CHAPTER
12 TEMPERATURE AND HEAT
CONCEPTUAL QUESTIONS
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5.
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  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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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  water Vg , where water 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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15. 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 one-quarter full. Therefore, the mug that is full
remains warmer longer.
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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.
PROBLEMS
5.
SSM 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
(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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8.
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,
1.90×103 V
T2  0.0 °C =
(110.0 °C  0.0 °C)
4.75×103 V
Solving for T2 yields
T2  44.0 C .
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10.
REASONING The change in length L of the pipe is proportional to the
coefficient of linear expansion  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   L0 T  1.2  105  C    65 m  18 C   45 C   4.9  102 m
(12.2)
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17.
REASONING AND SOLUTION L = L0T gives for the expansion of the
aluminum
LA = ALAT
(1)
and for the expansion of the brass
LB = BLBT
(2)
Adding (1) and (2) and solving for T gives
T 
L A  L B
 A L A   BL B

1.3  10 3 m
23  10 6 C 1 1.0 m  19  10 6 C 1 2.0 m
 21C
The desired temperature is then
T = 28 °C + 21 C° = 49 C
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28.
REASONING The change in volume of the cavity is proportional to the
coefficient of volume expansion  for aluminum, the original volume V0 of the cavity,
and the change in temperature T . The original volume of the spherical cavity is
V0  43  r 3 , where r is its radius. The coefficient of volume expansion for aluminum can
be found in Table 12.1.
SOLUTION
a.
The change in the volume of the spherical cavity is
V   V0 T
1
3
 6.9  105  C    43   5.0 cm    282 C  23 C   9.4 cm3



b.
Because the temperature increases, the cavity becomes larger .
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(12.3)
34.
REASONING The change in volume of the cylindrical mercury column is
proportional to the coefficient of volume expansion  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   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
 r 2L   V0 T
(12.3)
V
Solving for L gives


4
3


 V0 T 1.82  10  C   45 mm 1.0 C 
L 

 9.0 mm
2
 r2
 1.7  102 mm
1


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41.
SSM 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  coilmoil(Teq  T0oil ) .
SOLUTION
Qmetal  Qoil
c metal mmetal (T0metal  Teq )  c oil moil (Teq  T0oil )
Solving for T0metal , we have
T0metal 
or
c oil moil (Teq  T0oil )
c metal mmetal
 Teq
[2700 J/(kg C)](710 kg)(47 C  32 C)
 47 C = 940 C
[430 J/(kg C)](75 kg)
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T0metal 
51.
SSM 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 . 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
Qtotal  Q1  Q2  m(cT  Lf )
SOLUTION Substituting values, we obtain


Qtotal  (0.45 kg) [9.00 10 2 J/(kg C)](660 C – 130 C)  4.0 10 5 J/kg  3.9 10 5 J
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52.
REASONING AND SOLUTION The block of ice must undergo a change in
temperature, followed by a change in phase, and another change in temperature,


Q = Qice + Qice-to-water + Qwater
Q = (cmT)ice + mLf + (cmT)water
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
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63.
REASONING AND SOLUTION If we assume that all the heat generated by
kinetic friction goes into the block of ice, then from energy conservation,
Qfriction  Q gained
(1)
by ice
Also from energy conservation, Qfriction  Wfriction . From the work-energy theorem,
the work done by friction is given by
1
2
2
1
2
2
Wfriction  mv  mv0
Since the ice is at 0 °C, any heat that it absorbs will be used to change the phase
(i.e., melt) of the ice. Thus, Equation (1) becomes
  12 mblock v 2  12 mblock v02   mmelted Lf
(2)
Solving Equation (2) for the mass of the ice that melts gives
m melted 
1
m
(v 2
2 block 0
Lf
 v2 )

1
(35
2
kg)[(6.5 m/s) 2  (4.8 m/s) 2 ]
(33.5  10 4 J/kg)
 1.0  10 3 kg
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