each temperature, with values that are available from the vapor
pressure curve given with the problem statement. To determine the ratio R10/R40, we will
apply Equation 12.6 at each temperature.
SOLUTION Using Equation 12.6 and reading the values of PV, 10 and PV, 40 from the vapor
pressure curve given with the problem statement, we find
R10
R40
=
P / PV, 10
P / PV, 40
=
PV, 40
PV, 10
=
7200 Pa
= 5.5
1300 Pa
69. REASONING AND SOLUTION
⎛
⎞
Partial pressure
Percent relative humidity = ⎜
⎟ × 100 = 35%
⎝ Equilibrium vapor pressure ⎠
Using the vaporization curve from Problem 68, we see that at 27 °C, the equilibrium vapor
pressure is 3800 Pa. The partial pressure is, therefore,
Chapter 12 Problems
Partial pressure =
=
651
(Percent relative humidity )(Equilibrium vapor pressure )
100
(35% )(3800 Pa )
100
= 1300 Pa
From the vaporization curve accompanying Problem 68 we see that the dew point (100%
relative humidity at equilibrium temperature) is
Dew Point = 10 °C
______________________________________________________________________________
70. REASONING The definition of relative humidity is given by Equation 12.6 as:
Percent relative humidity =
Partial pressure of water vapor
×100
Equilibrium vapor pressure of
water at the existing temperature
The partial pressure of water vapor is given in the problem statement. The equilibrium vapor
pressure can be found by consulting the vapor pressure curve for water that accompanies
problem 68.
SOLUTION By using the vapor pressure curve for water given in problem 68, we estimate
that at a temperature of 37 °C the water vapor in the lungs has an equilibrium vapor pressure
of 6.3 × 103 Pa. The relative humidity is, then,
⎛ 5.5 ×103 Pa ⎞
Percent relative humidity = ⎜
⎟ ×100 = 87%
⎝ 6.3 ×103 Pa ⎠
______________________________________________________________________________
71. REASONING To bring the water to the point where it just begins to boil, its temperature
must be increased to the boiling point. The heat Q that must be added to raise the
temperature of a substance of mass m by an amount DT is given by Equation 12.4 as
Q = cmDT, where c is the specific heat capacity. The amount DT by which the temperature
changes is the boiling temperature minus the initial temperature of 100.0 ºC. The boiling
temperature is the temperature at which the vapor pressure of the water equals the external
pressure of 3.0 × 105 Pa and can be read from the vapor pressure curve for water given in
Figure 12.33.
SOLUTION Using Equation 12.4, with TBP being the boiling temperature and T0 being the
initial temperature, we have
Q = cm∆T = cm (TBP − T0 )
652 TEMPERATURE AND HEAT
According to Figure 12.33, an external pressure of 3.0 × 105 Pa corresponds to a boiling
point temperature of TBP = 134 ºC. Using this value in Equation 12.4 and taking the specific
heat capacity for water from Table 12.2, we determine the heat to be
Q = cm (TBP − T0 ) = ⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ ( 2.0 kg )(134 °C − 100.0 °C ) = 2.8 ×105 J
72. REASONING AND SOLUTION Equation 12.6 defines the relative humidity as
Partial pressure
of water vapor
Percent relative humidity =
× 100
Equilibrium vapor pressure of
water at the existing temeprature
According to the vapor pressure curve that accompanies Problem 68, the equilibrium vapor
pressure of water at 36°C is 5.8 x 103 Pa. Since the water condenses on the coils when the
temperature of the coils is 30 °C, the relative humidity at 30 °C is 100 percent. From
Equation 12.6 this implies that the partial pressure of water vapor in the air must be equal to
the equilibrium vapor pressure of water at 30 °C. From the vapor pressure curve, this
pressure is 4.4 x 103 Pa. Thus, the relative humidity in the room is
4.4 × 10 3 Pa
× 100 = 76%
5.8 × 10 3 Pa
______________________________________________________________________________
Percent relative humidity =
73. REASONING AND SOLUTION At 10 °C the equilibrium vapor pressure is 1250 Pa. At
25 °C, the equilibrium vapor pressure is 3200 Pa. To get the smallest possible value for the
relative humidity assume that the air at 10 °C is saturated. That is, take the partial pressure
to be 1250 Pa (see the vapor pressure curve for water that accompanies Problem 68). Then
we have,
⎛
⎞
⎛ 1250 Pa ⎞
Partial pressure
Percent relative humidity = ⎜
⎟ × 100 = ⎜
⎟ × 100 = 39%
⎝ Equilibrium vapor pressure ⎠
⎝ 3200 Pa ⎠
______________________________________________________________________________
74. REASONING AND SOLUTION The pressure inside the container is due to the weight on
the piston in addition to the pressure of the atmosphere. The 120-kg mass produces a
pressure of
P=
F mg (120 kg ) ( 9.80 m/s 2 )
=
=
= 1.0 × 105 Pa
2
A π r2
π ( 0.061 m )
If we include atmospheric pressure (1.01 × 105 Pa), the total pressure inside the container is
Ptotal = P + Patm = 2.0 × 105 Pa
Chapter 12 Problems
653
Examination of the vaporization curve for water in Figure 12.32 shows that the temperature
corresponding to this pressure at equilibrium is T = 120 °C .
______________________________________________________________________________
75.
REASONING We must first find the equilibrium temperature Teq of the iced tea.
Once this is known, we can use the vapor pressure curve that accompanies Problem 68 to
find the partial pressure of water vapor at that temperature and then estimate the relative
humidity using Equation 12.6.
SSM
According to the principle of energy conservation, when the ice is mixed with the tea, the
heat lost by the tea is gained by the ice, or Qtea = Qice . The heat gained by the ice is used to
melt the ice at 0.0 °C; the remainder of the heat is used to bring the water at 0.0 °C up to the
final equilibrium temperature Teq .
SOLUTION
Qtea = Qice
c water mtea (30.0 °C − Teq ) = m ice Lf + cwater mice ( Teq – 0.00 °C )
The specific heat capacity of water is given in Table 12.2, and the latent heat of fusion Lf of
water is given in Table 12.3. Solving for Teq , we have
Teq =
=
c water m tea (30.0 °C) − m ice Lf
c water (m tea + mice )
[4186 J/(kg ⋅C°) ] (0.300 kg)(30.0 °C) − (0.0670 kg)(33.5 × 10 4
[4186 J/(kg ⋅C° )] (0.300 kg + 0.0670 kg)
J/kg)
= 9.91 °C
According to the vapor pressure curve that accompanies Problem 68, at a temperature of
9.91 °C, the equilibrium vapor pressure is approximately 1250 Pa. At 30 °C, the
equilibrium vapor pressure is approximately 4400 Pa. Therefore, according to Equation
12.6, the percent relative humidity is approximately
⎛ 1250 Pa ⎞
Percent relative humidity = ⎜
⎟ × 100 = 28%
⎝ 4400 Pa ⎠
______________________________________________________________________________
76. REASONING AND SOLUTION The water will boil if the vapor pressure of the water is
equal to the ambient pressure. The pressure at a depth h in the water can be determined from
Equation 11.4: P2 = P1 + ρ gh . When h = 10.3 m,
P2 = (1.01× 105 Pa ) + ⎡⎣(1.000 × 103 kg/m3 ) ( 9.80 m/s 2 ) (10.3 m ) ⎤⎦ = 2.02 × 105 Pa
654 TEMPERATURE AND HEAT
The vapor pressure curve in Figure 12.32 shows that the vapor pressure of water is equal to
2.02 × 105 Pa at a temperature of 123 °C. Thus, the water at that depth has a temperature of
T = 123 °C .
______________________________________________________________________________
77. REASONING AND SOLUTION Using the value for the coefficient of thermal expansion
of steel given in Table 12.1, we find that the linear expansion of the aircraft carrier is
∆L = αL0 ∆T = (12 × 10–6 C°–1)(370 m)(21 °C − 2.0 °C) = 0.084 m
(12.2)
______________________________________________________________________________
78. REASONING AND SOLUTION The heat required to evaporate the water is Q = mLv, and
to lower the temperature of the jogger we have Q = mjc∆T. Equating these two expressions
and solving for the mass m of the water, we have
m=
m=
m jc ∆T
Lv
( 75 kg ) [3500 J/(kg ⋅ C°)] (1.5 C° )
= 0.16 kg
2.42 ×106 J/kg
______________________________________________________________________________
79.
REASONING From the conservation of energy, the heat lost by the mercury is
equal to the heat gained by the water. As the mercury loses heat, its temperature decreases;
as the water gains heat, its temperature rises to its boiling point. Any remaining heat gained
by the water will then be used to vaporize the water.
SSM
According to Equation 12.4, the heat lost by the mercury is Qmercury = (cm∆T )mercury . The
heat required to vaporize the water is, from Equation 12.5, Qvap = (m vap Lv )water . Thus, the
total amount of heat gained by the water is Qwater = (cm∆T )water + (mvap Lv ) water .
SOLUTION
Qlost by = Qgained by
mercury
water
( cm∆T ) mercury = ( cm∆T ) water + ( mvap Lv ) water
where ∆Tmercury = (205 °C − 100.0 °C) and ∆ Twater = (100.0 °C − 80.0 °C). The specific
heats of mercury and water are given in Table 12.2, and the latent heat of vaporization of
water is given in Table 12.3. Solving for the mass of the water that vaporizes gives
Chapter 12 Problems
m vap =
=
655
c mercury mmercury ∆Tmercury − c water mwater ∆ Twater
( Lv ) water
[139 J/(kg ⋅C°)](2.10 kg)(105 C°) − [4186 J/(kg ⋅C°)](0.110 kg)(20.0 C°)
22.6 ×10 5 J/kg
= 9.49 × 10 –3 kg
______________________________________________________________________________
80. REASONING Since the container is being ignored and since we are assuming negligible
heat exchange with the environment, the principle of conservation of energy applies in the
following form: heat gained equals heat lost. In reaching equilibrium the colder aluminum
gains heat in warming to 0.0 ºC, and the warmer water loses heat in cooling to 0.0 ºC. In
either case, the heat Q that must be supplied or removed to change the temperature of a
substance of mass m by an amount DT is given by Equation 12.4 as Q = cmDT, where c is the
specific heat capacity. In using this equation as we apply the energy-conservation principle,
we must remember to express the change in temperature DT as the higher minus the lower
temperature. The water that freezes into ice also loses heat. The heat Q lost when a mass m
of water freezes is given by Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion.
By including this amount of lost heat in the energy-conservation equation, we will be able to
calculate the mass of water that is frozen.
SOLUTION Using the energy-conservation principle and Equations 12.4 and 12.5 gives
cAluminum mAluminum ∆TAluminum = cWater mWater ∆TWater +
mIce Lf, Water
Heat gained by aluminum
Heat lost by water
Heat lost by water that freezes
Solving for mIce, taking values for the specific heat capacities from Table 12.2, and taking
the latent heat for water from Table 12.3, we find that
mIce =
cAluminum mAluminum ∆TAluminum − cWater mWater ∆TWater
Lf, Water
⎡9.00 × 102 J/ ( kg ⋅ C° ) ⎤ ( 0.200 kg ) ⎡⎣0.0 °C − ( −155 °C ) ⎤⎦
⎣
⎦
=
33.5 ×104 J/kg
⎡ 4186J/ ( kg ⋅ C° ) ⎤⎦ (1.5 kg )( 3.0 °C − 0.0 °C )
−⎣
= 0.027 kg
33.5 ×104 J/kg
656 TEMPERATURE AND HEAT
81. REASONING AND SOLUTION The value for the coefficient of thermal expansion of
steel is given in Table 12.1. The relation, ∆L = α L0∆T, written in terms of the diameter d
of the rod, is
∆d
0.0026 cm
∆T =
=
= 110 C°
(12.2)
α d0 ⎡12 × 10−6 ( C° )−1 ⎤ ( 2.0026 cm )
⎣
⎦
______________________________________________________________________________
82. REASONING To determine the fractional decrease in length
∆L
, we need
L0,Silver + L0, Gold
the decrease DL in the rod’s length. It is the sum of the decreases in the silver part and the
gold part of the rod, or DL = DLSilver + DLGold. Each of the decreases can be expressed in
terms of the coefficient of linear expansion α, the initial length L0, and the change in
temperature DT, according to Equation 12.2.
SOLUTION Using Equation 12.2 to express the decrease in length of each part of the rod,
we find the total decrease in the rod’s length to be
∆L = αSilver L0, Silver ∆T + α Gold L0, Gold ∆T
∆LSilver
∆LGold
The fractional decrease in the rod’s length is, then,
αSilver L0, Silver ∆T + α Gold L0, Gold ∆T
∆L
=
L0,Silver + L0, Gold
L0,Silver + L0, Gold
⎛
⎞
⎛
⎞
L0, Silver
L0, Gold
= αSilver ⎜
⎟ ∆T + α Gold ⎜
⎟ ∆T
⎜L
⎟
⎜L
⎟
+
+
L
L
0,Silver
0,
Gold
0,Silver
0,
Gold
⎝
⎠
⎝
⎠
Silver fraction =
1
3
Gold fraction =
2
3
Recognizing that one third of the rod is silver and two thirds is gold and taking values for the
coefficients of linear expansion for silver and gold from Table 12.1, we have
∆L
⎡
⎛1⎞
⎛2⎞
⎛1⎞
⎛ 2 ⎞⎤
= αSilver ⎜ ⎟ ∆T + α Gold ⎜ ⎟ ∆T = ⎢αSilver ⎜ ⎟ + α Gold ⎜ ⎟ ⎥ ∆T
L0,Silver + L0, Gold
⎝3⎠
⎝3⎠
⎝3⎠
⎝ 3 ⎠⎦
⎣
⎧
−1 ⎛ 1 ⎞
−1 ⎛ 2 ⎞ ⎫
= ⎨ ⎡19 × 10−6 ( C° ) ⎤ ⎜ ⎟ + ⎡14 ×10−6 ( C° ) ⎤ ⎜ ⎟ ⎬ ( 26 C° ) = 4.1× 10−4
⎣
⎦
⎣
⎦ ⎝ 3 ⎠⎭
⎝3⎠
⎩
Chapter 12 Problems
83.
657
REASONING AND SOLUTION The cider will expand according to Equation
12.3; therefore, the change in volume of the cider is
SSM
∆V = βV0 ∆T = ⎡⎣ 280 ×10 –6 (C°) −1 ⎤⎦ (1.0 gal)(22 C°) = 6.2 ×10−3 gal
At a cost of two dollars per gallon, this amounts to
⎞
( 6.2×10−3 gal ) ⎛⎜⎝ $12.00
⎟ = $ 0.01
gal ⎠
or
one penny
______________________________________________________________________________
84. REASONING Since there is no heat lost or gained by the system, the heat lost by the water
in cooling down must be equal to the heat gained by the thermometer in warming up. The
heat Q lost or gained by a substance is given by Equation 12.4 as Q = cm∆T, where c is the
specific heat capacity, m is the mass, and ∆T is the change in temperature. Thus, we have
that
cH O mH O ∆TH O = ctherm mtherm ∆Ttherm
2
2
2
Heat lost by water
Heat gained by thermometer
We can use this equation to find the temperature of the water before the insertion of the
thermometer.
SOLUTION Solving the equation above for ∆TH2O , and using the value of cH O from Table
2
12.2, we have
∆TH O =
2
ctherm mtherm ∆Ttherm
cH O mH O
2
2
⎡815 J/ ( kg ⋅ C° ) ⎤⎦ ( 31.0 g )( 41.5 °C − 12.0 °C )
=⎣
= 1.50 C°
⎡⎣ 4186 J/ ( kg ⋅ C° ) ⎤⎦ (119 g )
The temperature of the water before the insertion of the thermometer was
T = 41.5 °C + 1.50 C° = 43.0 °C
______________________________________________________________________________
85.
REASONING AND SOLUTION From the vapor pressure curve that accompanies
Problem 68, it is seen that the partial pressure of water vapor in the atmosphere at 10 °C is
about 1400 Pa, and that the equilibrium vapor pressure at 30 °C is about 4200 Pa. The
relative humidity is, from Equation 12.6,
SSM
658 TEMPERATURE AND HEAT
Percent ⎛ 1400 Pa ⎞
relative = ⎜
⎟ × 100 = 33%
humidity ⎝ 4200 Pa ⎠
______________________________________________________________________________
86. REASONING AND SOLUTION If the voltage is proportional to the temperature
difference between the junctions, then
V1
V
= 2
∆T1 ∆T2
or
∆T2 =
V2
∆T
V1 1
Thus,
T2 − 0.0 °C =
−3
1.90×10 V
(110.0 °C − 0.0 °C)
−3
4.75×10 V
T2 = 44.0 °C .
Solving for T2 yields
______________________________________________________________________________
87. REASONING The amount Q of heat required to melt an iceberg at 0 °C is equal to mLf,
where m is its mass and Lf is the latent heat of fusion for water (see Table 12.3). The mass is
related to the density ρ and the volume V of the ice by Equation 11.1, m = ρ V.
SOLUTION
a. The amount of heat required to melt the iceberg is
Q = mLf = ρ V Lf
(
= 917 kg/m3
(12.5)
× 10 m ) ( 230 m ) ( 3.35 × 10
) (120 × 10 m )(35
3
3
5
J/kg
)
Volume
= 3.0 × 1020 J
b. The number of years it would take to melt the iceberg is equal to the energy required to
melt it divided by the energy consumed per year by the U.S.
3.0 × 1020 J
= 3.2 years
Number of years =
9.3 × 1019 J/y
______________________________________________________________________________
88. REASONING The mass mremaining of the liquid water that remains at 100 °C is equal to the
original mass m minus the mass mvaporized of the liquid water that has been vaporized. The
heat Q required to vaporize this mass of liquid is given by Equation 12.5 as
Q = m Lv, where Lv is the latent heat of vaporization for water. Thus, we have
Chapter 12 Problems
659
Q
Lv
The heat required to vaporize the water comes from the heat that is removed from the water
at 0 °C when it changes phase from the liquid state to ice. This heat is also given by
Equation 12.5 as Q = mLf, where Lf is the latent heat of fusion for water. Thus, the
remaining mass of liquid water can be written as
mremaining = m − mvaporized = m −
mremaining = m −
mLf
L ⎞
⎛
Q
=m−
= m ⎜1 − f ⎟
Lv
Lv
Lv ⎠
⎝
SOLUTION Using the values of Lf and Lv from Table 12.3, we find that the mass of liquid
water that remains at 100 °C is
⎛
⎛
Lf ⎞
3.35 × 105 J/kg ⎞
mremaining = m ⎜⎜1 −
⎟ = ( 2.00 g ) ⎜1 −
⎟ = 1.70 g
Lv ⎟⎠
2.26 × 106 J/kg ⎠
⎝
⎝
______________________________________________________________________________
89. REASONING AND SOLUTION We wish to convert 2.0% of the heat Q into gravitational
potential energy, i.e., (0.020)Q = mgh. Thus,
mg =
( 0.020 ) Q =
⎛ 4186 J ⎞
⎟
⎝ 1 Calorie ⎠
( 0.020 )(110 Calories ) ⎜
= 4.4 × 103 N
2.1 m
h
______________________________________________________________________________
90. REASONING The rod contracts upon cooling, and we can use Equation 12.2 to express the
change ∆T in temperature as
∆L
(1)
∆T =
α L0
where ∆L and L0 are the change in length and original length, respectively, and α is the
coefficient of linear expansion for brass. Originally, the rod was stretched by the 860-N
block that hangs from the lower end of the rod. This change in length can be found from
Equation 10.17 as
F L0
∆L =
(2)
YA
where F is the magnitude of the stretching force, Y is Young’s modulus for brass, and A is
the cross-sectional area of the rod. Substituting Equation (2) into Equation (1) and noting
that L0 is algebraically eliminated from the final result, we have
660 TEMPERATURE AND HEAT
F L0
∆T =
F
YA
=
α L0 α Y A
(Note: The length L0 that appears in Equation (1) is slightly larger than the length L0 that
appears in Equation (2). This difference is extremely small, however, so we assume that
they are the same. Thus, the two lengths appearing in the equation above can be
algebraically eliminated.)
SOLUTION The stretching force F is the weight of the block (860 N). Young’s modulus Y
for brass can be obtained from Table 10.1, and the coefficient of linear expansion α can be
found in Table 12.1. Thus, the change in temperature of the brass rod is
F
860 N
=
= 39 C°
−
1
α Y A ⎡⎣19 ×10−6 ( C° ) ⎤⎦ ( 9.0 ×1010 N/m 2 )(1.3 ×10−5 m 2 )
______________________________________________________________________________
∆T =
91.
REASONING AND SOLUTION As the rock falls through a distance h, its initial
potential energy mrock gh is converted into kinetic energy. This kinetic energy is then
converted into heat when the rock is brought to rest in the pail. If we ignore the heat
absorbed by the pail, the principle of conservation of energy indicates that
SSM
mrock gh = c rockmrock ∆T + cwater mwater ∆T
where we have used Equation 12.4 to express the heat absorbed by the rock and the water.
Table 12.2 gives the specific heat capacity of the water. Solving for ∆T yields
∆T =
mrock gh
crock mrock + cwater mwater
Substituting values yields
(0.20 kg)(9.80 m/s2 )(15 m)
= 0.016 C°
[1840 J/(kg ⋅ C°)](0.20 kg) + [4186 J/(kg ⋅C°)](0.35 kg)
______________________________________________________________________________
∆T =
92. REASONING AND SOLUTION Let L0 = 0.50 m and L be the true length of the line at
40.0 °C. The ruler has expanded an amount
∆Lr = L − L0 = arL0∆Tr
The copper plate must shrink by an amount
(1)
Chapter 12 Problems
∆Lp = L0 − L = apL∆Tp
661
(2)
Eliminating L from Equations (1) and (2), solving for ∆Tp, and using values of the
coefficients of thermal expansion for copper and steel from Table 12.1, we find that
∆Tp =
=
−α r ∆Tr
α p (1 + α r ∆Tr )
⎡17 × 10−6
⎣
−1
− ⎡12 ×10−6 ( C° ) ⎤ ( 40.0 °C − 20.0 °C )
⎣
⎦
= −14 °C
−1 ⎤
−1 ⎤
−6
⎡
C
1
12
10
C
40.0
C
20.0
C
°
+
×
°
°
−
°
( ) ⎦ ⎣
( ) ⎦(
)
{
}
Therefore, Tp = 40.0 °C − 14 C° = 26 °C .
______________________________________________________________________________
93. REASONING AND SOLUTION The heat lost by the steel rod is Q = cm∆T =
cρ V0 ∆T. Table 12.2 gives the specific heat capacity c of steel. The rod contracts according
to the equation ∆L = α L0∆T. Table 12.1 gives the coefficient of thermal expansion of steel.
We also know that F = AY(∆L/ L0), Equation 10.17, so that F = AYα ∆T. Table 10.1 gives
Young’s modulus Y for steel.
Combining expressions yields,
−1
α QY ⎡⎣12 × 10−6 ( C° ) ⎤⎦ (3300 J)(2.0 × 1011 Pa)
F=
=
= 1.1 × 103 N
c ρ L0
[452 J/(kg ⋅ C°)(7860 kg/m3 )(2.0 m)
______________________________________________________________________________
94. REASONING AND SOLUTION The figure below (at the left) shows the forces that act on
the middle of the aluminum wire for any value of the angle θ. The figure below (at the
right) shows the same forces after they have been resolved into x and y components.
T
T sin θ
T
T sin θ
θ
θ
T cos θ
W
T cos θ
W
662 TEMPERATURE AND HEAT
Applying Newton's second law to the vertical forces in figure on the right gives
2T (sin θ ) − W = 0 . Solving for T gives:
T=
W
2 ( sin θ )
(1)
The following figures show how the angle is related to the initial length L0 of the wire and
to the final length L after the temperature drops.
L0
L
θ0
θ
x
x
If the distance between the supports does not change, then the distance x is the same in both
figures. Thus, at the original temperature,
x = L0 (cos θ0)
(2)
x = L (cos θ )
(3)
while at the lower temperature
Equating the right hand sides of Equations (2) and (3) leads to
cos θ =
L0 cos θ0
(4)
L
Now, L = L0 + ∆L, and from Equation 12.2, it follows that
∆L
= α ∆T
L0
Thus, Equation (4) becomes
cos θ =
L0 cos θ0
L0 + ∆L
=
cos θ0
1 + (∆L / L0 )
=
cos θ0
1 + α ∆T
From the figure in the text θ0 = 3.00°. Noting that the temperature of the wire drops by
20.0 C° (DT = -20.0 C°) and taking the coefficient of thermal expansion of aluminum from
Table 12.1, we find that the wire makes an angle θ with the horizontal that is
Chapter 12 Problems
⎛ cos θ0
⎝ 1 + α ∆T
θ = cos −1 ⎜
663
⎞
cos 3.00°
⎫⎪
−1 ⎧
⎪
⎟ = cos ⎨
⎬ = 2.446°
−
1
−6 (
⎡
⎤
)
⎠
+
×
−
1
23
10
C°
20.0
C°
(
)
⎦
⎩⎪ ⎣
⎭⎪
Using this value for θ in Equation (1) gives
W
85.0 N
=
= 996 N
2sin θ 2 sin 2.446°
______________________________________________________________________________
T=
95. CONCEPT QUESTIONS
a. 1 A° is larger than 1 B°, because there are 90 A° between the ice and boiling points of
water, while there are 110 B° between these points.
b. +20 °A is hotter than +20 °B, because, for example, when +20 °A is 50 A° above the ice
point of water, +20 °B is at the ice point.
SOLUTION
a. Since there are 90.0 A° and 110.0 B° between the ice and boiling points of water, we
have that
⎛ 110.0 ⎞
1 A° = ⎜
⎟ B° = 1.22 B°
⎝ 90.0 ⎠
b. +40.0 °A is 70.0 A° above the ice point of water. On the B thermometer, this is
⎛ 1.22 B° ⎞
⎟ = 85.4 B° above the ice point.
⎝ 1 A° ⎠
( 70.0 A° ) ⎜
The temperature on the B scale is
T = + 20.0 °B + 85.4 B° = 105.4 °B
______________________________________________________________________________
96. CONCEPT QUESTIONS
a. According to ∆L = α L0 ∆T (Equation 12.2), the factors that determine the amount ∆L by
which the length of a rod changes are the coefficient of linear expansion a, its initial length
L0, and the change in temperature ∆T.
b. The materials from which the rods are made have different coefficients of linear
expansion. Also, the change in length is the same for each rod when the change in
temperature is the same. Therefore, the initial lengths must be different to compensate for
the fact that the expansion coefficients are different.
664 TEMPERATURE AND HEAT
SOLUTION The change in length of the lead rod is (from Equation 12.2)
∆LL = α L L0, L ∆T
(1)
Similarly, the change in length of the quartz rod is
∆LQ = α Q L0, Q ∆T
(2)
where the temperature change ∆T is the same for both. Since both rods change length by
the same amount, ∆LL = ∆LQ. Equating Equations (1) and (2) and solving for L0, Q yields
L0, Q
⎛α
=⎜ L
⎜α
⎝ Q
⎡ 29 × 10−6 (C°) −1 ⎤
⎞
0.10 m ) = 5.8 m
⎟ L0, L = ⎢
−1 ⎥ (
−6
⎟
0.50
10
C
×
°
)
(
⎢
⎥
⎠
⎣
⎦
Values for the coefficients of thermal expansion for lead and quartz have been taken from
Table 12.1.
______________________________________________________________________________
97. CONCEPT QUESTIONS
a. According to ∆V = β V0 ∆T (Equation 12.3), the change ∆V in volume depends on the
coefficient of volume expansion β, the initial volume V0, and the change in temperature ∆T.
b. The liquid expands more, because its coefficient of volume expansion is larger and the
change in volume is directly proportional to that coefficient.
c. The volume of liquid that spills over is equal to the change in volume of the liquid minus
the change in volume of the can.
SOLUTION The volume of liquid that spills over the can is the difference between the
increase in the volume of the liquid and that of the aluminum can:
∆V = β LV0 ∆T − β AV0 ∆T
Therefore,
βL =
=
∆V
+ βA
V0 ∆T
3.6 × 10−6 m3
(3.5 × 10
−4
m
3
) ( 78 °C − 5 °C )
+ 69 × 10−6 ( C° )
−1
= 2.1 × 10−4 ( C° )
−1
______________________________________________________________________________
98. CONCEPT QUESTIONS
Chapter 12 Problems
665
a. If there is no phase change, the change in temperature is determined by the amount Q of
heat added, the specific heat capacity c and mass m of the material (see Equation 12.4).
b. The heat and the mass are the same, but the changes in temperature are different. The
only factor that can account for the different temperature changes is the specific heat
capacities, which must be different.
SOLUTION The identity of the second bar can be made by determining its specific heat
capacity and making a comparison with the values in Table 12.2. The heat supplied to each
bar is given by Q = cm∆T (Equation 12.4). The amount of heat QG supplied to the glass is
equal to the heat QS supplied to the other substance. Thus,
QG = QS
or
cG m ∆TG = cS m ∆TS
We know that cG = 840 J/(kg⋅C°) from Table 12.2. Solving for cS, we obtain
⎛ ∆T
cS = cG ⎜ G
⎜ ∆T
⎝ S
⎞
⎛ 88 °C – 25 °C ⎞
⎟⎟ = [840 J/(kg ⋅ C°) ] ⎜
⎟ = 235 J/(kg ⋅ C°)
⎝ 250.0 °C – 25 °C ⎠
⎠
______________________________________________________________________________
99. CONCEPT QUESTIONS
a. The amount of heat required to melt an object is directly proportional to the latent heat of
fusion, according to Equation 12.5. Since each has the same mass and more heat is required
to melt B, it has the larger latent heat of fusion.
b. The amount of heat required to melt an object is directly proportional to its mass,
according to Equation 12.5. If the mass is doubled, the heat required to melt the object also
doubles.
SOLUTION
a. According to Equation 12.5, the latent heats of fusion for A and B are
Lf,A =
QA 3.0 × 104 J
4
=
= 1.0 × 10 J/kg
m
3.0 kg
Lf,B =
QB 9.0 × 104 J
4
=
= 3.0 × 10 J/kg
m
3.0 kg
b. The amount of heat required to melt object A when its mass is 6.0 kg is
(
)
Q = mA Lf ,A = ( 6.0 kg ) 1.0 × 10 J/kg = 6.0 × 10 J
4
4
(12.5)
______________________________________________________________________________
666 TEMPERATURE AND HEAT
100. CONCEPT QUESTIONS
a. It does not mean that the partial pressure of water vapor in the air equals atmospheric
pressure. The partial pressure is less than atmospheric pressure.
b. No, the humidity is not 100%. The vapor pressure at the higher temperature is greater
than that at the lower temperature, but the partial pressure of the water vapor in the air has
remained the same. According to Equation 12.6, the humidity has fallen below 100%.
SOLUTION
a. The percentage of atmospheric pressure is
⎛ 2500 Pa ⎞
⎜
⎟ × 100% = 2.5%
5
⎝ 1.013 × 10 Pa ⎠
b. The percentage is 2.5% .
c. The relative humidity at 35 °C is
⎛ 2500 Pa ⎞
(12.6)
⎜
⎟ × 100% = 45%
⎝ 5500 Pa ⎠
______________________________________________________________________________
101. CONCEPT QUESTIONS
a. When the ball and the plate are both heated to a higher common temperature, the ball
passes through the hole. Since the ball’s diameter is greater than the hole’s diameter to start
with, this must mean that the hole expands more than the ball for the same temperature
change. The hole expands as if it were filled with the material that surrounds it. We
conclude, therefore, that the coefficient of linear expansion for the plate is greater than that
for the ball.
b. In each arrangement, the ball’s diameter exceeds the hole’s diameter by the same amount,
the diameters of the holes are the same, the diameters of the balls are the same, and the initial
temperatures are the same. The only difference between the various arrangements, then, is in
the coefficients of linear expansion. The ball and the hole are both expanding. However, the
hole is expanding more than the ball, to the extent that the coefficient of linear expansion of
the material of the plate exceeds that of the ball. Thus, we need to examine the difference
between the two coefficients, in order to decide the order in which the balls fall through the
holes as the temperature increases. Referring to Table 12.1, we find the following
differences for each of the arrangements:
Arrangement A
αLead - αGold = 29 × 10-6 (Cº)-1 - 14 × 10-6 (Cº)-1 = 15 × 10-6 (Cº)-1
Arrangement B
αAluminum - αSteel = 23 × 10-6 (Cº)-1 - 12 × 10-6 (Cº)-1 = 11 × 10-6 (Cº)-1
Chapter 12 Problems
Arrangement C
667
αSilver - αQuartz = 19 × 10-6 (Cº)-1 - 0.50 × 10-6 (Cº)-1 = 18.5 × 10-6 (Cº)-1
In Arrangement C the coefficient of linear expansion of the plate-material exceeds that of the
ball-material by the greatest amount. Therefore, the quartz ball will fall through the hole first.
Next in sequence is Arrangement A, so the gold ball will fall second. Last is Arrangement B,
so the steel ball will be the last to fall.
SOLUTION According to Equation 12.2, the diameter increases by an amount DD = αD0DT
when the temperature increases by an amount DT, where D0 is the initial diameter and α is
the coefficient of linear expansion. Thus, we can write the final diameter as
D = D0 + αD0DT. Since the diameters of the ball and the hole are the same when the ball
falls through the hole, we have
D0, Ball + α Ball D0, Ball ∆T = D0, Hole + α Hole D0, Hole ∆T
Final diameter of ball
Final diameter of hole
Solving for the change in temperature, we obtain
∆T =
D0, Ball − D0, Hole
α Hole D0, Hole − α Ball D0, Ball
668 TEMPERATURE AND HEAT
Arrangement A
∆T =
=
D0, Gold − D0, Lead
α Lead D0, Lead − α Gold D0, Gold
1.0 × 10−5 m
= 6.7 C°
⎡ 29 ×10−6 ( C° )−1 ⎤ ( 0.10 m ) − ⎡14 × 10−6 ( C° )−1 ⎤ 0.10 m + 1.0 ×10−5 m
⎣
⎦
⎣
⎦
(
)
Arrangement Β
∆T =
=
D0, Steel − D0, Aluminum
α Aluminum D0, Aluminum − αSteel D0, Steel
1.0 × 10−5 m
= 9.1 C°
⎡ 23 ×10−6 ( C° )−1 ⎤ ( 0.10 m ) − ⎡12 × 10−6 ( C° )−1 ⎤ 0.10 m + 1.0 × 10−5 m
⎣
⎦
⎣
⎦
(
)
Arrangement C
∆T =
=
D0, Quartz − D0, Silver
αSilver D0, Silver − α Quartz D0, Quartz
1.0 × 10−5 m
= 5.4 C°
⎡19 × 10−6 ( C° )−1 ⎤ ( 0.10 m ) − ⎡ 0.50 × 10−6 ( C° )−1 ⎤ 0.10 m + 1.0 ×10−5 m
⎣
⎦
⎣
⎦
(
)
Since each arrangement has an initial temperature of 25.0 ºC, the temperatures at which the
balls fall through the holes are as follows:
Arrangement A
T = 25.0 ºC + 6.7 Cº = 31.7 ºC
Arrangement B
T = 25.0 ºC + 9.1 Cº = 34.1 ºC
Arrangement C
T = 25.0 ºC + 5.4 Cº = 30.4 ºC
These results are consistent with our answer to Concept Question (b).
102. CONCEPT QUESTIONS
a. Two portions of the same liquid that have the same mass, but different initial
temperatures, when mixed together, will yield an equilibrium mixture that has a temperature
lying exactly midway between the two initial temperatures. That is, the final temperature is
T = 12 (T0A + T0B ) . This occurs only when the mixing occurs without any exchange of heat
with the surroundings. Then, all of the heat lost from the warmer portion is gained by the
cooler portion. Since each portion is identical except for temperature, the warmer portion
Chapter 12 Problems
669
cools down by the same number of degrees that the cooler portion warms up, yielding a
mixture whose final equilibrium temperature is midway between the two initial temperatures.
b. There are two ways to apply the logic described in Concept Question (a). Portions A and
B have the same mass m, so they will yield a combined mass of 2m. Thus, we can imagine
portions A and B mixed together to yield an equilibrium temperature of
1 T
+ T0B ) = 12 ( 94.0 °C + 78.0 °C ) = 86.0 °C . This mixture has a mass 2m, just like the
2 ( 0A
mass of portion C, so when it is mixed with portion C, the final equilibrium temperature will
be 12 ( 86.0 °C + 34.0 °C ) = 60.0 °C . The other way to apply the logic from Concept
Question (a) is to mix half of portion C with portion A and half with portion B. This will
produce two mixtures with different temperatures and each having a mass 2m, which can
then be combined. The final equilibrium temperature of this combination is again 60.0 ºC.
SOLUTION Since we are assuming negligible heat exchange with the surroundings, the
principle of conservation of energy applies in the following form: heat lost equals heat
gained. In reaching equilibrium the warmer portions lose heat and the cooler portions gain
heat. In either case, the heat Q that must be supplied or removed to change the temperature of
a substance of mass m by an amount DT is given by Q = cmDT (Equation 12.4), where c is
the specific heat capacity. The final temperature is 50.0 ºC. We have, then,
cC mC ( 50.0 °C − 34.0 °C ) = cA mA ( 94.0 °C − 50.0 °C ) + cB mB ( 78.0 °C − 50.0 °C )
Heat lost
Heat gained
The specific heat capacities for each portion have the same value c, while mA = mB = m.
With these substitutions, we find
cmC ( 50.0 °C − 34.0 °C ) = cm ( 94.0 °C − 50.0 °C ) + cm ( 78.0 °C − 50.0 °C )
mC ( 50.0 °C − 34.0 °C ) = m ( 94.0 °C − 50.0 °C ) + m ( 78.0 °C − 50.0 °C )
Solving this equation for the mass mC of portion C gives
mC =
m ( 94.0 °C − 50.0 °C ) + m ( 78.0 °C − 50.0 °C )
50.0 °C − 34.0 °C
= 4.50m