Chapter 14
HEAT
Conceptual Questions
1. Heat flows from the hotter to the colder object.
2. The moisture forms an insulating layer between the hot iron and the skin, so that brief contact is safe. The water
temperature will not go over 100°C and evaporating the water takes a lot of heat due to its large heat of
vaporization. When the iron is sufficiently hot, it vaporizes some of the water on the finger upon contact, forming
an additional insulating layer of vapor between the iron and moistened finger.
3. Water is most dense at 3.98°C and becomes less dense as the temperature is lowered further. The coldest water
(below 3.98°C) therefore floats to the surface, which is why ice forms there.
4. The water can then cool by evaporation.
5. The air between the layers acts as very good thermal insulation.
6. The water is slow to cool in the winter due to the large specific heat of water, so it helps keep the vineyard warmer
than if there were no large body of water nearby.
7. No, the metal has a higher thermal conductivity than the wood, so the rate at which it takes in heat from the hand
is greater.
8. The breeze blows from the cooler land toward the warmer water.
9. The fins increase the area of the surface responsible for radiating heat from the engine to the air so that the engine
cools more efficiently.
10. The air under the bridge cools faster than the earth under the roadway. Thus, the bridge cools faster than the
roadway.
11. At higher elevations the atmospheric pressure is lower so that water boils at a lower temperature and cooking
times need to be lengthened.
12. The boiling temperature of water increases as its pressure increases. Food therefore cooks more quickly in a
pressure cooker because the increased pressure heightens the temperature at which the food cooks.
13. Water contained in the sauce and cheese of a pizza has a higher specific heat and thermal conductivity than the
crust. Thus, for a given time interval, the amount of heat energy transferred to the roof of the mouth by the sauce
and cheese is greater than that which is transferred to the hand by the crust.
14. The internal energy of a gas of monatomic molecules is a function of the average molecular translational energy.
In addition to this translational mode, the internal energy of a gas of diatomic molecules is also affected by the
average molecular rotational and vibrational energies. Some of the heat energy put into a diatomic gas goes into
exciting these rotational and vibrational modes and not into the translational modes. Since temperature is only a
function of the average translational energy, the energy put into these other modes does not affect the temperature.
As a result, the molar specific heat of a diatomic gas is greater than that for a monatomic gas.
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Chapter 14: Heat
College Physics
15. At very low temperatures, the internal energy of a diatomic gas is too low to excite the rotational and vibrational
energy modes of its constituent molecules. At room temperature, some of the energy put into the system goes into
exciting these modes and not toward increasing the temperature of the gas. Thus, the molar specific heat is greater
for higher temperatures.
16. No, Stefan’s law may be used to compare the radiation rates of two bodies only if the absolute temperature (in
kelvins) of each body is used.
17. The rate at which heat flows via conduction, convection, and radiation depends upon the temperature difference
between the hot and cold bodies. By adding the milk to the coffee immediately, the temperature difference
between the coffee and the surrounding air is minimized, so the rate of heat flow is smaller. Additionally, when
the color of the coffee is lighter, the radiative heat loss will be further reduced.
18. The air gap inside a double-paned window is designed to reduce the flow of heat through the window via
conduction. If the gap size were larger, convective currents could be set up that would increase the rate of energy
flow via convection.
19. A greenhouse effect is set up within the packages of meat that causes the temperature inside to be larger than that
of the surrounding air. This could be remedied by turning the clear plastic wrapping face-down away from the
lights.
20. The average translational kinetic energy of a single molecule in the pot of hot tea is larger than that of a molecule
in the partially frozen lake. The total internal energy of the water in the lake is however much greater than that of
the water in the teapot because the mass of the water in the lake is orders of magnitude larger than the mass of
water in the teapot.
21. In winter, the walls of a room are at a lower temperature than the air within the room. In summer, the walls are
warmer than the inside air. The amount of heat lost from a person standing within the room via conduction and
convection is constant throughout the year since the temperature of the air is held fixed. The temperature
difference between the body and the walls is greatest in winter. As a result, more heat is lost via radiative transfer
in winter and the room therefore feels cooler.
22. A dark-toned and flat-finished object has a higher emissivity than a light-toned and shiny-finished object. Stefan’s
law therefore tells us that a radiator painted with a dark finish is a more efficient radiator of heat than one finished
with a light tone.
23. The two objects must be in thermal equilibrium with each other and the walls of the evacuated chamber in which
they reside. If an object were to absorb more heat energy than it emitted, its temperature would change and the
system would no longer be in thermal equilibrium. Each body must therefore emit exactly as much radiation as it
absorbs—a good absorber must be a good emitter.
24. (a) In fluid flow, the pressure difference between ends of pipe = volume flow rate × fluid flow resistance; in heat
conduction, the temperature difference between ends of thermal conductor = rate of heat flow × thermal
resistance.
(b) Yes; the pressure at the end of pipe 1 is equal to the pressure at the beginning of pipe 2, and so forth, so the
same reasoning that leads to Eq. (14-13) holds for fluid flow.
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College Physics
Chapter 14: Heat
Problems
1. (a) Strategy The gravitational potential energy of the 1.4 kg of water is converted to internal energy in the
6.4-kg system.
Solution Compute the increase in internal energy.
U = mgh = (1.4 kg)(9.80 m s 2 )(2.5 m) = 34 J
(b) Strategy and Solution Yes; the increase in internal energy increases the average kinetic energy of the water
molecules, thus the temperature is slightly increased.
2. Strategy The decrease in gravitational potential energy per unit mass is equal to the increase in internal energy
per unit mass.
Solution Compute the internal energy produced per kg.
U
= gh = (9.80 N kg)(105 m) = 1.03 kJ kg
m
3. Strategy The amount of internal energy generated is equal to the decrease in kinetic energy of the bullet.
Solution Compute the amount of internal energy generated.
1
1
∆K = mvi2 = (0.0200 kg)(7.00 × 102 m s)2 = 4.90 kJ
2
2
4. Strategy The amount of internal energy generated is equal to the decrease in kinetic energy of the ball.
Solution Compute the amount of internal energy generated.
2
∆K =
1 2 1
⎛ 1h ⎞
mvi = (0.1475 kg)(162,000 m h)2 ⎜
⎟ = 149 J
2
2
⎝ 3600 s ⎠
5. (a) Strategy The decrease in gravitational potential energy of the child is equal to the amount of internal energy
generated.
Solution Compute the amount of internal energy generated.
U = mgh = (15 kg)(9.80 m s 2 )(1.7 m) = 250 J
(b) Strategy and Solution Friction warms the slide and the child, and the air molecules are deflected by the
child’s body. The energy goes into all three .
6. Strategy According to conservation of energy, ∆E = 0.
Solution Find the amount of energy dissipated by air resistance.
1
1
0 = ∆U + ∆K + U air = mg ∆y + m(vf2 − vi2 ) + U air = −mgh + mvf2 + U air , so
2
2
1 2
1
⎡
⎤
2
3
U air = mgh − mvf = (64 kg) ⎢ (9.80 m s )(0.90 × 10 m) − (5.8 m s)2 ⎥ = 560 kJ .
2
2
⎣
⎦
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Chapter 14: Heat
College Physics
7. Strategy The total change in energy is zero. The total energy of the system is conserved.
Solution Find the amount of energy dissipated by air drag.
0 = ∆U + ∆K + U air
1
U air = −∆U − ∆K = −mg ∆y − m(vf2 − vi2 )
2
1
⎧
⎫
2
= −(0.60 kg) ⎨(9.80 m s )(3.0 m − 2.0 m) + [(4.5 m s) 2 − (7.6 m s) 2 ]⎬ = 5.4 J
2
⎩
⎭
8. Strategy The potential energy of the mass is converted into heat in the water.
Solution Determine the descending mass.
∆U g
1.00 × 103 J
∆U g = mg ∆y, so m =
=
= 2.72 kg .
g ∆y (9.80 m s 2 )(30.0)(1.25 m)
9. Strategy The conversion factor is 1 kW ⋅ h = 3.600 MJ.
Solution Convert 1.00 kJ to kilowatt-hours.
1 kW ⋅ h
(1.00 × 103 J)
= 2.78 × 10−4 kW ⋅ h
3.600 × 106 J
10. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of 20.0 kg of silver.
C = mc = (20.0 kg)[0.235 kJ (kg ⋅ K)] = 4.70 kJ K
11. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of the 5.00-g gold ring.
C = mc = (0.00500 kg)[0.128 kJ (kg ⋅ K)] = 6.40 × 10−4 kJ K
12. Strategy Solve Eq. (14-4) for the final temperature.
Solution Find the final temperature of the water.
Q
125.6 kJ
+ Ti =
+ 22°C = 82°C .
Q = mc∆T = mc(Tf − Ti ), so Tf =
(0.500 kg)[4.186 kJ (kg ⋅ K)]
mc
13. Strategy Use Eq. (14-4).
Solution Find the amount of heat that must flow into the water.
Q = mc∆T = (2.0 × 10−3 m3 )(1.0 × 103 kg m3 )[4186 J (kg ⋅ K) ](80.0 − 20.0) K = 0.50 MJ
14. Strategy The final kinetic energy equals the energy content of the banana.
Solution Find the man’s speed.
1 2
mv = U , so vf =
2 f
2U
=
m
2(1.00 × 102 kcal)(4186 J kcal)
= 100 m s .
83 kg
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College Physics
Chapter 14: Heat
15. Strategy The 3.3% of the energy from the food is converted to gravitational potential energy of the high jumper.
Solution Find the height the athlete could jump.
U
(3.00 × 106 cal)(4.186 J cal)(0.033)
U = mgh, so h =
=
= 700 m .
mg
(60.0 kg)(9.80 m s 2 )
16. Strategy The heat capacity of an object is equal to its mass times its specific heat.
Solution Find the heat capacity of the 30.0-kg block of ice.
C = mc = (30.0 kg)[2.1 kJ (kg ⋅ K)] = 63 kJ K
17. Strategy The heat capacity of an object is equal to its mass times its specific heat. The mass of an object is equal
to its density times its volume.
Solution Find the heat capacities.
(a) C = mc = ρVc = (2702 kg m3 )(1.00 m3 )[0.900 kJ (kg ⋅ K)] = 2430 kJ K
(b) C = mc = ρVc = (7860 kg m3 )(1.00 m3 )[0.44 kJ (kg ⋅ K)] = 3500 kJ K
18. Strategy The heat capacity of an object is equal to its mass times its specific heat. Find the heat capacities of the
systems by adding the individual heat capacities.
Solution Find the heat capacities.
(a) C = mb cb + mw cw = (0.450 kg)[0.384 kJ (kg ⋅ K) ] + (0.050 kg)[4.186 kJ (kg ⋅ K)] = 0.38 kJ K
(b) C = mw cw + mAl cAl = (7.5 kg)[4.186 kJ (kg ⋅ K)] + (0.75 kg)[0.900 kJ (kg ⋅ K)] = 32 kJ K
19. Strategy Use Eq. (14-4) to find the heat required.
Solution The heat capacity of the system is C = mAl cAl + mw cw .
Q = mc∆T = {(0.400 kg)[0.900 kJ (kg ⋅ K)] + (2.00 kg)[4.186 kJ (kg ⋅ K)]}(100.0°C − 15.0°C) = 742 kJ
20. Strategy Use Eq. (14-4).
Solution Find the heat required to raise the woman’s body temperature.
Q = mc∆T = (50.0 kg)[3.5 kJ (kg ⋅ K)](38.4°C − 37.0°C) = 250 kJ
21. Strategy Use Eq. (14-4).
Solution Find the specific heat of lead.
Q
0.88 kJ
c=
=
= 0.13 kJ (kg ⋅ K)
m∆T (0.35 kg)(20.0 K)
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Chapter 14: Heat
College Physics
22. Strategy The gravitational potential energy of the falling water is converted into internal energy of the system
(total water). Neglect heat flow into the vessel. Use Eq. (14-4).
Solution Find the mass of water that was in the vessel.
m1c∆T + mvc∆T = c∆T (m1 + mv ) = m1gh, so
⎡ (9.80 m s 2 )(1.00 × 102 m) ⎤
m gh
− 1⎥ = 1.34 kg .
mv = 1 − m1 = (1.00 kg) ⎢
c ∆T
⎣⎢ [4186 J (kg ⋅ K) ](0.100 K) ⎦⎥
23. Strategy The decrease in the internal energy of the mercury is equal to the amount of heat that flowed out of it.
Use Eq. (14-4).
Solution Find the amount of energy lost.
∆U = Q = mc∆T = (0.10 ×10−3 kg)[0.139 kJ (kg ⋅ K) ](8.5°C − 15.0°C) = −0.090 J
0.090 J of energy was lost by the mercury.
24. Strategy Use the definition of average power and Eq. (14-4).
Solution Solve for ∆t using P = ∆E ∆t = ∆Q ∆t .
∆Q mc∆T (0.50 kg)[4186 J (kg ⋅ K) ](100.0°C − 20.0°C)
∆t =
=
=
= 80 s .
P
P
2.1× 103 W
25. Strategy The energy required to increase the internal energy of the gas is equal to nCV ∆T where
CV = 20.4 J (mol ⋅ K) for H 2. Use the ideal gas law to find the number of moles of H 2.
Solution Find the energy required.
PV
(10.0 atm)(1.013 × 105 Pa atm)(250 L)(10−3 m3 L)[20.4 J (mol ⋅ K)](25.0 K)
CV ∆T =
∆U = nCV ∆T =
RT
[8.314 J (mol ⋅ K) ](273.15 K + 0.0 K)
= 57 kJ
26. Strategy Use the ideal gas law to find the number of moles of nitrogen gas in the container. Then, solve for the
final temperature of the gas in Eq. (14-6).
Solution Find the number of moles of gas in the container.
PV
PV = nRT , so n =
.
RT
Find the new temperature of the gas after the heat is added.
PV
Q = nCV ∆T =
C (T − Ti ), so
RTi V f
Tf = Ti +
RTi Q
PVCV
= 23°C +
[8.314 J (mol ⋅ K) ](23 K + 273.15 K)(26.6 × 103 J)
(3.5 atm)(1.013 × 105 Pa atm)(425 L)(10−3 m3 L)[20.8 J (mol ⋅ K)]
600
= 44°C .
College Physics
Chapter 14: Heat
27. Strategy The number of moles of air is given by the ideal gas law, n = PV ( RT ) , where these are the initial
quantities. The power generated by the crowd is 501Pav = ∆Q ∆t and ∆Q = 52 nR∆T for a diatomic gas.
Solution Find ∆T .
5
5 ⎛ PV ⎞
nR∆T = ⎜
R∆T = ∆Q = 501Pav ∆t , so
2
2 ⎝ RT ⎟⎠
2(501)TPav ∆t 2(501)(273.15 K + 20.0 K)(110 W)(2.0 h)(3600 s h)
∆T =
=
= 58°C .
5PV
5(1.01× 105 Pa)(8.00 × 103 m3 )
28. Strategy The gas is monatomic, so the molar specific heat is
3
2
Q=
3
2
R = 3Nk (2n). Therefore, the heat is
Nk ∆T .
Solution Find N.
3
2Q
2(10.0 J)
Nk ∆T = Q, so N =
=
= 4.83 × 1021 molecules .
23
−
2
3k ∆T 3(1.38 × 10
J K)(1.00 × 102 K)
29. (a) Strategy Phase changes are indicated by the graph where the temperature is constant while heat is added.
Solution Initially, the substance is solid. As the temperature increases, the substance changes from the solid
to the liquid phase, then from the liquid phase to the gas phase. There are two phase changes shown by the
graph: from B to C, solid to liquid ; and from D to E, liquid to gas .
(b) Strategy and Solution The beginning of the phase change from solid to liquid indicates the melting point of
the substance. That point is labeled by the letter B .
(c) Strategy and Solution The beginning of the phase change from liquid to gas indicates the boiling point of
the substance. That point is labeled by the letter D .
30. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the heat of vaporization of water.
0 = Qs + Qw + Qc = −ms Lv + mscw ∆Ts + mw cw ∆Tw + mccc∆Tc , so
c (m ∆T + mw ∆Tw ) + mccc∆Tc
Lv = w s s
ms
[4.186 J (g ⋅ K)][(18.5 g)(−38.0 K) + (2.00 × 102 g)(47.0 K)] + (3.00 × 102 g)[0.380 J (g ⋅ K)](47.0 K)
18.5 g
= 2260 J g
=
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Chapter 14: Heat
College Physics
31. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the heat of fusion of water.
0 = Qice + Qw + Qc = mice Lf + micecw ∆Tice + mw cw ∆Tw + mccc∆Tc , so
c (m ∆T + mw ∆Tw ) + mccc∆Tc
Lf = − w ice ice
mice
[4.186 J (g ⋅ K)][(30.0 g)(8.5 K) + (2.00 × 102 g)( − 11.5 K)]+ (3.00 × 102 g)[0.380 J (g ⋅ K)](−11.5 K)
30.0 g
= 330 J g
=−
32. Strategy The sum of the heat flows is zero. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the liquid nitrogen that is converted to gas.
0 = Qb + QN = mb cb ∆Tb + mN Lv , so
m c ∆T
(25.0 g)[0.384 kJ (kg ⋅ K)][(77.2 − 293) K]
mN = − b b b = −
= 10.4 g .
Lv
199.1 kJ kg
33. Strategy The sum of the heat flows is zero. Find the mass of the water required to melt the ice and leave the
temperature of the drink at 5.0°C. Use Eqs. (14-4) and (14-9).
Solution Find the mass of water to be added to the cup.
0 = Qw + Qice
0 = mw cw ∆Tw + mice Lf + mice cice ∆T1 + mice cw ∆T2
− mw cw ∆Tw = mice ( Lf + cice ∆T1 + cw ∆T2 )
m ( L + cice ∆T1 + cw ∆T2 )
mw = ice f
−cw ∆Tw
(50.0 g + 50.0 g){333.7 J g + [2.1 J (g ⋅ K)](15.0 K) + [4.186 J (g ⋅ K)](5.0 K)}
= 461 g
mw = −
[4.186 J (g ⋅ K)](−20.0 K)
34. Strategy Find the sum of the heats required to raise the temperature of the ice to 0.0°C, melt the ice, raise the
resulting water to 100.0°C, evaporate the water, and raise the temperature of the resulting steam to 110.0°C. Use
Eqs. (14-4) and (14-9).
Solution Find the required heat.
Q = Qice + Qw + Qs
= mLf + mcice ∆Tice + mcw ∆Tw + mLv + mcs ∆Ts
= (1.0 kg){333.7 kJ kg + [2.1 kJ (kg ⋅ K)](20.0 K) + [4.186 kJ (kg ⋅ K)](100.0 K) + 2256 kJ kg
+ [2.01 kJ (kg ⋅ K)](10.0 K)}
= 3100 kJ
35. Strategy Heat flows from the water to the ice, melting some of it. Find the mass of ice required to lower the
temperature of the water to 0.0°C. Use Eqs. (14-4) and (14-9).
Solution Find the required mass of ice.
0 = Qice + Qw = mice Lf + mw cw ∆Tw , so
mice = −
mw cw ∆Tw
Lf
=−
(5.00 × 102 mL)(1.00 g mL)[4.186 J (g ⋅ K)](−25.0 K)
= 157 g .
333.7 J g
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College Physics
Chapter 14: Heat
36. Strategy The sum of the heat flows is zero. The tea is basically water. The mass of the tea is found by
multiplying the density of water by the volume of the tea. Neglect the temperature change of the glass.
Use Eqs. (14-4) and (14-9).
Solution Find the mass of the ice required to cool the tea to 10.0°C.
0 = Qt + Qice = ρ wVt cw ∆Tt + mice Lf + mice cice ∆T1 + mice cw ∆T2 = ρ wVt cw ∆Tt + mice ( Lf + cice ∆T1 + cw ∆T2 ), so
mice =
− ρ wVt cw ∆Tt
Lf + cice ∆T1 + cw ∆T2
=
−(1.00 × 103 kg m3 )(2.00 × 10−4 m3 )[4.186 kJ (kg ⋅ K)](−85.0 K)
= 179 g .
333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K) + [4.186 kJ (kg ⋅ K)](10.0 K)
37. Strategy The sum of the heat flows is zero. The tea is basically water. The mass of the tea is found by
multiplying the density of water by the volume of the tea. Do not neglect the temperature change of the
glass. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the ice required to cool the tea to 10.0°C. Let ∆T be the temperature change of the tea
and the glass.
0 = Qt + Qice + Qg
0 = ρ wVt cw ∆T + mice Lf + mice cice ∆T1 + mice cw ∆T2 + mg cg ∆T
0 = ( ρ wVt cw + mg cg )∆T + mice ( Lf + cice ∆T1 + cw ∆T2 )
mice =
−( ρ wVt cw + mg cg )∆T
Lf + cice ∆T1 + cw ∆T2
−{(1.00 × 103 kg m3 )(2.00 × 10−4 m3 )[4.186 kJ (kg ⋅ K)] + (0.35 kg)[0.837 kJ (kg ⋅ K)]}(−85.0 K)
333.7 kJ kg + [2.1 kJ (kg ⋅ K)](10.0 K) + [4.186 kJ (kg ⋅ K)](10.0 K)
= 242 g
=
The percentage change from the answer for Problem 40 is
242 g − 179 g
× 100% = 35% .
179 g
38. Strategy Use Eq. (14-9) for fusion.
Solution The heat required to melt the ice is 12.0 kJ − 4.0 kJ = 8.0 kJ. Find the mass of the ice.
Q
8.0 kJ
m=
=
= 24 g
Lf 333.7 kJ kg
39. Strategy Heat flows from the aluminum into the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of aluminum required to melt 10.0 g of ice.
QAl + Qice = mAlcAl∆TAl + mice Lf = 0, so
m L
(10.0 g)(333.7 J g)
mAl = − ice f = −
= 46.3 g .
cAl∆TAl
[0.900 J (g ⋅ K)](0.0°C − 80.0°C)
40. Strategy Heat flows from the aluminum into the ice. Use Eqs. (14-4) and (14-9).
Solution Find the temperature of the aluminum at which it will melt a mass of the ice equal to its own.
L
333.7 J g
+ 0.0°C = 371°C .
mAlcAl∆TAl + mice Lf = mAlcAl (Tf − Ti ) + mice Lf = 0, so Ti = f + Tf =
cAl
0.900 J (g ⋅ K)
Since 371°C < 660°C (the melting point of aluminum), the answer is yes .
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Chapter 14: Heat
College Physics
41. Strategy Use Eq. (14-4) for vaporization. Q is equal to the rate of heat loss per square meter times the area times
the time (1 h = 3600 s).
Solution Find the mass of water lost through transpiration.
Q (250 W m 2 )(0.005 m 2 )(3600 s)
m=
=
= 2g
Lv
2256 J g
42. Strategy The rate of heat loss is ∆Q ∆t = Lv ∆m ∆t since Q = mLv for evaporation and where ∆m represents
the mass of water evaporated.
Solution Compute the rate of heat lost through transpiration.
∆Q
⎛ 1 min ⎞
= (0.618 g min) ⎜
⎟ (2256 J g) = 23.2 W
∆t
⎝ 60 s ⎠
43. Strategy The heat flows from the coffee to the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of ice required to cool the coffee.
Qice + Qc = mice Lf + micec∆Tice + mcc∆Tc = mice ( Lf + c∆Tice ) + mcc∆Tc = 0, so
−mcc∆Tc
(0.25 kg)[4186 J (kg ⋅ K) ](−20.0 K)
=−
= 36 g .
mice =
Lf + c∆Tice
333.7 × 103 J kg + [4186 J (kg ⋅ K) ](60.0 K)
44. (a) Strategy Follow each line segment and consider what happens to the substance represented by the diagram.
Solution For each segment, the pressure, temperature, and phase changes are given in the table.
Segment
Pressure
Temperature
Phase Change
AB
Decreases
Constant
Fluid to solid
BC
Constant
Increases
Solid to liquid
CD
Decreases
Constant
Liquid to vapor
DE
Constant
Decreases
Vapor to solid
(b) Strategy and Solution a is the critical point: If the path for changing a liquid to a gas goes around the
critical point without crossing the vapor pressure curve, no phase change occurs. At temperatures above the
critical temperature or pressures above the critical pressure, it is impossible to make a clear distinction
between the liquid and gas phases. b is the triple point: The three states―solid, liquid, and vapor―can
coexist in equilibrium.
45. Strategy The heat supplied heats the substance to its melting point, melts it, then raises the temperature of the
resulting liquid to 327°C. Use Eqs. (14-4) and (14-9).
Solution Compute the heat of fusion.
Q
31.15 kJ
Q = mLf + mc∆T , so Lf = −c∆T + = −[0.129 kJ (kg ⋅ K)](327 − 21) K +
= 22.8 kJ kg .
m
0.500 kg
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College Physics
Chapter 14: Heat
46. Strategy The rate of heat loss is ∆Q ∆t = Lv ∆m ∆t since Q = mLv for evaporation and where ∆m represents
the mass of water evaporated.
Solution Compute the rate of heat lost by the dog through panting.
∆Q
⎛ 1 min ⎞
= (670 min −1)(0.010 g)(2256 J g) ⎜
⎟ = 250 W
∆t
⎝ 60 s ⎠
47. Strategy The R-factor is equal to the quotient of the length of a material d and its thermal conductivity κ .
Solution
(a) Form a proportion.
Rcork
Rair
dcork
=1=
κ cork
dair
, so dcork =
κ air
κ cork
0.046 W (m ⋅ K)
dair =
(1.0 cm) = 2.0 cm .
κ air
0.023 W (m ⋅ K)
(b) Use the result from part (a). Replace cork with tin.
κ
66.8 W (m ⋅ K)
d tin = tin dair =
(0.010 m) = 29 m
κ air
0.023 W (m ⋅ K)
48. Strategy Use the latent heat of fusion for water to determine the rate of heat flow. Then, use Fourier’s law of heat
conduction to find the thermal conductivity of the rod and identify the metal.
Solution Determine the rate of heat flow. The temperature difference of 180°F is equal to 100 K.
Q mLv
=
∆t
∆t
Find the thermal conductivity.
dmLv
Q
∆T mLv
(1.10 m)(0.00132 kg)(333.7 × 103 J kg)
=κA
=
, so κ =
=
= 66.6 W (m ⋅ K) .
∆t
d
∆t
A∆T ∆t
π (0.0230 m 2) 2 (100 K)(175 s)
Referring to Table 14.5, we see that the thermal conductivity of tin is 66.8 W (m ⋅ K). No other value is close to
66.6 W (m ⋅ K) , so the metal rod is tin .
49. Strategy Use Eq. (14-12).
Solution Compute the thermal resistance for each material.
(a) R =
(b) R =
(c) R =
d
2.0 × 10−2 m
=
= 0.12 K W
κ A [0.17 W (m ⋅ K)](1.0 m 2 )
2.0 × 10−2 m
[80.2 W (m ⋅ K)](1.0 m 2 )
2.0 × 10−2 m
[401 W (m ⋅ K)](1.0 m 2 )
= 2.5 × 10−4 K W
= 5.0 × 10−5 K W
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Chapter 14: Heat
College Physics
50. Strategy Add the thermal resistances. Use Eqs. (14-12) and (14-13).
Solution Find the equivalent thermal resistance of the rods.
d
d
d⎛ 2
1 ⎞
ΣRn = Cu + Fe = ⎜
+
⎟ where d = d Fe .
κ Cu A κ Fe A A ⎝ κ Cu κ Fe ⎠
Find the rate of heat transfer.
∆T
A∆T
(6.0 × 10−6 m 2 )(100.0 K)
ᏼ=
=
=
= 0.14 W
2
1
⎡
⎤
ΣRn d 2 + 1
(0.25
m)
+
κ Cu κ Fe
⎢⎣ 401 W (m⋅K) 80.2 W (m⋅K) ⎥⎦
(
)
51. Strategy From the given information, ∆T = ᏼ1R1 = ᏼ 2 R2 = ᏼ( R1 + R2 ).
Solution Find the rate of heat flow per unit area.
R1 ᏼ 2
ᏼ R
ᏼ
ᏼ
20.0 W m 2
= 6.67 W m 2 .
=
and ᏼ = 2 2 . Thus, ᏼ = R 2 = ᏼ 2 =
20.0 + 1
1
2
R1 + R2
R2 ᏼ1
1
1
+
+
10.0
R
ᏼ
2
1
52. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution
(a) Find the temperature at the interface between the wood and cork if the cork is on the inside and the wood on
the outside. Set ᏼ w = ᏼ c.
∆T
∆T
κ w A w = κc A c
d
d
κ w (0.0°C − T ) = κ c (T − 20.0°C)
(0.0°C)κ w + (20.0°C)κ c = T (κ c + κ w )
(20.0°C)(0.046)
= 5.2°C
T=
0.046 + 0.13
(b) Find the temperature at the interface between the wood and cork if the wood is inside and the cork is outside.
κ w (20.0°C − T ) = κ c (T − 0.0°C)
(20.0°C)κ w = T (κ c + κ w )
(20.0°C)(0.13)
= 15°C
T=
0.046 + 0.13
(c)
The temperature at the interface differs for the two cases, but the total thermal resistance is
the same either way, so it doesn’t matter whether the cork is placed on the inside or the
outside of the wooden wall.
53. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Find the lowest temperature the dog can withstand without increasing its heat output.
T −T
dᏼ
(0.050 m)(51 W)
∆T
ᏼ=κA
= κ A i o , so To = Ti −
= 38°C −
= −37°C .
κA
d
d
[0.026 W (m ⋅ K)](1.31 m 2 )
54. Strategy Use Eq. (14-11).
Solution Find the required heat output.
37°C
∆T
ᏼ=
=
= 110 W
R
0.33 K W
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College Physics
Chapter 14: Heat
55. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Compute the rate of heat conduction for each situation.
(a) ᏼ = κ A
∆T
35°C − 4.0°C
= [0.040 W (m ⋅ K)](1.2 m 2 )
= 300 W
0.0050 m
d
(b) ᏼ = [0.60 W (m ⋅ K)](1.2 m 2 )
35°C − 4.0°C
= 4500 W
0.0050 m
56. Strategy Use Fourier’s law of heat conduction, Eq. (14-10), and Eqs. (14-11) and (14-12).
Solution The relevant quantities are
d
d
∆T
ᏼ g = κg A
, Rg =
, and Rf =
.
κg A
κf A
d
Find the effective thermal resistance.
d
d
d⎛ 1
1 ⎞
ΣRn =
+
= ⎜
+
⎟
κ g A κ f A A ⎜⎝ κ g κ f ⎟⎠
Form a proportion with the initial and final rates of heat flow.
ᏼ gf
ᏼg
=
∆T
d⎛ 1 1
⎜ +
A ⎜⎝ κ g κ f
κ g A ∆dT
⎞
⎟
⎟
⎠
=
1
κ g ⎜⎛ κ1 + κ1 ⎟⎞
⎝
g
f
=
⎠
1
κ
1 + κg
f
=
1
1.0
1 + 0.025
= 0.024
The conduction rate has been reduced by a factor of 0.024 .
57. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution
(a) ᏼ = κ A
(b)
∆T
104°C − 24°C
= [401 W (m ⋅ K)](1.0 × 10−6 m 2 )
= 0.32 W
d
0.10 m
∆T 104°C − 24°C
=
= 800 K m
d
0.10 m
(c) The effective length has doubled.
ᏼ = [401 W (m ⋅ K)](1.0 × 10−6 m 2 )
(d) The effective area has doubled.
ᏼ = [401 W (m ⋅ K)](2.0 × 10−6 m2 )
104°C − 24°C
= 0.16 W
0.20 m
104°C − 24°C
= 0.64 W
0.10 m
(e) Since the bars are identical, the temperature at the junction will be midway between the temperatures of the
baths.
104°C + 24°C
= 64°C
2
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Chapter 14: Heat
College Physics
58. (a) Strategy and Solution
The skier with the down jacket will stay warmer longer, since the jacket’s thermal
conductivity is lower and the jacket is thicker.
(b) Strategy Use Fourier’s law of heat conduction, Eq. (14-10). Form a proportion.
Solution
∆T
ᏼ down κ d A dd
κ d
(0.025)(0.50)
1
=
= d w =
=
∆
T
κ w dd (0.040)(2.0) 6.4
ᏼ wool κ w A d
w
The person with the down jacket can stay outside 6.4 times longer.
59. Strategy Use Wien’s law, Eq. (14-17).
Solution Compute the wavelength of the maximum intensity.
2.898 × 10−3 m ⋅ K
= 1.76 µm .
λmax T = 2.898 × 10−3 m ⋅ K, so λmax =
1650 K
60. Strategy Find a mathematical relationship that fits the set of data.
Solution When the temperature is 2000 K, the wavelength is 1.45 µm. When the temperature is 1500 K, the
wavelength is 1.9 µm. So, as the temperature decreases, the wavelength increases. This seems to imply an inverse
relationship, such as λ =
k
, where k is a constant given by k = λ1T1 = λ2T2 . Compute k.
T
λ1T1 = (1.45 × 10−6 m)(2000 K) = 2.9 × 10−3 m ⋅ K and λ2T2 = (1.9 × 10−6 m)(1500 K) = 2.9 × 10−3 m ⋅ K. The
products are the same, so the constant predicted by Wien is 2.9 × 10−3 m ⋅ K .
61. Strategy Use Stefan’s law of radiation, Eq. (14-16).
Solution Compute the power radiated by the bulb.
ᏼ = eσ AT 4 = 0.32[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.00 × 10−4 m 2 )(3.00 × 103 K) 4 = 150 W
62. Strategy Use Stefan’s law of radiation, Eq. (14-16).
Solution Find the surface area of the filament.
ᏼ
40.0 W
=
= 4.8 × 10−5 m 2 .
ᏼ = eσ AT 4 , so A =
4
−8
2
4
3
4
0.32[5.670 × 10 W (m ⋅ K )](2.6 × 10 K)
eσ T
63. Strategy The person must lose heat at the same rate as it is produced and absorbed to maintain a constant body
temperature. Power is equal to intensity times area.
Solution Compute the rate of heat loss required.
Produced: 90 W
Absorbed: ᏼ = IA = (7.00 × 102 W m 2 )(0.57)(1.80 m 2 )(0.42) = 300 W
Rate of heat loss: 90 W + 300 W = 390 W
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College Physics
Chapter 14: Heat
64. Strategy Use Eq. (14-18).
Solution Find the rate at which the student “burns” calories.
ᏼ net = eσ A(T 4 − Ts 4 )
⎛ 1 kcal ⎞ ⎛ 3600 s ⎞
= 1.0[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.7 m 2 )[(35 K + 273.15 K) 4 − (16 K + 273.15 K) 4 ] ⎜
⎟⎜
⎟
⎝ 4186 J ⎠ ⎝ 1 h ⎠
= 170 kcal h
65. Strategy Use Stefan’s law of radiation, Eq. (14-16). Form a proportion. Assume the cross-sectional area A is
constant.
Solution Find the temperature of the filament.
14
ᏼ = eσ AT 4 , so
4
⎛ᏼ ⎞
ᏼ 58 T58
= 4 and T58 = ⎜ 58 ⎟
ᏼ 60 T60
⎝ ᏼ 60 ⎠
14
⎛ 58.0 ⎞
T60 = ⎜
⎟
⎝ 60.0 ⎠
(2820 K) = 2800 K .
66. Strategy Use Wien’s law, Eq. (14-17).
Solution Compute the temperature of the blackbody.
2.898 × 10−3 m ⋅ K
λmax T = 2.898 × 10−3 m ⋅ K, so T =
= 1090 K .
2.65 × 10−6 m
67. Strategy Assume that the black wood stove is a blackbody (e = 1). Use Eq. (14-18).
Solution Find the net rate at which heat is radiated into the room.
ᏼ net = eσ A(T 4 − Ts4 ) = (1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](1.20 m 2 )[(175 K + 273.15 K)4 − (20 K + 273.15 K)4 ]
= 2.24 kW
68. (a) Strategy Power is equal to intensity times area and the heat absorbed is given by Eq. (14-4).
Solution Find the rate of increase of the lizard’s temperature ∆T ∆t .
Q mc∆T
IA
∆T
=
= IA, so
=
=
∆t
∆t
∆t mc
1 (1.4 × 103
2
W m 2 )(1.6 × 10−4 m 2 )
(3.0 g)[4.2 J (g ⋅ °C)]
= 8.9 × 10−3 °C s .
(b) Strategy The time required to raise the temperature of the lizard is equal to the temperature difference
divided by the rate of temperature increase.
Solution
∆T1
5.0°C
∆t1 =
=
= 9.4 min
∆T ∆t 8.9 × 10−3 °C s
609
Chapter 14: Heat
College Physics
69. Strategy Approximate the pots as cubes of similar volume. Use Eq. (14-18).
Solution Find the net rate of radiative heat loss from the two pots.
s3 = V , so s = V 1/ 3 and 6s 2 = A = 6V 2 / 3.
Coffeepot:
ᏼ net = eσ A(T 4 − Ts4 )
= 0.12[5.670 × 10−8 W (m 2 ⋅ K 4 )][6(1.00 L)2/3 (10−3 m3 L) 2 / 3 ][(98 K + 273.15 K) 4 − (25 K + 273.15 K)4 ]
= 4.5 W
Teapot:
ᏼ net = 0.65[5.670 × 10−8 W (m 2 ⋅ K 4 )][6(1.00 L) 2/3 (10−3 m3 L) 2 / 3 ][(98 K + 273.15 K) 4 − (25 K + 273.15 K)4 ]
= 24 W
70. Strategy Power is equal to intensity times area and the heat absorbed is given by Eq. (14-4).
Solution Find the rate of increase of the leaf’s temperature ∆T ∆t .
Q mc∆T
∆T
IA 0.700(9.00 × 102 W m 2 )(5.00 × 10−3 m 2 )
=
= IA, so
=
=
= 1.70 °C s
∆t
∆t
∆t mc
(0.500 g)[3.70 J (g ⋅ °C)]
71. (a) Strategy The power absorbed by the leaf must equal that radiated away. Power is equal to intensity times
area. Use Stefan’s law of radiation, Eq. (14-16).
Solution Absorbed:
I top A + eσ ATs4 = 0.700(9.00 × 102 W m 2 )(5.00 × 10−3 m 2 ) +
(1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](5.00 ×10−3 m 2 )(273.15 K + 25.0 K) 4
= 5.39 W
Find the temperature of the leaf. The area is now 2(5.00 × 10−3 m 2 ) = 10.0 × 10−3 m 2 (both sides of the leaf).
ᏼ = eσ AT 4 , so
1/ 4
⎛ ᏼ ⎞
T =⎜
⎟
⎝ eσ A ⎠
1/ 4
⎡
⎤
5.39 W
=⎢
−8
−3 2 ⎥
2
4
⎣⎢ (1)[5.670 × 10 W (m ⋅ K )](10.0 × 10 m ) ⎦⎥
= 312 K − 273 K = 39°C .
(b) Strategy Since the bottom of the leaf absorbs and emits at the same rate, it can be ignored.
Solution Find the power per unit area that must be lost by other methods.
ᏼ abs,Sun ᏼ rad ᏼ other
ᏼ
=
+
= eσ T 4 + other , so
A
A
A
A
ᏼ
ᏼ other
abs,Sun
=
− eσ T 4 = 0.700(9.00 × 102 W m 2 ) − (1)[5.670 × 10−8 W (m 2 ⋅ K 4 )](273.15 K + 25.0 K)4
A
A
= 182 W m 2
Thus, the power per unit area that must be lost by other methods is 182 W m 2 .
72. Strategy Heat must be added to the room at the same rate that it is lost. Use Fourier’s law of heat conduction, Eq.
(14-10).
Solution Find the approximate rate at which heat must be added to the room.
∆T
75°F − 32°F ⎛ 1 K ⎞
ᏼ=κA
= [0.63 W (m ⋅ K)](1.56 m)(0.762 m)
⎜
⎟ = 1.38 kW
d
0.0130 m ⎝ 1.8°F ⎠
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College Physics
Chapter 14: Heat
73. Strategy Use Eq. (14-4) and the relationship between power and intensity.
Solution The energy provided by the sunlight is converted to heat in the water. The energy provided is
ᏼ∆t = IA∆t. Compute the time to heat the water.
mc∆T (1.0 L)(1000 g L)[4.186 J (g ⋅ K) ](100.0 − 15.0) K
Q = mc∆T = IA∆t , so ∆t =
=
= 320 s .
IA
(750 W m 2 )(1.5 m 2 )
74. Strategy The temperature of the ice must be raised to 0°C. Next, is must be melted. Then, the resulting liquid
water must be raised to 100°C. Finally, the water must be vaporized. Use Eqs. (14-4) and (14-9).
Solution Find the heat energy required to convert the ice to steam.
Q = mcice ∆Tice + mLf + mcwater ∆Twater + mLv = m(cice ∆Tice + Lf + cwater ∆Twater + Lv )
= 5(0.0220 kg){[2.1 kJ (kg ⋅ K)](50.0 K) + 333.7 kJ kg + [4.186 kJ (kg ⋅ K)](100.0 K) + 2256 kJ kg}
= 342 kJ
75. (a) Strategy The change in the internal energy of the bullet is equal to the initial kinetic energy of the bullet. Use
Eq. (14-4).
Solution Calculate the temperature increase of the bullet.
1
(4.00 × 102 m s) 2
v2
mc∆T = mv 2 , so ∆T =
=
= 180°C .
2
2c 2[0.44 × 103 J (kg ⋅ K)]
(b) Strategy The change in the internal energy of the bullet and block system is equal to the initial kinetic
energy of the bullet. Use Eq. (14-4).
Solution Calculate the equilibrium temperature T.
1
mFecFe ∆T + mw cw ∆T = (mFecFe + mw cw )∆T = mFev 2 , so
2
Tf =
=
mFev 2
+ Ti
2(mFecFe + mw cw )
(10.0 × 10−3 kg)(4.00 × 102 m s)2
2{(10.0 × 10−3 kg)[0.44 × 103 J (kg ⋅ K)] + (0.500 kg)[1680 J (kg ⋅ K)]}
+ 20.0°C = 20.9°C
76. Strategy The person must lose heat at the same rate as it is produced and absorbed to maintain a constant body
temperature. Refer to Problem 75. Use Eq. (14-9).
Solution Find the perspiration rate.
Produced: 90 W
Absorbed: ᏼ = IA = (7.00 × 102 W m 2 )(0.57)(1.80 m 2 )(0.42) = 300 W
Rate of heat loss: 90 W + 300 W = 390 W
So, heat must be carried away from the body by perspiration at a rate of 390 W.
⎞⎛ 1 L ⎞
390 J s ⎛ 3600 s ⎞ ⎛
1
Q mLv
m ᏼ
=
=
=
= 0.58 L h .
Q = mLv , so ᏼ =
, or
⎟⎟ ⎜ 3
⎜
⎟ ⎜⎜
3
∆t
∆t
∆t Lv 2430 J g ⎝ 1 h ⎠ ⎝ 1.0 g cm ⎠ ⎝ 10 cm3 ⎠⎟
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Chapter 14: Heat
College Physics
77. (a) Strategy Use Eqs. (14-4) and (14-9).
Solution Find the heat given up by the steam.
Q = − mcw ∆T + mLv = (4.0 g){ − [4.186 J (g ⋅ K)](45.0 − 100.0) K + 2256 J g} = 9.9 kJ
(b) Strategy Use Eq. (14-4).
Solution Compute the mass of the tissue.
Q
9945 J
m=
=
= 360 g
c∆T [3.5 J (g ⋅ K)](45.0 − 37.0) K
78. (a) Strategy Use Eq. (14-4).
Solution Find the heat given up by the water.
Q = − mc∆T = −(4.0 g)[4.186 J (g ⋅ K)](45.0 − 100.0) K = 920 J
(b) Strategy Use Eq. (14-4).
Solution Compute the mass of the tissue.
Q
920 J
m=
=
= 33 g
c∆T [3.5 J (g ⋅ K)](45.0 − 37.0) K
From Problem 93b, m = 360 g.
33 g
1
= 0.093 < , so less than 1 10 as much skin was involved .
355 g
10
79. Strategy Use Eq. (14-4), substituting for Q the expression given in the problem statement.
Solution Compute the temperature rise.
Q 0.544 × 10−3 J + (1.46 × 10−3 J cm)(1.5 cm)
Q = mc∆T , so ∆T =
=
= 6.5 × 10−3°C .
mc
(0.10 g)[4.186 J (g ⋅ K)]
80. Strategy The heat loss is given by ᏼ∆t and it is equal to Q = mLv .
Solution Find the mass of water required to replenish the fluid loss.
ᏼ∆t (650 W)(30.0 min)(60 s min)
m=
=
= 480 g
Lv
2430 J g
81. Strategy The heat loss is proportional to the temperature difference.
Solution Compute the increase in heat loss.
∆T
−8.0°C − (−18°C)
ᏼ 2 = 2 ᏼ1 =
ᏼ1 = 0.25ᏼ1
∆T1
22°C − (−18°C)
Therefore, ᏼ1 = 4.0ᏼ 2 ; thus, the heat loss would be 4.0 times higher without the insulation.
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College Physics
Chapter 14: Heat
82. Strategy Use Eqs. (14-4) and (14-9).
Solution First check to see if the ice completely melts.
Water:
Qw = mc∆T = (186 g)[4.186 J (g ⋅ K) ](0°C − 24°C) = −19 kJ
Ice:
mc∆T = 2(62 g)[2.1 J (g ⋅ K) ][0°C − (−15°C)] = 3.9 kJ
mLf = 2(62 g)(333.7 J g) = 41 kJ
Since 41 kJ + 3.9 kJ = 45 kJ > 19 kJ, the ice does not completely melt, so the final temperature is 0°C .
83. Strategy Use Eq. (14-4).
Solution Find the specific heat of granite.
Q
2.93 kJ
Q = mc∆T , so c =
=
= 0.792 kJ (kg ⋅ K) .
m∆T (0.500 kg)(7.40 K)
84. Strategy The potential energy of the spring is equal to 12 kx 2. Use Eq. (14-4).
Solution Find the temperature change of the water.
1
kx 2 (8.4 × 103 N m)(0.10 m)2
Q = mc∆T = ∆U = kx 2 , so ∆T =
=
= 0.010°C .
2
2mc 2(1.0 kg)[4186 J (kg ⋅ K) ]
85. Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Determine the rate of heat flow.
∆T
327 K − 37 K
ᏼ=κA
= [67.5 W (m ⋅ K)]π (0.0130 m)2
= 10.4 W
d
1.00 m
86. Strategy Heat flows from the iron to the water. Use Eq. (14-4).
Solution Find the mass of 20.0°C water required to cool the iron to 23.0°C.
0 = Qw + QFe = mw cw ∆Tw + mFecFe∆TFe , so
m c ∆T
(0.38 kg)[0.44 kJ (kg ⋅ K) ](23.0°C − 498°C)
= 6.3 kg .
mw = − Fe Fe Fe = −
[4.186 kJ (kg ⋅ K) ](23.0°C − 20.0°C)
cw ∆Tw
87. Strategy Use Fourier’s law of heat conduction, Eq. (14-10). The rate the student “burns” calories is equal to the
rate of heat conduction through the 3.0-mm thickness of water right next to his skin.
Solution Compute the rate of heat flow.
∆T
⎡ (35 − 16) K ⎤ ⎛ 1 kcal ⎞ ⎛ 3600 s ⎞
ᏼ=κA
= [0.58 W (m ⋅ K)](1.7 m 2 ) ⎢
⎟⎜
⎟ = 5400 kcal h
⎥⎜
d
⎣ 0.0030 m ⎦ ⎝ 4186 J ⎠ ⎝ 1 h ⎠
613
Chapter 14: Heat
College Physics
88. (a) Strategy Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Find the temperature at the copper-steel interface.
Set ᏼ st = ᏼ Cu .
κ st A
∆Tst
∆T
κ d
(401)(0.350)
= κ Cu A Cu , so ∆Tst = Cu st ∆TCu =
∆TCu = 20.3∆TCu .
κ st dCu
dst
dCu
(46.0)(0.150)
Since ∆Tst + ∆TCu = ∆T = 4.00°C, ∆TCu (20.3406 + 1) = 4.00°C, or ∆TCu = 0.187°C.
Thus, T = 104.00°C − 0.1874°C = 103.81°C.
∆m
. The rate at which heat enters the
∆t
∆Q ∆mLv
∆m κ A∆T
∆Q
∆T
⎛ ∆m ⎞
=
= Lv ⎜
water is
=ᏼ=κA
, and
⎟ . So, ∆t = dL . Use the values for steel.
∆t
∆t
∆t
d
⎝ ∆t ⎠
v
(b) Strategy and Solution The rate at which the water evaporates is
(
)
2
0.180 m (3.8126°C)
∆m [46.0 W (m ⋅ K)]π
2
=
= 0.565 g s
∆t
(0.00350 m) ( 2256 J g )
89. Strategy Gravitational potential energy is converted into internal energy. Use Eq. (14-9) and U = mgh.
Solution Find the mass of the ice melted by friction.
0.75mgh 0.75(75 kg)(9.80 m s 2 )(2.43 m)
Q = mm Lf = 0.75U = 0.75mgh, so mm =
=
= 4.0 g .
Lf
333,700 J kg
90. Strategy The basal metabolic rate is the minimal energy intake necessary to sustain life in a state of complete
inactivity.
Solution Calculate the BMR kg of body mass and BMR m 2 of surface area for each animal.
Animal
(a) BMR/kg
(b) BMR/m 2
Mouse
210
1200
Dog
51
1000
Human
32
1000
Pig
18
1000
Horse
11
960
(a) According to the table, since BMR kg is larger for smaller animals, it is true that smaller animals must
consume more food per kilogram of body mass.
(c) When an animal is resting, the food energy metabolized must be shed as heat (no work). Since
radiative loss depends upon surface area , BMR m 2 must be approximately the same for different-sized
animals.
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College Physics
Chapter 14: Heat
91. Strategy Use Eq. (14-18).
Solution
(a) Compute the net rate of heat loss through radiation.
ᏼ net = eσ A(T 4 − Ts4 )
= 0.97[5.670 × 10−8 W (m 2 ⋅ K 4 )](2.20 m 2 )[(273.15 K + 37.0 K)4 − (273.15 K + 23.0 K)4 ] = 190 W
(b) Find the skin temperature such that the net heat loss due to radiation is equal to the basal metabolic rate.
eσ A(T 4 − Ts4 ) = BMR, so
1/ 4
⎛ BMR
⎞
T =⎜
+ Ts4 ⎟
⎝ eσ A
⎠
(
)
1/ 4
⎡ (2167 kcal day)(4186 J kcal) 1 day
⎤
86,400 s
⎢
4⎥
=⎢
+ (273.15 K + 23.0 K) ⎥
2
4
2
−8
⎢⎣ 0.97[5.670 × 10 W (m ⋅ K )](2.20 m )
⎥⎦
− 273.15 K
= 31°C
(c) Wearing clothing slows heat loss by radiation because air layers trapped between clothing layers act as
insulation and thus reduce the net radiative heat loss. (T 4 − Ts4 is reduced.)
92. Strategy Set the rates of heat conduction equal. Use Fourier’s law of heat conduction, Eq. (14-10).
Solution Compute the ratio of the depths.
∆T
∆T
d
κ
3.1
= κs A
ᏼ = κb A
, so b = b =
= 1.3, or d b = 1.3ds .
db
ds
ds κ s 2.4
93. Strategy W = Fd (work) and P = Fv (power) so W = Pd v . Use Eq. (14-4).
Solution Find the distance that the cheetah can run before it overheats.
0.700W = Q
0.700 Pd
= mc∆T
v
3
1h
vmc∆T (110 × 10 m h) 3600 s (50.0 kg)[3500 J (kg ⋅ °C)](41.0°C − 38.0°C)
d=
=
= 140 m
0.700P
0.700(160,000 W)
(
)
94. Strategy The shaking will heat the water, although very slowly. The work done by the scientist is converted to
heat in the water. The rate at which energy is supplied by the shaking is equal to the gravitational potential energy
lost by the water during each fall times the frequency of the shaking. Use Eq. (14-4).
Solution Find the time it will take to heat the water.
mc∆T
c∆T
[4186 J (kg ⋅ K) ](87°C − 12°C)
= mghf , so ∆t =
=
= 2 days .
∆t
ghf
(9.80 m s 2 )(0.333 m)(30 min −1)(1440 min day)
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Chapter 14: Heat
College Physics
95. Strategy Heat flows from the copper block to the water and iron pot. Use Eq. (14-4).
Solution Find the final temperature of the system.
0 = Qw + QCu + QFe
Qw = −QCu − QFe
mw cw (Tf − Ti ) = −mCu cCu (Tf − TCu ) − mFecFe (Tf − Ti )
Tf (mw cw + mCu cCu + mFecFe ) = mCu cCuTCu + (mFecFe + mw cw )Ti
Solve for Tf .
mCu cCuTCu + (mFecFe + mw cw )Ti
mw cw + mCu cCu + mFecFe
(2.0 kg)[385 J (kg ⋅ K) ](100.0°C) + {(2.0 kg)[440 J (kg ⋅ K) ] + (1.0 kg)[4186 J (kg ⋅ K) ]}(25.0°C)
=
= 35°C
(1.0 kg)[4186 J (kg ⋅ K) ] + (2.0 kg)[385 J (kg ⋅ K) ] + (2.0 kg)[440 J (kg ⋅ K) ]
Tf =
96. Strategy Heat flows from the gold to the water and copper pot. Use Eq. (14-4).
Solution Find the final temperature of the system.
0 = Qw + QAu + QCu
= mw cw (Tf − Ti ) + mAu cAu (Tf − TAu ) + mCu cCu (Tf − Ti )
= Tf (mw cw + mAu cAu + mCu cCu ) − mAu cAuTAu − (mw cw + mCu cCu )Ti
m c T + (mw cw + mCu cCu )Ti
Tf = Au Au Au
mw cw + mAu cAu + mCu cCu
(0.250 kg)[128 J (kg ⋅ K) ](75.0°C) +
{(0.500 L)(1.00 kg L)[4186 J (kg ⋅ K) ] + (1.500 kg)[385 J (kg ⋅ K) ]}(22.0°C)
= 22.6°C
=
(0.500 L)(1.00 kg L)[4186 J (kg ⋅ K) ] + (0.250 kg)[128 J (kg ⋅ K) ] + (1.500 kg)[385 J (kg ⋅ K) ]
97. (a) Strategy The work done by each animal is proportional to the heat generated. Use Eq. (14-4). Form a
proportion.
Solution Compare the temperature changes.
Qc 2.00mc∆Tc
0.700
=
=
= 14.0, so ∆Tc = 7.00∆Td .
Qd
mc∆Td
0.0500
The temperature change of the cheetah is 7.00 times higher than that of the dog.
(b) Strategy Use the result from part (a).
Solution Find the final temperature of the dog.
Tc − Ti = 7.00(Td − Ti )
Tc + 6.00Ti = 7.00Td
T + 6.00Ti 40.0°C + 6.00(35.0°C)
Td = c
=
= 35.7°C
7.00
7.00
The dog is a much better regulator of temperature and, as a result, has more endurance.
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College Physics
Chapter 14: Heat
98. Strategy If all the kinetic energy of the bullet is converted into heat, then Q = 12 mv 2 , where Q is the minimum
heat required to raise the temperature of the bullet to its melting point and melt it and, therefore, v is the minimum
speed of the bullet. Use Eqs. (14-4) and (14-9).
Solution Find the minimum required speed.
1
Q = mLf + mc∆T = mv 2 , so
2
v = 2( Lf + c∆T ) = 2{22.9 J g + [0.13 J (g ⋅ K)](327°C − 87.0°C)}(103 g kg) = 330 m s .
99. Strategy Heat flows from the tetrachloromethane to the water. Use Eq. (14-4).
Solution Find the specific heat of CCl4 , ct .
0 = Qw + Qt = mw cw ∆Tw + mt ct ∆Tt , so
m c ∆T
(2.00 kg)[4.186 kJ (kg ⋅ K)](18.54 − 18.00) K
ct = − w w w = −
= 0.84 kJ/(kg ⋅ K) .
mt ∆Tt
(2.50 × 10−1 kg)(18.54 − 40.00) K
100. Strategy Heat flows from the drink to the ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of the drink required to just melt the ice.
0 = Qdrink + Qice = mdrink c∆T + mice Lf , so
m L
(0.10 kg)(333.7 kJ kg)
mdrink = − ice f = −
= 0.32 kg .
c∆T
[4.186 kJ (kg ⋅ K) ](0°C − 25°C)
101. Strategy Multiply the energy required to melt the urethane by the molar mass and divide by the total mass to find
the latent heat.
Solution Find the latent heat of fusion of urethane.
(17.10 kJ)[3(12.011) + 7(1.00794) + 2(15.9994) + 14.00674] g mol
1.00 × 102 g
= 15.2 kJ mol
102. Strategy Heat flows from the bullet to the ice. Assume that all the kinetic energy goes into heating the bullet and
ice. Use Eqs. (14-4) and (14-9).
Solution Find the mass of ice that melts.
1
0 = mice Lf + mPbcPb∆T + mPb (vf2 − vi2 ), so
2
mPb ⎛ 1 2
0.0200 kg ⎡ 1
⎞
⎤
mice =
vi − cPb ∆T ⎟ =
(5.00 × 102 m s)2 − [130 J (kg ⋅ K)](0°C − 47.0°C) ⎥ = 7.86 g .
⎜
⎢
Lf ⎝ 2
⎠ 333,700 J kg ⎣ 2
⎦
617