12
THERMAL PROPERTIES OF MATTER
Q12.1. Reason: The mass of a mole of a substance in grams equals the atomic or molecular mass of the
substance. Since neon has an atomic mass of 20, a mole of neon has a mass of 20 g. Since N2 has a molecular
mass of 28, a mole of N2 has a mass of 28 g. Thus a mole of N2 has more mass than a mole of neon.
Assess: Even though nitrogen atoms are lighter than neon atoms, nitrogen molecules are more massive, so a
mole of nitrogen has more mass than a mole of neon.
Q12.2. Reason: Hydrogen and helium can escape the atmosphere while heavier elements such as oxygen
and nitrogen cannot. This is because hydrogen and helium, being lighter, move faster. From Chapter 11, we
know that the root-mean-square speed of a particle is given by the following:
vrms 
3kBT
m
It follows that the lighter a particle is, the greater is its rms speed.
Assess: We can be thankful that our Earth’s gravity holds on to the vital nutrients oxygen and nitrogen. A
smaller body than the earth might not be able to hold on to these elements. For example, the moon’s gravity is
too weak for it to have an atmosphere of oxygen or nitrogen.
Q12.3. Reason: Since there is almost pure helium in a helium balloon and almost no helium in the outside
air, helium tends to diffuse out of the balloon. Similarly, with almost no oxygen or nitrogen in the balloon
initially and high concentrations of oxygen and nitrogen in the air, these molecules tend to diffuse into the
balloon. However, since helium atoms travel about three times faster than oxygen or nitrogen molecules and
since helium atoms are smaller, they diffuse much faster so gas leaves the balloon faster than it enters. An airfilled balloon has the same particles inside as out and so the stated effect does not contribute to the deflation of
such balloons. Instead, there is a weaker effect which is also at work for helium balloons: Higher pressure inside
the balloon than outside makes the interior air molecules diffuse faster.
Assess: Because of the advantages helium atoms have in diffusion, helium balloons deflate faster than air-filled
ones.
Q12.4. Reason: From the ideal gas law, the lower the pressure of a container of gas, the higher the volume.
The pressure inside the bag equals the pressure outside. So when the atmospheric pressure goes down, the
pressure in the bag goes down and the volume goes up.
Q12.5. Reason: From Equation 12.7, we know that the temperature of a gas is directly proportional to the
square of the rms speed of the molecules. Thus, doubling the typical speed of molecules in a gas increases the
temperature by a factor of four. We also know, from Equation 12.11, that the pressure of a gas is directly
proportional to its temperature, so doubling the typical speed of the molecules must also increase the pressure by
12-1
12-2 Chapter 12
a factor of four. Increasing the speed of molecules in a gas increases the amount of force a molecule exerts on the
wall of the container and the rate of collisions with the walls.
Assess: It makes sense that the pressure depends on the square of the average velocity of molecules in a gas
since the force and collision rate are proportional to the velocity of a molecule.
Q12.6. Reason: Note that N /V and vrms are the same for both gases.
(a) In the process of deriving the ideal gas law we saw that
1N
2
p
mvrms
3V
or that p  m, so given the conditions above, the gas with the more massive molecules (gas 2) will have the higher
pressure.
p2 > p1
(b) The ideal gas law can be rearranged as
N
kBT
V
which shows that p  T , so given the answer to part (a) the temperature of gas 2 must be greater than the
temperature of gas 1.
T2 > T1
p=
Assess: We conclude that, other things being equal, the gas with the more massive molecules will have a
greater pressure and a greater temperature.
Q12.7. Reason: (a) As defined in the chapter, a mole is 6.02 1023 basic particles, regardless of which
chemical element we have. So there are equal numbers of particles in a mole of helium gas and a mole of
oxygen gas.
(b) A mole of helium gas has a mass of 4 g, while a mole of oxygen gas has a mass of 32 g, so one gram of
helium gas has 14  6.02 1023 particles, while one gram of oxygen gas has 321  6.02 1023 particles. Therefore, the
gram of helium gas has more particles than the gram of oxygen gas.
Assess: We note that the basic particles for the helium gas are helium atoms while the basic particles for the
oxygen gas are diatomic oxygen molecules. That is why the O2 molecule has a mass of 32 u.
Q12.8.
Reason:
From Table 12.1, an atom of aluminum has a mass of 27 u  4.48 1026 kg and an atom of
lead has a mass of 207 u  3.44 1025 kg. The number of moles in 100 g of aluminum and lead is
nAl  NAl/NA  (0.100 kg)/(4.48 1026 kg/atom)/(6.02 1023 atom/mol)  3.71 mol
nPb  NPb /NA  (0.100 kg)/(3.44 1025 kg/atom)/(6.02 1023 atom/mol)  0.483 mol
The number of atoms in each is
NAl  (0.100 kg)/(4.48 1026 kg/atom)  2.23 1024 atom
NPb  (0.100 kg)/(3.44 1025 kg/atom)  2.911023 atom
Assess: This makes sense, since an atom of lead is much more massive than an atom of aluminum.
Q12.9. Reason: (a) The thermal energy of the gas does change. In Section 11.4 we are given
Eth 
1
2
Nmvrms
2
so if the speed of every molecule is doubled then the thermal energy is increased by a factor of 22  4.
Thermal Properties of Matter
12-3
(b) No, the molar specific heat doesn’t change. A careful reading of Section 12.7 shows that the molar specific
heat is only a function of the possible forms of thermal energy (translational, rotational, vibrational; these are
called degrees of freedom).
Assess: In reality the molar specific heat does actually change a little bit as a function of temperature. That is
why the temperature is specified at the top of Table 12.6.
Q12.10. Reason: Since the helium atoms are five times lighter than the neon atoms, they are traveling, on
average,
5 times faster. For the sake of argument, we will compare a sample neon atom to a sample helium
atom which is traveling faster by exactly 5 and in the same direction. You might think the helium atom would
exert greater pressure since it is faster. However, since momentum depends on velocity and mass, its momentum
is actually less by a factor of
5. This means that every time the helium atom collides with the walls of the
container, it imparts a force which is less than that exerted by the neon atom by
atoms travel faster, they collide more often, again by
5. Finally, since the helium
5. The combined effect of a force which is weaker by
5 with collisions which occur 5 times more often is that the time-averaged force is the same for both atoms.
Consequently, the pressures exerted by the two different gases are the same.
Assess: It is interesting to see how a number of effects have canceled so that the two gases have the same
pressure. Similarly, helium and neon atoms have the same average kinetic energy even though the helium atoms
are faster because they are correspondingly lighter.
Q12.11.
Reason: Equation 12.11 applies. The number of molecules in the gas is constant since the
container is sealed. Equation 12.11 can be written as p  NkB (T/V ).
(a) If the volume is doubled and the temperature tripled, the pressure increases by a factor of 3/2.
(b) If the volume is halved and the temperature tripled, the pressure increases by a factor of six.
Assess: This makes sense. Increasing the temperature increases the pressure in a gas as does decreasing the
volume of the container.
Q12.12. Reason: Assume the gas is an ideal gas, and use the ideal gas law pV  nRT. Since the number of
moles doesn’t change and R is a constant, then Equation 12.14 gives
pfVf
pV
= i i
Tf
Ti
In each case we want to solve for
Tf =
pfVf
Ti
pV
i i
(a) For Vf  2Vi and pf  3 pi
Tf =
(3 pi )(2Vi )
Ti = 6Ti
pV
i i
So the absolute temperature has increased by a factor of six.
(b) For Vf  12 Vi and pf = 3 pi
Tf 
(3 pi )( 12 Vi )
3
Ti  Ti
pV
2
i i
So the absolute temperature has increased by a factor of 3/2.
Assess: A shortcut is that if T  pV and we double V and triple p, then T is increased by a factor of 2  3  6.
12-4 Chapter 12
Q12.13. Reason: If the work done is equals the area under the pV graph, then there is no work done if V is
constant, because the graph would be a vertical straight line encompassing no area.
Stated another way, Equation 12.17 says Wgas  pV and if V is constant then V  0 and Wgas  0.
Assess: As the chapter says, “in order for a gas to do work, the volume must change.”
Q12.14. Reason: Thermal expansion will make your sun-drenched tape longer than the shaded tape, so if you
and your coworker measure the same object yours will read a smaller value.
Assess: To the extent that this is noticeable, this would be an error in the measurements made with the hot tape,
as they are calibrated to be correct at room temperature.
Q12.15. Reason: From Table 12.3, we see that water has a significantly higher coefficient of thermal
expansion than steel—about six times as much. As the water and steel get hotter, the water expands six times
more than the steel. Thus the water will overflow out of the radiator.
Assess: This seems reasonable since we expect gases to expand more than liquids and liquids to expand more
than solids when the temperature increases.
Q12.16.
Reason: Since A and B are in a well-insulated container, any heat leaving B goes entirely into
raising the temperature of A so
mcA (Tf  0 C)  mcB (Tf  200 C)
Solving for the final temperature,
Tf 
200 C
1  cA / cB
Since the specific heat of A is larger than the specific heat of B, the final temperature is less than 100C.
Assess: As expected, if the specific heats of the two materials are equal, the final temperature will be 100C.
Note that the temperature change depends only on the mass of material, not the density.
Q12.17. Reason: You are heating both containers (each with n moles of nitrogen gas) and thereby increase the
internal energy of each by Q, but the temperatures do not rise by the same amount. See Equations 12.25 and 12.26.
Q  nCVTA  nCPTB  10 J
Consulting Table 12.6 shows that CP > CV; therefore TA  TB and since (TA)i  (TB)i, then (TA)f  (TB)f .
Assess: Some of the energy in container B is used as work done in changing the volume. In container A the
volume did not change, so no work was done and all of the energy went into changing the temperature.
Q12.18.
Reason: From Equations 12.25 and 12.26, the heat required to raise the temperature of a gas
depends on the number of moles of the gas, the molar specific heat of the gas, and whether the gas is heated at
constant volume or constant pressure. Looking at Table 12.6, the molar specific heat of gasses is always smaller
for gasses heated at constant volume than those heated at constant pressure. In order to use the least amount of
heat, you should heat the gas at a constant volume.
Assess: The constant pressure molar specific heat of a gas is generally higher than the constant volume molar
specific heat because gasses do work when their volume increases and this work must be paid for with extra heat.
Q12.19. Reason: Since this is an isothermal process, the internal energy of the gas does not change.
Consequently, the first law of thermodynamics says: 0  Q  W . So W and Q are negatives of one another. Q
is the heat which flows into a system. Here, it is 600 J. So W  600 J. When W is negative, it means that the
system is doing work. In this case, the gas does 600 J of work.
Assess: If a gas does no work its thermal energy increases when heat is added. But by doing work equal in
amount to the heat added, the gas in this problem “spends” the energy given to it so that its thermal energy and
therefore its temperature stay constant.
Q12.20.
Reason: The chocolate starts at some temperature. As it’s heated the temperature rises to a point
where the chocolate changes phase. Assuming the chocolate started as a solid, the chocolate melts during the
second portion of the graph. During melting, the temperature is constant. After all the chocolate melts, the
temperature rises as the liquid chocolate is heated.
Thermal Properties of Matter
12-5
Assess: Compare to Figure 12.21.
Q12.21.
Reason: At higher elevations the air pressure is lower. The amount of leavening agent should be
reduced, since bubbles will form more readily in the lower pressure. Water will also vaporize at a lower
temperature, so the baking temperature should be decreased. Decreasing the temperature requires an increase in
the baking times.
Assess: Most non-physicist cooks will suggest increasing the amount of water in a recipe instead of increasing
baking time.
Q12.22. Reason: Review Example 12.11 to see how Q  McT is used in these calorimetry problems.
Qw  Qm  M wcw Tw  M mcmTm  0
In our case we want to consider Tw in both the aluminum case and the iron case. Note that all the variables that
can be the same for the two cases are.
The aluminum has more “thermal mass” and so can give up more energy to the water without lowering its own
temperature as much, whereas the iron with less “thermal mass” lowers its temperature more as it gives up
energy to the water. The aluminum raises the temperature of the water more than the iron raises the temperature
of the water. The aluminum block ends up with the warmer water.
Assess: It should be noted in connection with this question that in raising the two blocks of metal to the 100 C
temperature from room temperature in the first place, the aluminum block absorbed more energy for the same
temperature change; therefore it had more thermal energy to give up to the water.
Q12.23.
Reason: Part (a) of the figure represents a constant pressure, or isobaric, expansion of the gas. Part
(b) represents a constant volume reduction of pressure of the gas. During part (b), the temperature also decreases,
from Equation 12.12. Part (c) represents a decrease in volume along with an increase in pressure. However, part
(c) is not isothermal since the graph is a straight line. Isothermal processes are hyperbolae on pV diagrams.
Assess: A correct diagram would look like the following figure.
Q12.24. Reason: According to Table 12.7 silver has a much greater thermal conductivity than stainless steel, so
the silver spoon will feel hot first.
Assess: The same atomic crystalline structure and electrons free to move in the metal make silver a good
conductor of both thermal energy and electricity.
Q12.25. Reason: The trees help prevent the energy from being radiated out into space on a cold clear night;
the trees reflect back down some of the infrared radiation and keep the ground under them warmer. In contrast,
the open ground radiates its thermal energy into space without the “blanket” of the trees or clouds to keep the
energy in.
Assess: Gardeners in northern climes know to cover their plants on clear fall nights to keep the radiation in and
keep the plants from freezing. These early first frosts in the fall are even called radiation frosts. They take place
on clear nights with calm winds. Another type, called advective freeze, occurs when very cold air moves in (by
convection); advective freezes can take place with winds and clouds present, and are much harder to protect
plants against.
Reason: From the equation defining gauge pressure, p  pg  1 atm. The absolute pressure is higher
inside the tire. The correct choice is C.
Assess: Gauge pressure is the pressure above atmospheric pressure.
Q12.26.
12-6 Chapter 12
Q12.27. Reason: Apply the ideal gas law in Equation 12.11: pV  NkBT. We are told that T is kept constant
and assume that V is also. So p  N ; if the number of atoms is doubled then the pressure is also.
The correct answer is D.
Assess: This makes sense because with twice as many atoms there will be twice as many atomic collisions in a
given amount of time, so the pressure is doubled.
Q12.28. Reason: If a gas is compressed isothermally, then from the ideal gas law, we know that PV is
constant. Thus, if the volume decreases by a factor of two, the pressure must increase by a factor of two. The
answer is C.
Assess: We see that for an isothermal process, pressure and volume are inversely proportional. You may have
heard this result called “Boyle’s law.”
Q12.29. Reason: In an adiabatic compression, the temperature rises. From the ideal gas law, we can say:
P  nRT / V . If the temperature were constant, halving the volume would double the pressure. But since the
temperature rises and temperature is in the numerator of the pressure formula, the pressure will more than
double. The answer is D.
Assess: It makes sense that an adiabatic compression would raise the temperature more than an isothermal
compression such as we saw in the previous question. This is because adiabatic processes are generally processes
which occur rapidly so that heat does not have time to escape. After the heat starts to escape, the gas particles
have less energy and exert less pressure on the walls of the container.
Q12.30. Reason: To measure the specific heat of a material by putting it in water, you need to know the
mass of the sample, the mass of the water, the initial temperatures and the final temperature. The mass of the
pennies was measured, but one was lost so that the mass of the sample was overestimated. When you solve the
calorimetry equation, you obtain the following:
mpcp Tp  mw cw Tw  0 J 
cp  
mw cw Tw
mp Tp
Since you are using a value of mp which is too high, and since mp is in the denominator, your value of cp will
be too low. Thus you will underestimate the specific heat. The answer is A.
Assess: Another way to look at this is that since you are using too few pennies, the temperature of the water
will rise less than it would have if you had used all of them. In a calorimetry problem, the greater the specific
heat of a sample, the more influence it has on the final temperature. Since the sample will change the temperature
of the water less than it would have with all the pennies, you will underestimate its specific heat.
Q12.31. Reason: We’ll use Equation 12.22 Q  McT and solve for M. We are given
T  20C  20 K and Table 12.4 provides c  4190 J/kg  K.
 60 s 
Q  energy  power  time  (100 W)(1 min) 
  6000 J
 1 min 
Q
6000 J
M

 0.072 kg  72 g
cT (4190 J/kg  K)(20 K)
The correct answer is A.
Assess: For the water (with a large specific heat) to rise in temperature by that much in one minute the mass
must be pretty small (72 g of water has a volume of about 1/3 of a cup).
To boil a liter (just over a quart, and almost 14 times the answer to this question) of water on your stove in just a
few minutes you need to deliver much more than 100 W to it.
Q12.32.
Reason: Assume that the beakers are well-insulated. The specific heat of water is much higher than
the specific heat of aluminum, so more heat will be taken from the beaker with the water added than the beaker
with the aluminum added. In the beaker with ice added, the ice first melts and then the meltwater will have its
temperature raised. Since the melting takes extra heat, the second beaker will end up with the lowest temperature.
The correct choice is B.
Assess: This is as expected from experience.
Thermal Properties of Matter
12-7
Q12.33. Reason: To condense the steam to liquid at 100C would require extracting an amount of energy of
Qcond  MLv  (0.1 kg)(22.6 105 J/kg)  226000 J.
The amount of energy required to melt the ice is Qmelt  MLf  (0.1 kg)(3.33 105 J/kg)  33300 J.
The amount of energy required to raise the temperature of the melted ice (now a liquid) from 0 C to 100C is
QT  McT  (0.1 kg)(4190 J/kg  K)(100 K)  41900 J.
The steam has more than enough energy to melt the ice and raise its temperature by 100C. Some of the steam
would not even need to condense to a liquid; it would stay as steam.
The correct answer is E.
Assess: The key to this question is the large difference between Lv and Lf for water. Had the difference been so
large the other way the answer would have been A.
Q12.34.
Reason: The amount of heat needed can be calculated with Equation 12.21. The heat the
microwave oven provides is given by the power absorbed multiplied by the time it has been on, Q  Pt. Setting
this equal to the heat needed to raise the water’s temperature and solving for the time,
McT (0.25 kg)(4190 J/kg  K)(60 K)
t 

 100 s
P
(600 W)
The correct choice is B.
Assess: This makes sense, from common experience.
Q12.35. Reason: We’ll first bring the ice up to 0C from –10C and then see how much energy is left to melt
some of the ice.
Q  MciceT  (1.00 kg)(2090 J/kg  K)(10K)  20900 J
Now we subtract: 40000J  20900J  19100J; this is the energy left to melt the ice.
M
Q
19100 J

 0.0574 kg
Lf 3.33105 J/kg
The correct answer is B.
Assess: There was enough energy to raise the water to 0 , but not enough to melt it all.
Q12.36.
Reason: Choice A can’t be right because the question explicitly states both the steam and liquid
water are at 100C. But as the steam condenses on the skin it transfers a lot of heat to the skin; this causes the
more severe burn.
The correct answer is B.
Assess: The specific heat of steam is actually less than that of liquid water.
Problems
P12.1. Prepare: For each element, one mole of atoms has a mass in grams equal to the atomic mass number;
for example, since the atomic mass number of carbon is 12 then there is one mole of carbon atoms in 12 grams;
likewise, there is one mole of argon atoms in 40 grams of argon. See Table 12.1.
Solve: The only catch is that hydrogen gas is diatomic, so one mole of diatomic hydrogen gas molecules has a
mass of 2 g.
For hydrogen: 10 g 1 mol/2 g   5 mol
For carbon: 100 g 1 mol/12 g   8.3 mol
For lead:  50 g 1 mol/207 g   0.24 mol
The answer is that 100 g of carbon has the most moles.
Assess: Notice that we count atoms for the solids, but molecules for the diatomic gas.
P12.2. Prepare: A water molecule has a single oxygen atom, while an oxygen molecule is made up of two
oxygen atoms.
12-8 Chapter 12
Solve: A single mole of oxygen gas will contain two moles of oxygen atoms, since there are two oxygen atoms
per molecule. There is one oxygen atom per molecule of water. In order to have the same number of oxygen
atoms in the water as in one mole of gas, there must be two moles of water. A molecule of water consists of two
hydrogen atoms and one oxygen atom and has a mass of 1 u  1 u  16 u  18 u. Two moles of water has a mass of
m  (18 u)(1.66 1027 kg/u)(2 mol)(6.02 1023 mol 1)  3.60 102 kg
The mass of water required is 36.0 g.
Assess: Note that since an oxygen molecule has a mass of 16 u  16 u  32 u, the mass of one mole of oxygen is
a little less than the mass of water required.
P12.3. Prepare: We’ll first compute how many moles of hydrogen peroxide molecules there are in 100 g and then
use Equation 12.1 to find how many particles that is. The molecular mass number for H 2O2 is 2  1  2  16  34.
Solve: (100 g)(1 mol/34 g)  2.94 mol of hydrogen peroxide molecules. However, there are two hydrogen atoms
in each molecule of hydrogen peroxide, so there are 2  2.94 mol  5.88 mol of hydrogen atoms.
N  nNA  (5.88 mol)(6.02 1023 mol1)  3.54 1024
Assess: Three trillion trillion hydrogen atoms is a lot, but one gets used to huge numbers in these types of
problems.
P12.4. Prepare: One liter is 1000 cubic centimeters.
Solve:
3
3
3
 10 mm 
3  10 mm 
6
3
1 L  1000 cm3 
  10 mm
  1000 cm 
3
1
cm
1
cm




Assess: Note that the entire conversion factor must be cubed.
P12.5. Prepare: The volume is clearly the product of the three length measurements; the issue is converting
the units. First multiply L W  H to get the number of cm3, then convert to m3.
Solve:
V  (200 cm)(40 cm)(3.0 cm)  24,000 cm3
Now remember that while 1 m  100 cm, 1 m3  100 cm3. Instead, 1 m3  1,000,000 cm3.


1 m3
24,000 cm3  (24,000 cm3 ) 
 0.024 m3
3 
 1,000,000 cm 
Assess: The answer is small—not a very big fraction of one cubic meter; however, this is reasonable given the
small height. The conversion factor comes from (1 m/100 cm)3.
P12.6. Prepare: We can solve this problem using the ideal gas law. But first we need to find the number of
moles in 1.0 kg of dry ice. Since carbon has an atomic mass of 12 and oxygen has an atomic mass of 16, the
molecular mass of CO2 is 44. Thus 1 mole of CO2 has a mass of 44 g. Finally, 1.0 kg of dry ice is
(1000 g)(1 mol/44 g)  22.7 mol. We are also given the temperature: T  20  293 K and we will use
P  1 atm  1.013 105 Pa.
Solve: Solving Equation 12.11 for V, we have:
V
nRT (22.7 mol)(8.315 J/mol  K)(293 K)

 0.55 m3
P
1.013 105 Pa
Assess: From the chapter, we know that 1 mole of gas at STP has a volume of about 20 L. The conditions of
this problem are not STP but it is a good approximation. We would expect 20 moles to have a volume of about
400 L. This is close to the value we obtained, which converts to 550 L.
P12.7. Prepare: The absolute pressure is the gauge pressure plus one atmosphere at sea
level. 1atm  14.7psi.
Thermal Properties of Matter
12-9
Solve:
p  pg  1atm  35.0 psi  14.7 psi  49.7 psi
Assess: The difference between p and pg is due to the fact that your tire gauge measures pressure differences.
P12.8. Prepare: We’ll assume that air is an ideal gas so we can use the ideal gas law, pV  nRT .
We are given V  5.0 L  0.0050 m,3 p  1atm  101.3 kPa, and T  37C  310 K.
Also recall that R  8.31J/(mol  K) and oxygen makes up 20% of the air.
Solve: Solve Equation 12.12 for n, the number of moles of air.
n
pV (101.3 kPa)(0.0050 m3 )

 0.20 mol
RT (8.31 J/(mol  K))(310 K)
Multiply the number of moles of air by 20% to get the number of moles of oxygen: (0.20 mol)(0.20)  0.040 mol
of oxygen.
Assess: The answer is a small number of moles of oxygen, but a large number of molecules of oxygen.
P12.9. Prepare: Equation 12.9 gives the force due to a pressure applied over an area. The preliminary
calculation is to compute the cross section area of the tube.
2
 0.015 m 
2
4
2
AR  
  1.77  10 m
2


We are given p  6.0 kPa.
Solve:
F  pA  (6.0 kPa)(1.77 104 m2 )  1.1 N
Assess: 1.1 N is not a large force, but it is pushing a light dart, so the dart achieves a respectable acceleration.
P12.10. Prepare: Equation 12.9 gives the force due to a pressure applied over an area. The preliminary
calculation is to compute the cross-section area of the tube.
2
 0.0084 m 
5
2
A   R2   
  5.54  10 m
2


We are given p  45 kPa.
Solve:
F  pA  (45 kPa)(5.54 105 m2 )  2.5 N
Assess: 2.5 N is not a large force, but it is pushing on delicate tissue, so it pays to be careful.
F  PA  (45,000 Pa) (.0042 m)2  2.5 N
P12.11. Prepare: In order to use the ideal gas law (Equation 12.11) we need to know the number of helium
atoms in the gas.
N  nNA  (7.5 mol)(6.02 1023 mol1)  4.52 1024
 1m3 
3
V  15L 
  0.015m
1000L


As a further preliminary calculation add 1 atm to the gauge pressure to give the absolute pressure and convert the
pressure to SI units.
 1 atm  101.3 kPa 
p  pg  1 atm  65 psi  14.7 psi  79.7 psi 

  549 kPa
 14.7 psi  1 atm 
Solve: (a) Solve Equation 12.11 for T.
12-10 Chapter 12
T
pV
(549 kPa)(0.015 m3 )

 132 K  105 C
NkB (4.52 1024 )(1.38 1023 J/K)
(b) Now use Equation 12.5 for K ave.
Kave 
3
3
kBT  (1.38 1023 J/K)(132 K)  2.7 10 21 J
2
2
Assess: The answer to part (a) is a cold temperature, but it needs to be to get that much gas in that volume.
P12.12.
Prepare: We need the mass of a mole of CO2. Since carbon has an atomic mass of 12 and oxygen has an
atomic mass of 16, the molecular mass of CO2 is 44. Hence a mole of CO2 has a mass of 44 g or 0.044 kg.
Solve: We use Equation 12.10 for rms speed but modify it by multiplying numerator and denominator by
Avogadro’s number.
vrms  3kT / m  3RT / (mmol )  3(8.315 J/(mol  K))(210 K)/(44 103 kg/mol)  350 m/s
Assess: This is a typical speed for a gas molecule with a temperature in the hundreds of Kelvins. For example,
the rms speed of an O2 molecule at room temperature is about 480 m/s.
P12.13.
Prepare: Equation 12.12 applies. We must convert all quantities to SI units.
Solve: Converting units,
T  120  273  153 K
 103 m3 
3
3
V  (2.0 L) 
  2.0  10 m
 1L 
Using Equation 12.12,
p
nRT (3.0 mol)(8.31 J/(mol  K))(153 K)

 1.9 106 Pa
V
2.0 103 m3
Assess: Note that this is about twenty times atmospheric pressure.
P12.14.
Prepare: The atomic mass of a nitrogen molecule is 28.0 so 1 mole of nitrogen gas has a mass of
28.0 g or 0.0280 kg. We are given the rms speed of the nitrogen gas: 265 m/s.
Solve: We solve Equation 12.10 for T to obtain the following:
T
mvrms 2 mmolvrms 2 (0.0280 kg)(265 m/s) 2


 78.8 K
3kB
3R
3(8.315 J/mol  K)
This converts to (78.8  273.15) C  194 C.
Assess: It is remarkable that at such a low temperature, the molecules are still moving as fast as a jet. One way
to put this in perspective is to see that, in Kelvins, this temperature is about one-fourth of room temperature.
Consequently, from Equation 12.10, the rms speed of the molecules will be about half that at room temperature.
Indeed, the rms speed of N2 molecules at 20 C is about 510 m/s or around twice the speed of the jet.
P12.15. Prepare: We assume that the steam is an ideal gas and solve the ideal gas law for V. We are
given T  100C  373 K and p  101.3 kPa, but we don’t yet know n. In the sealed bag the number of water
molecules won’t change as the liquid is boiled into steam, so we’ll compute the number of water molecules
(with molecular mass number 18) in 10 g.
n  (10 g)(1 mol/18 g)  0.556 mol
Solve:
Thermal Properties of Matter
V
12-11
nRT (0.556 mol)(8.31 J/(mol  K))(373 K)

 0.017 m3  17 L
p
101.3 kPa
Assess: The answer (over half a cubic foot) is 17,000 times the original volume of the liquid water. It is a good
thing the bag was very flexible. The lesson is that a gas occupies a whole lot more volume than the same mass of
liquid.
P12.16. Prepare: The gas is assumed to be ideal. As a general rule, we must convert all quantities into SI
units. In the present case, however, we will be dealing with the ratio of the final and the initial value of V, so we
do not have to convert L into m3.
Solve: The before-and-after relationship of an ideal gas is
 3.0 L  600 K 
p1V1 p2V2
V T

 p2  p1 1  2  (2.4 atm) 

  1.6 atm
T1
T2
V2 T1
 9.0 L  300 K 
P12.17. Prepare: The gas is assumed to be ideal and it expands isothermally.
Solve: (a) Isothermal expansion means the temperature stays unchanged. That is T2  T1.
(b) The before-and-after relationship of an ideal gas under isothermal conditions is
V  p
p1V1 p2V2
V

 p2  p1 1  p1  1   1
T1
T1
V2
 2V1  2
Assess: The gas has a lower pressure at the larger volume, as we would expect.
P12.18. Prepare: In an isochoric process, the volume of the container stays unchanged. Argon gas in the
container is assumed to be an ideal gas. We must first convert the volumes and temperatures to SI units with
V1  50 cm3  50  10–6 m3, T1  20C  (273 + 20)293 K, and T2  300C  (300  273)K  573 K.
Solve: (a) The container has only argon inside with n  0.1 mol. The pressure before heating is
p1 
nRT (0.10 mol)(8.31 J/(mol  K))(293 K)

 4.87  106 Pa  4870 kPa
V1
50 106 m3
An ideal gas process has p2V2/T2  p1V1/T1. Isochoric heating to a final temperature T2 has V2  V1, so the final
pressure is
V T
573 K
p2  1 2 p1  1
 4870 kPa  9500 kPa
V2 T1
293 K
(b)
Assess: Note that it is essential to express temperatures in Kelvins. Increase in temperature at a constant
volume leads to increased pressure, as would be expected.
P12.19. Prepare: The isobaric heating means that the pressure of the argon gas stays unchanged. Argon
gas in the container is assumed to be an ideal gas. We must first convert the volumes and temperatures to SI units
with V1  50 cm3  50  10–6 m3, T1  20C = (273 + 20)K293 K, and T2  300C  (300  273)K  573 K.
Solve: (a) The container has only argon inside with n  0.10 mol. This produces a pressure
nRT1 (0.10 mol)(8.31 J/(mol  K))(293 K)
p1 

 4.87 106 Pa  4870 kPa
V1
50 106 m3
12-12 Chapter 12
An ideal gas process has p2V2/T2  p1V1/T1. Isobaric heating to a final temperature T2  300C  573 K has p2 
p1, so the final volume is
p T
573
V2  1 2 V1  1
 50 cm3  97.8 cm3
p2 T1
293
(b)
P12.20. Prepare: In an isothermal expansion, the temperature stays the same. The argon gas in the container is
assumed to be an ideal gas. We must first convert the volumes and temperature to SI units: V1  50 cm3  50  10–6
m3, V2  200 cm3  200  10–6 m3, T1  20C = (273 + 20)K293 K.
Solve: (a) The container has only argon inside with n  0.10 mol, V1  50 cm3  50 10–6 m3, and T1  20°C 
293 K. This produces a pressure
nRT1 (0.10 mol)(8.31 J/(mol  K)) (293 K)
p1 

 4.87 106 Pa  48.07 atm
V1
50 106 Pa
An ideal gas process obeys p2V2/T2  p1V1/T1. Isothermal expansion to V2  200 cm3 gives a final pressure
p2 
T2 V1
50
p1  1
 48.07 atm  12 atm
T1 V2
200
(b)
P12.21. Prepare: Assume the gas to be an ideal gas. Please refer to Figure P12.21. We will make use of the
following conversions: 1 atm  1.013  105 Pa and 1 cm3  1  10–6 m3.
Solve: (a) Because the volume stays unchanged, the process is isochoric.
(b) The ideal-gas law pV
1 1  nRT1 gives
T1 
p1V1 (3 1.013 105 Pa)(100 106 m3 )

 914 K
nR
(0.0040 mol)(8.31 J/(mol  K))
The final temperature T2 is calculated as follows for an isochoric process:
 1 atm 
p
p1 p2
 T2  T1 2  (914 K) 

  300 K
p1
T1 T2
 3 atm 
P12.22. Prepare: Assume that the gas is an ideal gas. Please refer to Figure P12.22. We will make use of
the following conversions: 1 atm  1.013  105 Pa and 1 cm3  1  10–6 m3.
Solve: (a) The graph shows that the pressure is inversely proportional to the volume. The process is isothermal.
Thermal Properties of Matter
12-13
(b) From the ideal-gas law,
T1 
p1V1 (3 1.013 105 Pa)(100 106 m3 )

 914 K
nR
(0.0040 mol)(8.31 J/(mol  K))
T2 is also 914 K, because the process is isothermal.
(c) The before-and-after relationship of an ideal gas under isothermal conditions is
p1V1  p2V2  V2  V1
 3 atm 
p1
 (100 cm3 ) 
  300 cm3
p2
 1 atm 
P12.23. Prepare: Assume that the gas is ideal. Please refer to Figure P12.23. We will make use of the
following conversions: 1 atm  1.013  105 Pa and 1 cm3  1  10–6 m3.
Solve: (a) Because the process is at a constant pressure, it is isobaric.
(b) For an ideal gas at constant pressure,
V
100 cm3
V2 V1
 391 K
 T2  T1 2  [(273  900) K]

V1
300 cm3
T2 T1
(c) Using the ideal-gas law p2V2  nRT2,
n
p2V2 (3 1.013 105 Pa)(100 106 m3 )

 9.4 103 mol
RT2
(8.31 J/(mol  K))(391 K)
P12.24. Prepare: Please refer to Figure P12.24. Since the process is occurring at constant pressure, it is
isobaric and the work done by a gas is the area under the p-versus-V curve or given by Equation 12.17. The gas is
compressing, so we expect the work by the gas to be negative or the work on the gas to be positive.
Solve: The work done by the gas is
W  pV  (area under the pV curve)  (200 cm3 )(200 kPa)  (200 106 m3 )(2.0 105 Pa)  40 J
Thus the work done on the gas is 40 J.
Assess: The area under the curve is negative because the gas is compressed. Thus, the environment does
positive work on the gas to compress it.
P12.25. Prepare: For a gas in a sealed container (n is constant) we use Equation 12.14.
pfVf
pV
 i i
Tf
Ti
In this case we want to solve for Tf,
Tf 
pfVf
Ti
pV
i i
but we get to cancel the volumes since they are the same.
We need to take the usual steps of converting temperatures to the absolute scale and pressures from gauge
pressures to absolute pressures.
Ti  0.00C  273 K
pi  55.9 kPa  101.3 kPa  157.2 KPa
pf  65.1 kPa  101.3 kPa  166.4 KPa
Solve:
Tf 
pf
166.4 KPa
Ti 
273 K  289 K  16C
pi
157.2 KPa
Assess: We expected the final temperature to be higher than the initial temperature as the pressure rose. The
answer is a reasonable real-life temperature.
12-14 Chapter 12
P12.26. Prepare: We are given the volume and gauge pressure at the bottom of the sea: Vb  1.0 cm3 and
pb g  1.5 atm. The absolute pressure is: pb  pb g  1 atm  2.5 atm. Since the process is isothermal, from the
ideal gas law, pV is constant.
Solve: When the bubble reaches the surface, its pressure is 1 atm. Since pV is constant, we can equate its
value at the bottom of the sea with its value at the top as follows:
(2.5 atm)(1.0 cm3 )  (1.0 atm)Vt
This equations can be solved to yield: Vt  2.5 cm3.
(b) As the bubble rises it expands. If no heat were exchanged with the sea, it would cool by adiabatic expansion.
Since it does not cool down, we know heat is flowing into the bubble.
Assess: We can understand why heat flows into the bubble if we realize that as the bubble expands it does work
on its surroundings and does this work at the expense of its thermal energy. The sea responds to the loss in
thermal energy of the bubble by giving it heat. You can imagine the bubble doing an infinitesimal amount of
work, which causes its temperature to decrease infinitesimally, which in turn causes heat to flow from the sea to
the slightly cooler bubble until they have the same temperature again.
P12.27. Prepare: We are given the initial volume, temperature and pressure of the gas; Vi  4.0 m3 ,
Ti  20 C  293 K and pi  1.0 atm, as well as the final volume and temperature;
Vf  1 2
3
mand
Tf  10 C  263 K. We can use the ideal gas law.
Solve:
that
We can rewrite the ideal gas law as nR 
pV
. Since the number of moles of gas is constant, it follows
T
pV
is constant, that is,
T
pV
pV
i i
 f f
Ti
Tf
This equation, in turn, can be solved for pf .
pf 
pVT
(1.0 atm)(4.0 m3 )(263 K)
i i f

 0.30 atm
Vf Ti
(12 m3 )(293 K)
Assess: This is in agreement with what we know, that pressure decreases with increasing altitude.
P12.28. Prepare: Equation 12.19 applies. Table 12.3 has coefficients of expansion for various materials.
Note that the aluminum and the steel have different coefficients of linear thermal expansion.
Solve: The total expansion in the rod will be the expansion in the steel added to the expansion of the
aluminum.
L  Laluminum  Lsteel  steel ( Lsteel )i T   aluminum ( Laluminum )i T
 (12 106 K 1 )(0.020 m)(20 K)  (23 10 6 K 1)(0.012 m)(20 K)  1.0  105 m
Assess: This is a very small expansion, as expected.
P12.29. Prepare: We are given much of the data needed in Equation 12.19 except the coefficient of linear
expansion for steel, which we look up in Table 12.3. L  0.73 mm, T  13 K, steel  12 106 K 1.
Thermal Properties of Matter
12-15
We want to know the original length, so we solve Equation 12.19 for that quantity.
Solve:
Li 
L
 T

 1m 
(0.73 mm)

  4.7 m
(12 106 K 1 )(13 K)  1000 mm 
Assess: This seems to be a reasonable answer—in the realm of daily life, about the width of a room. It would
have taken an aluminum beam only about half that long to produce the same L under the same T .
P12.30. Prepare: The gaps must be at least the length of the thermal expansion in the rails. Equation 12.19
and Table 12.3 apply.
Solve: The expansion of a single rail during this temperature change is
L   Li T  (12 106 K)(12 m)(50 K  16 K)  4.9 103 m
The rails should be laid with gaps of about half a centimeter.
Assess: This result makes sense. Gaps on real tracks are about this size or larger.
P12.31. Prepare: We need to rearrange Equation 12.2 to give the fractional volume expansion (the
percentage expansion is then just 100 times that fractional expansion).
Look up Al  69 106 K 1 in Table 12.3. T  120 K.
Solve:
V
  T  (69 106 K 1 )(120 K)  0.0083  0.83%
Vi
Assess: The expansion is small, but greater than it would have been for steel. While 120C is a larger
temperature swing than we might see on a daily basis, there are situations (engines, etc.) where there are
significant temperature differences, and people who design and build precise things must take such expansions
into account.
P12.32. Prepare: The heat needed to change an object’s temperature by T is Q  McT. The mass of the
ice cube is 0.200 kg and its specific heat from Table 12.4 is cice  2090 J/(kg · K).
Solve:
Q  (0.200 kg)(2090 J/(kg · K))(243 K – 273 K)  –12,500 J
Thus, the energy removed from the ice block is 12,500 J.
Assess: The negative sign with Q means loss of energy because removal of heat from the ice reduces its
thermal energy and its temperature.
P12.33. Prepare: The mass of the mercury is M  20 g  2.0  102 kg, the specific heat from Table
12.4 cmercury  140 J/kg K, the boiling point Tb  357C, and the heat of vaporization from Table 12.5 LV  2.96  105
J/kg. We will use Equation 12.22 for obtaining heat needed to raise its temperature to boiling and Equation 12.24
for obtaining heat needed to boil mercury into vapors at the boiling temperature. Note that heating the mercury at
its boiling point changes its thermal energy without a change in temperature. We also note that T is the same
whether we calculate it in Kelvins or in C, so we don’t have to convert C into K.
Solve: The heat required for the mercury to change to the vapor phase is the sum of two steps. The first step is
Q1  McmercuryT  (2.0  102 kg)(140 J/(kg · K))(357C  20C)  944 J
The second step is
Q2  MLV  (2.0  102 kg)(2.96  105 J/kg)  5920 J
The total heat needed is 6870 J, which will be reported as 6900 J.
Assess: More energy is needed to vaporize mercury (86%) than to warm it to its boiling temperature (14%), as
we would expect.
The mass of the mercury is 20 g  2.0  10 2 kg and its specific heat from Table
12.4 is cHg  140 J/kg K. The mass of the water is 20 g  2.0  10 2 kg and its specific heat from Table 12.4
is cwater  4190 J/kg K. We will use Equation 12.22 to obtain the needed heats.
Solve: (a) The heat needed to change the mercury’s temperature is
P12.34. Prepare:
12-16 Chapter 12
Q  McHg T  T 
Q
100 J

 35.7 K  35.7C, which will be reported as 36C.
McHg (0.020 kg)(140 J/(kg  K))
(b) The amount of heat required to raise the temperature of the same amount of water by the same number
of degrees is
Q  Mcwater T  (0.020 kg)(4190 J/(kg · K))(35.7 K)  3000 J
Assess: Q is directly proportional to cwater and the specific heat for water is much higher than the specific heat
for mercury. This explains why Qwater  Qmercury.
P12.35. Prepare: We’ll compute the energy necessary to evaporate 3.5 L and then divide by an hour to get the
rate. One liter of water has a mass of one kilogram.
Solve:
Q  MLf  (3.5 kg)(24 105 J/K)  8.4 106 J
The rate is the energy divided by the time.
rate 
Q 8.4  106 J  1 h 


  2300 W
t
1h
 3600 s 
Assess: That is an impressive power output, but necessary to keep cool in tropical climates. Notice that
the value given for Lf at body temperature is different from the one given in Table 12.5 for standard
temperature (0C).
P12.36. Prepare: We are asked to find the time it takes for the alligator’s temperature to rise from 25 C to
30 C, that is by 5 K. It is absorbed at a rate of 200 W. We use the mammalian specific heat:
c  3400 J/(kg  K). We will need to use the definition of power, which in this case can be written as
P  Q / t , as well as the formula for heat absorbed in connection with an increase in temperature, Q  mcT .
Solve: The time needed for the alligator to reach its final temperature can be obtained by solving the power
equation: t  Q / P. This, coupled with the formula for heat gives us the following:
t 
Q mcT (300 kg)(3400 J/(kg  K))(5 K)


 25,500 s
P
P
200 W
This works out to a time of (25,500 s)(1 hr)/(3600 s) or 7 hr.
Assess: This seems like a long time. But consider that a cup of water of about 1/ 5 kg takes about 2 min to
heat in a 600 W microwave oven. The alligator is about 1500 times more massive and the temperature increase
of the alligator is about 15 times less. So we might expect the alligator to take around 200 min or about 3 hr to
heat up. The difference between 3 and the actual value of 7 comes from the different wattages in the two cases as
well as the different specific heats.
P12.37. Prepare: Water in the body is converted to water vapor. Equation 12.22 applies.
Solve: Each breath converts 25 mg of water to water vapor. The heat required for this is
Q  ML v  (2.5 105 kg)(24 105 J/kg)  60 J
At 12 breaths/min, there are 0.2 breaths/s. Multiplying by the above result of 60 J/breath we obtain that P  12 J/s
or P  12 W is the rate of heat loss.
Converting to Calories/day, we have
 1 Calorie   60 s  60 min  24 h 
P  12 W  



  250 Calorie
 4190 J   min  h  d 
Assess: This seems reasonable since it is a small fraction of a person’s daily caloric intake o f about 2000
calories.
P12.38. Prepare: We need to find the temperature increase and we are given the metabolic power
P  1000 W, the time of exercise t  30 min  (30 min)(60 s)/(1 min)  1800 s, and the man’s mass
Thermal Properties of Matter
12-17
m  70 kg. We need the formula for power, which in this case is P  Eth / t and Equation 12.20 for heat
absorbed. Here the heat produced by the exercise goes to increasing the thermal energy of the man, so we can
write the following: Eth  mcT . For the specific heat, we use the mammalian specific heat from Table 12.4:
c  3400 J/kg  K.
Solve: First we will find the thermal energy produced by the exercise by solving the power equation.
Eth  Pt  (1000 W)(1800 s)  1.8 106 J
We can solve the equation Eth  mcT for T .
T  Eth / mc 
(1.8 106 J)
 7.56 K
(70 kg)(3400 J/(kg  K))
To two significant figures, his temperature increases by 7.6 K.
Assess: Even though he only exercised for 30 min, the man’s temperature has increased by 7.6 C or
9 F
(7.56 C) 
  14 F. This would bring a temperature of 98.6 F up to about 113 F, which would be very dangerous.
5 C
P12.39. Prepare: The water first needs to be raised from 20C to 100C (assuming normal atmospheric
pressure), and then changed from liquid to gas at the boiling temperature.
We’ll use Equation 12.22 Q  McT for the first term and Equation 12.23 Q  ML for the second.
We look up the specific heat of water in Table 12.4 as follows: c  4190 J/(kg  K). We look up the heat of
vaporization of water in Table 12.5: Lv  22.6  105 J/kg.
We are given M  0.10 kg. T  100  20  80 K.
Solve:
Q  McT  MLv  (0.10 kg)(4190 J/(kg  K))(80 K)  (0.10 kg)(22.6 105 J/kg)
 33,000 J  230,000 J  260,000 J
Assess: That’s quite a few joules, but notice that the vast majority is needed to change the water from liquid to
solid rather than raise it from 20C to 100C.
Pay attention to the significant figures. The terms that have factors multipled together each end up with two
significant figures, but when those terms are added, the ten-thousand’s place is the last significant figure (see
Tactics Box 1.1, rule 2).
P12.40. Prepare: We have a thermal interaction between the copper pellets and the water. The initial
temperatures of both copper pellets and the water are known and we denote the common final temperature by Tf.
The specific heats of copper and water from Table 12.4 are as follows: cc  385 J/(kg  K) and cw  4190 J/
(kg  K). While the mass of copper pellets is Mc  0.030 kg, the mass of water is obtained from Mw  V  : Mw 
(100 mL)(10–6 m3/1 mL) (1000 kg/m3)  0.10 kg. We can determine the common final temperature using
Equations 12.22 and 12.25.
Solve: The conservation of energy equation Qc  Qw  0 is
M ccc (Tf  300C)  M wcw (Tf  20C)  0 J
Solving this equation for the final temperature Tf gives
Tf 
M ccc (300C)  M w cw (20C) (0.030 kg)(385 J/(kg K )) (300C)  (0.10 kg)(4190 J/(kg  K) (20C)

 28C
M ccc  M w cw
(0.030 kg)(385 J/(kg  K)  (0.10 kg)(4190 J/(kg  K)
The final temperature of the water and the copper is 28°C.
Assess: Due to the large specific heat of water compared to copper and three times more water in our system,
we expected a small temperature increase.
12-18 Chapter 12
P12.41. Prepare: A thermal interaction between the copper block and the water leads to a common final
temperature denoted by Tf. The initial temperatures of both the copper block and the water are known. The
specific heats of copper and water from Table 12.4 are as follows: cc  385 J/(kg K) and cw  4190 J/(kg  K).
While the mass of the water is known, we can determine the mass of the copper block using Equations 12.22
and 12.25.
Solve: The conservation of energy equation Qcopper  Qwater  0 J is
Mcopperccopper (Tf – Ti copper)  Mwatercwater(Tf – Ti water)  0 J
Both the copper and the water reach the common final temperature Tf  25.5C. Thus
Mcopper(385 J/(kg  K)(25.5C – 300C)  1.0 kg(4190 J/(kg  K))(25.5C – 20C)  0 J  M copper  0.220 kg
Assess: Due to the large specific heat of water compared to copper, a smaller value obtained for the mass of the
copper block is reasonable.
P12.42. Prepare: We have a thermal interaction between the aluminum pan and the water that leads to the
common final temperature denoted by Tf. The initial temperature of the aluminum pan is unknown, but that of the
water is known. The specific heats of aluminum and water from Table 12.4 are cAl  900 J/(kg  K) and cw  4190 J/(kg
 K). The mass of the aluminum pan is MAl  0.750 kg and the mass of water is Mwater  10 kg. We can determine
the initial temperature of the pan using Equations 12.22 and 12.25.
Solve: The conservation of energy equation QAl  Qwater  0 J is
MAlcAl(Tf – Ti Al)  Mwater cwater(Tf – Ti water)
The pan and water reach a common final temperature Tf  24.0C
(0.750 kg)(900 J/(kg  K))(24.0C  Ti Al )  (10 kg/m3 )(4190 J/(kg  K))(24.0C  20.0C)
 (675.0 J/K)(24.0C  Ti Al )  167,600 J  0 J  Ti Al  270C
Assess: Due to (1) the large specific heat of water compared to aluminum and (2) the large mass of water
compared to aluminum, an increase of temperature of 4C needed high initial temperature of the pan.
P12.43. Prepare: We have a thermal interaction between the metal sphere and the mercury that leads to the
common final temperature Tf  99.0C. The initial temperatures of the metal sphere and the mercury are known.
The specific heat of mercury from Table 12.4 is cHg  140 J/(kg  K). The mass of the metal sphere is Mmetal 
0.500 kg and the mass of water is MHg  4.08 kg. Our strategy is to determine cmetal using Equations 12.22 and
12.25.
Solve: The conservation of energy equation Qmetal  QHg  0 J is
Mmetalcmetal(Tf – Ti metal)  MHgcHg(Tf – Ti Hg)  0 J
(0.500 kg)cmetal(99C – 300C)  (4.08 kg)(140 J/(kg  K))(99C – 20C)  0 J
We find that cmetal  449 J/ (kg K). The metal is iron.
P12.44. Prepare: This is a basic calorimetry problem. We are given the mass and initial temperature of the
aluminum mAl  200 g and Ti Al  20 C, as well as the mass and initial temperature of the coffee mc  500 g
and Ti c  85 C . We obtain the specific heats from Table 12.4, according to which cAl  900 J/(kg  K) and
cc  4190 J/(kg  K).
Solve: The basic calorimetry equation becomes mAlcAlTAl  mcccTc  0 which we can solve for the final
temperature:
Thermal Properties of Matter
12-19
(0.200 kg)(900 J/kg)( Tf  20 C)  (0.500 kg)(4190 J/(kg  K))(Tf  85 C)  0 J 
(2280 J/K)Tf  174000 J  0 
Tf  76 C
Assess: Notice that even though the aluminum was comparable in mass to the coffee, its temperature change of
96 C was far more than the temperature change of the coffee, 9 C. This is a result of water having a much
higher specific heat than metal.
P12.45. Prepare: Problem-Solving Strategy 12.1 will be useful. Assume the heat is transferred entirely from
the coffee to the ice since the container is well-insulated. All the ice melts and is converted to water at 0C,
otherwise the leftover ice would continue to cool the coffee. The coffee transfers additional heat to the water to
raise its temperature to 30C.
Solve: The ice is at its melting temperature when it’s put in the coffee. The heat required to melt the ice is
Qmelt  MiceLf. Once the ice melts, heat goes into the water that results in raising its temperature to 30 C. The heat
required is Qwater  M icec water(Tf  Ti )  M icec water(Tf  0C)  M icec waterTf .
The temperature of the coffee decreases as a result of the ice melting and the heating of the resulting water. The
mass of the coffee is M coffee  V. The density of the coffee is   1000 kg/m3. The volume of the coffee is
 1 m3 
4
3
V  200 mL  (200 103 L) 
  2.00 10 m
1000
L


Three significant figures have been assumed in the volume of coffee.
Energy conservation gives
M ice Lf  M icecwaterTf  Vcwater (Tf  Ti )  0
Where Ti is the initial temperature of the coffee. Solving for the mass of ice required,
M ice 
 Vcwater (Tf  Ti ) (1000 kg/m3 )(2.00 104 m3 )(4190 J/(kg  K))(  50 K)

 9.1 102 kg
Lf  cwaterTf
(3.33 105 J/kg)+(4190 J/(kg  K))(30 K)
It takes 91 g of ice to cool the coffee.
Assess: About half the mass of the coffee in ice has to be added to cool the coffee, as we might expect.
P.12.46. Prepare: Heat will leave the patient and enter the ice, melting it. We need the formula for heat absorbed
in melting Qice  mice Lf , and the formula for the heat lost by the patient Qp  mpcp T . For the specific heat, we will
use the value for mammals from Table 12.4, cp  3400 J/kg  K. We need to solve for mice .
Solve: The calorimetry equation is as follows:
mice (3.33  105 J/kg)  (60 kg)(3400 J/(kg  K))(39 C  40 C)  0 J 
mice 
(3400 J/(kg  K))
(60 kg)(1 K)  0.61 kg  610 g
(3.33 105 J/kg)
It takes 610 g of ice to reduce the fever by 1 C.
Assess: Since the ratio of the mammalian specific heat to the latent heat of fusion of ice is about
1
100
K 1 , it
follows that a rule of thumb is that the amount of ice needed to cool a person by 1 C is one hundredth of their
body mass. This is seen to be the case here.
P12.47. Prepare: We will use Equations 12.26 and 12.27 to find the heat needed at constant volume and
constant pressure. Since these equations involve the number of moles of the gas, we will calculate it from the
mass of the gas and its molar mass. From Table 12.6, CP  20.8 J/(mol · K) and CV  12.5 J/(mol · K). Note that
the change in temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale.
Solve: (a) The atomic mass number of argon is 40. That is, Mmol  40 g/mol. The number of moles of argon gas
in the container is
12-20 Chapter 12
n
M
1.0 g

 0.025 mol
M mol 40 g/mol
The amount of heat is
Q  nCV T  (0.025 mol)(12.5 J/(mol · K))(100C)  31.25 J
which will be reported as 31 J.
(b) For the isobaric process Q  nCP T becomes
31.25 J  (0.025 mol)(20.8 J/(mol · K))T  T  60C
P12.48. Prepare: The heating processes are isobaric (in part (a)) and isochoric (in part (b)). O2 is a diatomic
ideal gas. We will use Equations 12.26 and 12.27 to find the heat needed at constant volume and constant
pressure. Since these equations involve the number of moles of the gas, we will calculate it from the mass of the
gas and its molar mass. From Table 12.6, CP  29.2 J/(mol · K) and CV  20.9 J/(mol · K). Note that the change in
temperature on the Kelvin scale is the same as the change in temperature on the Celsius scale.
Solve: (a) The number of moles of oxygen is
n
M
1.0 g

 0.03125 mol
M mol 32 g/mol
For the isobaric process,
Q  nCP T  (0.03125 mol)(29.2 J/(mol · K))(100C)  91.2 J
which will be reported as 91 J.
(b) For the isochoric process,
Q  nCV T  91.2 J  (0.03125 mol)(20.9 J/(mol · K))T  T  140C
P12.49. Prepare: From the first law of thermodynamics, Q  Eth – W, where W is the work done by the
gas. W  0 at constant volume, so, using Equation 12.26, Eth  Q  nCVT. From Table 12.6, the value of CV for
a monatomic gas is 12.5 J/(mol · K) (which is equal to 3R/2). For a diatomic gas, we take CV to be 20.8 J/(mol ·
K) (which is equal to 5R/2).
Solve: For a monatomic gas,
Eth  nCVT  1.0 J  (1.0 mol)(12.5 J/ (mol  K))T  T  0.0800C or 0.080 K
P12.50. Prepare: The heating is an isochoric process. From Equation 12.26, Q  nCV T, and from Table
12.6, the value of CV for oxygen is 20.9 J/(mol · K) and for helium is 12.5 J/(mol · K). The number of moles n of
each gas is obtained from the mass of the gas and its molar mass.
Solve: The number of moles of helium is
n
M
2.0 g

 0.50 mol
M mol 4 g/mol
For the isochoric processes,
QHe  nCVT  (0.50 mol)(12.5 J/ (mol  K ))T


M
QO2  nCV T  
 (20.9 J/ (mol  K ))T
 32 g/mol 
Because QHe  QO2,


M
(0.50 mol)(12.5 J/(mol  K))  
 (20.9 J/(mol  K))  M  9.6 g
 32 g/mol 
P12.51. Prepare: Please refer to the following figure. The work done by a gas is equal to the area under the
pV graph between Vi and Vf. The work done by a gas is positive when V > 0, negative when V < 0, and zero
when V  0. W12  0, W23  Area (I)  Area (II), and W31  –Area (II). Thus, the net work done is equal to
Area (I). That is, the work done by the gas per cycle is the area inside the closed p-versus-V curve. We also need
to convert the units of pressure from atm to Pa using the conversion: 1 atm  1.013  105 Pa.
Thermal Properties of Matter
Solve:
12-21
The area inside the triangle is
Wby gas 
1
1
1.013 105 Pa 
6
3
(3 atm  1 atm)(600 106 m3  200 106 m3 )   2 atm 
 (400 10 m )  41 J
2
2
1 atm

P12.52. Prepare: Please refer to the next figure. The work done by a gas is equal to the area under the pV
graph between Vi and Vf. The work done by a gas is positive when V > 0, negative when V < 0, and zero when
V  0. W12 = I + III, W23 = II + IV, and W31 = –III –IV. Thus, the net work done is equal to I + II. That is,
the work done by the gas per cycle is the area inside the closed p-versus-V curve. We also need to convert the
units of pressure from atm to Pa using the conversion 1 atm  1.013  105 Pa and the units of volume from cm3 to
m3 using 1 cm3  10–6 m3.
Solve:
The work done by the gas per cycle is the area enclosed within the pV curve. We have
60 J 
1
2(60 J)
( pmax  100 kPa)(800 cm3  200 cm 3 ) 
2
600 106 m3
 pmax  1.0  105 Pa  pmax  3.0 105 Pa  300 kPa
P12.53. Prepare: The rate of conduction across a temperature difference is given in Equation 12.32.
Q  kA 

T
t  L 
where A  4.0 m  5.5 m  22 m2 is the area, L  0.018 m is the thickness of the flooring, and k  0.2W/(m  K) is
the thermal conductivity of wood given in Table 12.7. T  19.6C  16.2C  3.4C.
Solve:
Q  kA 
(0.2 W/(m  K))(22 m2 )
   T 
(3.4C)  830 J/s  830 W
t  L 
0.018 m
Assess: 830 W is about as much as a dozen incandescent light bulbs. In the winter when you are trying to keep
the room warm this energy is being wasted; you could do drastic things like increase the thickness of the wood,
12-22 Chapter 12
or simpler, cheaper things like cover the floor with carpet, which has a much smaller k. In the summer you might
be grateful to have this energy conducted from the room if the subfloor can stay at a cooler temperature.
P12.54. Prepare: The bottom of the interior of the kettle is the same temperature of the boiling water,
100C. Equation 12.31 can be used.
Solve: Solving Equation 12.31 for the temperature difference,


3.0 103 m
 Q  L 
T   
 (800 W) 
 1.3 101 K

2 

t
kA
(400
W/(m

K))

(0.
12m)
 



The bottom of the kettle is only 0.13 K hotter than the interior.
Assess: This result makes sense.
P12.55. Prepare: The rate of energy loss by radiation is given by Equation 12.33.
Q
 e AT 4
t
We are given e  0.20, T  700C  973 K, A  6  (2.0 cm  2.0 cm)  24 cm2  0.0024 m2. The textbook gives
Stefan’s constant as   5.67  10–8W/(m2  K4).
Solve:
Q
 e AT 4  (0.20)(5.67 108 W/(m 2  K 4 ))(0.0024 m 2 )(973 K) 4  24 W
t
Assess: 700C is quite hot, so the cube radiates a reasonable amount of energy, but even at 700C the radiation
is mostly infrared, not visible. If we double the (absolute) temperature the total radiation would increase by a
factor of 16 (due to the T 4 ) and also a greater portion of the radiation would be in the visible range.
P12.56. Prepare: Equation 12.32 applies. For maximum possible radiated power, assume the
emissivity is 1.0.
Solve:
Q
 e AT 4  (1.0)(5.67 108 W/(m2  K 4 ))4 (0.025 m) 2 (373 K) 4  8.6 W
t
Assess: This result makes sense for such a small sphere.
P12.57. Prepare: The rate of net energy loss by radiation is given by Equation 12.34.
Qnet
 e A(T 4  T 04 )
t
where T0 is the termperature of the surroundings.
We are given T  30C  303 K, T0  –10C  263 K, and A  0.030 m2. We are told to assume the emissivity of
seal skin is the same as human skin; the text gives this vaule as e  0.97.
The textbook gives Stefan’s constant as   5.67 108 W/(m2  K 4 ).
Solve:


Qnet
 e A T 4  T 04  (0.97)(5.67 108 W/(m 2  K 4 ))(0.030 m 2 )[(303 K) 4  (263 K)4 ]  6.0 W
t
Assess: 6 W isn’t a lot, but it is sufficient to cool the seal when the surroundings are very cool. If there were no
thermal windows the seal would have difficulty regulating its temperature.
Thermal Properties of Matter
12-23
P12.58. Prepare: To use the formula for the rate of emission of radiated energy we need to know the area of the
panels. Since they emit radiation from both sides, the area is given by the following:
A  2LW  2(3.6 m)(1.8 m)  13.0 m2. The temperature is 279 K. Since these panels are designed to emit
radiation, they probably have an emissivity close to 1 so we will use e  1 .
Solve: We are ready to find the power using Equation 12.32.
P
Q
 e AT 4  (1)(5.67 108 W/(m2  K 4 ))(13.0 m2 )(279 K) 4  4500 W
t
Assess: This seems reasonable since this is about five times the power emitted by a typical person, 900 W,
and the plates have an area about seven times a typical value for a person, 2 m2 . The fact that the power is not
quite seven times as great is owing to the somewhat cooler temperature of the panels.
P12.59. Prepare: The rate of energy loss by radiation is given by Equation 12.33.
Q
 e AT 4
t
We are given e  0.23, T  1500C  1773 K, and Q/t  60 W. We are asked to find A.
The textbook gives Stefan’s constant as   5.67 108 W/(m2  K 4 ).
Solve: Solve the equation for A.
A
Q/t
60 W

 4.7 104 m2
4
8
e T
(0.23)(5.67 10 W/(m 2  K 4 ))(1773 K) 4
Assess: We knew that light bulb filaments have a small surface area, so we are not concerned to get a small
answer. The units work out properly.
P12.60. Prepare: Since you are lying on the ground, your back does not emit radiation or absorb radiation
from the sky. We might guess that a little more than half of a person’s area is off the ground. A typical person’s
surface area from the book is 1.8 m2 . So we will use 1 m2 for the area in contact with the air. We will use
e  0.97. The temperature of your clothing is Tc  303 K and the temperature of the sky is Ts  233 K.
Solve: The net rate that your body loses energy to the sky is given by Equation 12.33.
Qnet
 e A(Tc 4  Ts 4 )  (0.97)(5.67 108 W/(m2  K 4 ))(1 m2 )((303 K) 4  (233 K)4 )  300 W
t
Assess: This is about three times the net rate of heat loss you would experience if you were in a room at room
temperature (see Example 12.22). This is true despite the lower surface area here. The big difference is that heat
is absorbed by your body much slower in this case.
P12.61. Prepare: Treat the gas in the sealed container as an ideal gas and use Equation 12.14.
Solve: (a) From the ideal gas law equation pV  nRT, the volume V of the container is
V
nRT (2.0 mol)(8.31 J/(mol  K))[(273  30) K]

 0.050 m3
p
1.013 105 Pa
12-24 Chapter 12
(b) The before-and-after relationship of an ideal gas in a sealed container (constant volume) is
p1V p2V
T
(273  130) K

 p2  p1 2  (1.0 atm)
 1.3 atm
T1
T2
T1
(273  30) K
Assess: Note that gas-law calculations must use T in kelvins and pressure must be in Pa.
P12.62. Prepare: Treat the air in the compressed-air tank as an ideal gas and use Equation 12.12 to find n.
We will, however, need to convert pressure and temperature to SI units using 1 atm  1.013  105 Pa and T (in K)
 T (in C)  273. Also note that  r 2h is the volume of a cylinder.
Solve: (a) From the ideal-gas law pV  nRT, the number of moles n is
pV p( r 2h) (150 atm)(1.013 105 Pa/1 atm)[ (0.075 m) 2 (0.50 m)]


 55.1 mol
RT
RT
(8.31 J/(mol  K))[(273  20) K]
which will be reported as 55 mol.
(b) At STP, the ideal-gas law yields
n
V
nRT (55.1 mol)(8.31 J/(mol  K))(273 K)

 1.2 m3
p
1.013 105 Pa
Assess: The volume of the compressed air tank is (r 2 )h  8.84  103 m3. The volume at STP is 140 times the
volume of the tank. That is, the air is compressed 140 times compared to STP values, which does not look
unreasonable.
P12.63. Prepare: Treat the helium gas in the sealed cylinder as an ideal gas. The volume of the cylinder
is
V  r 2h  (0.05 m)2(0.30 m)  2.356  10–3 m3
The gauge pressure of the gas is
120 psi 
1 atm 1.013 105 Pa

 8.269  105 Pa
14.7 psi
1 atm
so the absolute pressure of the gas is 8.269  105 Pa  1.013  105 Pa  9.282  105 Pa. The temperature of the
gas is T  (273 20) K  293 K.
Solve: (a) The number of moles of the gas in the cylinder is
n
pV (9.282 105 Pa)(2.356 103 m3 )

 0.898 mol
RT
(8.31 J/(mol  K))(293 K)
So, the number of atoms is
N  nNA  (0.898 mol)(6.02 1023 mol1)  5.4 1023 atoms
(b) The mass of the helium is
M  nM mol  (0.898 mol)(4 g/mol)  3.6 g  3.59 103 kg
P12.64. Prepare: We are given a mass of helium that is to be used to fill balloons of a known volume. We
need to find the mass of helium that goes into a balloon. This we can do by first finding the number of moles
from the ideal gas law. For the temperature, we will use 20 C or 293 K and for the pressure we will use 1 atm
or 1.013 105 Pa.
Solve: We can rewrite the ideal gas law as n  pV / RT to obtain the following:
n
(1.013 105 Pa)(0.010 m3 )
 0.416 mol
(8.315 J/(mol  K))(293 K)
Helium has an atomic mass of 4 so a mole of helium has a mass of 4 g. Thus each balloon holds a mass of
helium given by the following:
Thermal Properties of Matter
12-25
m  (0.416 mol)(4 g/mol)  1.66 g
Finally, the number of balloons that can be filled is found by dividing this number into the total mass of helium
in the tank.
N
30 g
 18
1.66 g
You can fill 18 balloons with the helium.
Assess: The value we got for the number of moles of helium, 0.42 mol, is reasonable since a mole of ideal gas
at STP is about 22 L and in this problem we have 10 L of helium, or about half as much.
P12.65. Prepare: We find the heat which must be supplied to the air from Equation 12.25 for constant
pressure processes. The value of Cp is given in Table 12.6: Cp  20.8 J/(mol  K). The number of moles of air
inhaled can be obtained from the ideal gas law. We also find the final volume from the ideal gas law and subtract
the initial volume to obtain V .
Solve: (a) Solving the ideal gas law for the number of moles using the initial volume and temperature, we obtain
the following:
n
pVi (1.013 105 Pa)(4.0 103 m3 )

 0.134 mol
RTi
(8.315 J/(mol  K))(273 K)
The amount of heat which must be supplied is
Q  nCpT  (0.134 mol)(29.1 J/(mol  K))(37 K)  144 J  140 J. s
(b) From the ideal gas law as stated in Equation 12.14, the final volume of the air is given by the following:
Vf 
VT
(3.0 L)(310 K)
i f

 3.4 L
Ti
273 K
The increase in volume is V  Vf  Vi  0.4 L.
Assess: We could have estimated the change in volume using Equation 12.18, V  Vi T , but this would not
have been as accurate since  depends on temperature (it is given for T  20 C in Table 12.3) and here the
temperature varies from 0 C to 37 C.
P12.66. Prepare: The carbon dioxide in the cube is an ideal gas and we will use the ideal gas Equation
12.12 at STP with n  M/Mmol.
Solve: Using the ideal gas equation
pV  nRT  V 
nRT
MRT

p
pM mol
The molar mass of CO2 is 44 g/mol or 0.044 kg/mol. Thus,
V
(10,000 kg)(8.31 J/(mol  K))(273 K)
 5090 m3
(1.013 105 Pa)(0.044 kg/mol)
The length of the cube is L  (V ) 1/3  17 m.
Assess: The length of the cube is large, but so is the mass of the carbon dioxide emission.
P12.67. Prepare: We will use the ideal gas Equation 12.14 and assume that the volume of the tire and that
of the air in the tire is constant. That is, the gas undergoes an isochoric (constant-volume) process. Because the
gas equation needs absolute rather than gauge pressure, a gauge pressure of 30 psi corresponds to an absolute
pressure of (30 psi)  (14.7 psi)  44.7 psi.
Solve: Using the before-and-after relationship of an ideal gas for an isochoric process,
T
pi pf
 273  45 
 pf  f pi  

 (44.7 psi)  49.4 psi
Ti
Ti Tf
 273  15 
12-26 Chapter 12
Your tire gauge will read a gauge pressure pf  49.4 psi  14.7 psi  34.7 psi, which is to be reported as 35 psi.
Assess: A 5 psi increase in gauge pressure due to an increase in temperature by 30C is reasonable.
P12.68. Prepare: For a gas in a sealed container (n is constant) we use Equation 12.14.
pfVf
pV
 i i
Tf
Ti
In our constant volume case Vi and Vf cancel.
The 35 psi is the gauge pressure. We take the usual steps of converting the gauge pressure to absolute pressure
and converting the temperatures to absolute temperatures.
pi  35 psi  14.7 psi  49.7 psi
Ti  20C  293 K
Tf  0C  273 K
Solve: Solve the top equation for pf.
T 
 273 K 
pf   f  pi  
 49.7 psi  46.3 psi
T
 293 K 
 i
This answer is the absolute final pressure. To report the answer as a gauge pressure, we subtract 1 atm  14.7 psi.
46.3psi  14.7psi = 31.6psi, which should be reported to two significant figures as 32 psi.
Assess: We expected the pressure to decrease with decreasing temperature. On the absolute scale the
temperature didn’t decrease a lot, however, so neither did the pressure.
We did not need to convert all the pressure data to SI units because we wanted the answer in the same units as
the original data, and the formula shows the ratio in any units (using absolute pressure, not gauge pressure) is the
ratio of the absolute temperatures.
P12.69. Prepare: The gas’s pressure does not change, so this is an isobaric process. We will use Equation
12.14 with pi  pf.
Solve: The triple point of water is 0.01C or 273.16 K, so Ti  273.16 K. Because the pressure is a constant,
 1638 mL 
V
Vi Vf
 Tf  Ti f  (273.16 K) 

  447.44 K  174C
Vi
Ti Tf
 1000 mL 
P12.70. Prepare: Assume that the compressed air in the cylinder is an ideal gas. The volume of the air in
the cylinder is a constant. We will use Equation 12.14 to calculate the new pressure in atm and compare it with
the maximum pressure (in atm) of the compressed gas that the cylinder can withstand.
Solve: Using the before-and-after relationship of an ideal gas,
 1223K  Vi
pfVf
pV
T V
 i i  pf  pi f i  (25 atm) 
 104 atm

Tf
Ti
Ti Vf
 293K  Vi
where we have converted to the Kelvin temperature scale. Because the pressure does not exceed 110 atm, the
compressed air cylinder does not blow.
P12.71. Prepare: The rigid sphere’s volume does not change, so this is an isochoric process. The air is
assumed to be an ideal gas according to Equation 12.14.
Solve: (a) When the valve is closed, the air inside is at pi  1 atm and Ti  100C. The before-and-after
relationship of an ideal gas in the closed sphere (constant volume) is
T 
pV
pV
(273  0) K
i
 f  pf  pi  f   (1.0 atm)
 0.73 atm
Ti
Tf
T
(273
 100) K
 i
(b) Dry ice is CO2. Cooling the sphere to –78.5C gives
T 
(273  78.5) K
pf  pi  f   (1.0 atm)
 0.52 atm
373 K
 Ti 
Thermal Properties of Matter
12-27
Assess: Under constant-volume conditions, pressure decreases with decrease in temperature, as we would have
expected.
(b) The final pressure is 0.52 atm.
P12.72. Prepare: Please refer to Figure P12.72. Since the process is occurring at constant pressure, it is
isobaric and the work done by a gas is the area under the p-versus-V curve. The gas is compressing, so the work
done by the gas is negative or the work done on the gas is positive.
Solve: The work done by gas in an isobaric process is
W  pV  p (Vf  Vi )
Substituting into this equation,
80 J  (200 103 Pa)(V1  3V1)  Vi  2.0 104 m3  200 cm3
Assess: The work done on a gas during compression is positive.
P12.73. Prepare:
Please refer to Figure P12.73. The work done by a gas is the area under the p-versus-V
curve. The gas is expanding, so the work done by the gas is positive or the work done on the gas is negative.
Solve: The area under the pV curve is the area of the rectangle and triangle. The work done by the gas is as
follows:
1
(200 106 m3 )(200 103 Pa)  (200 10 6 m 3 )(200 103 Pa)  60 J
2
Thus, the work done on the gas is W   60 J.
Assess: The environment does negative work on the gas as it expands, so we expected the work to be negative.
P12.74. Prepare: Assume the gas is an ideal gas. We will use Equation 12.12, convert all quantities to SI
units, and refer to Figure P12.74 for initial and final pressure and volume.
Solve: (a) We can find the temperatures directly from the ideal gas law.
T1 
p1V1 (3.0 atm 101,300 Pa/atm)(1000 cm3 10 6 m3/cm3 )

 366 K  93C
nR
(0.10 mol)(8.31 J/(mol  K))
T2 
p2V2 (1.0 atm 101,300 Pa/atm)(3000 cm3 10 6 m3/cm3 )

 366 K  93C
nR
(0.10 mol)(8.31 J/(mol  K))
(b) T2  T1, so this is an isothermal process.
(c) A constant volume process has V3  V2. Because p1  3p2, restoring the pressure to its original value means
that p3  3p2. From the ideal gas law,
 p  V 
p3V3 p2V2

 T3   3  3  T2  3  1 T2  3  366 K  1098 K  825C
T3
T2
 p2  V2 
The temperature to be reported is thus 830C.
P12.75. Prepare: Assume CO2 gas is an ideal gas and Equation 12.12 is applicable. The molar mass for CO 2
is Mmol  44 g/mol, so a 10 g piece of dry ice is 0.2273 mol. This becomes 0.2273 mol of gas at 0C.
Solve: (a) With V1  10,000 cm3  0.010 m3 and T1  0C  273 K, the pressure is
p1 
nRT1 (0.2273 mol)(8.31 J/(mol  K))(273 K)
 5.156  104 Pa  0.509 atm

V1
0.010 m3
or 0.51 atm to two significant figures.
(b) From the isothermal compression, that is, at constant temperature,
p2V2  p1V1  V2  V1
 0.509 atm 
p1
3
3
3
 (0.010 m3 ) 
  1.70 10 m  1700 cm
p2
 3.0 atm 
From the isobaric compression, that is at constant pressure,
T3  T2
 1000 cm3 
V3
 (273 K) 
  161 K  112C
3
V2
 1700 cm 
12-28 Chapter 12
or –110C to two significant figures.
(c)
Assess: When volume and pressure appear as a ratio of before and after values, we do not have to convert them
to SI units. Temperature, however, must always be converted to Kelvin in the gas equation.
P12.76. Prepare: Please refer to the following figure. The work done in expanding the swim bladder is given
by W  pV . To use this, we need to find the pressure at the lower depth. The temperature and number of
moles of gas remain constant as the fish descends, so from the ideal gas law, pV is constant, that is pV
i i  pfVf
or, rearranging factors, pf 
pV
i i
. Since the ratio of the final volume to the initial volume is 60% and the initial
Vf
pressure was 3.0 atm, we can say that Vf  (0.60)Vi and pf 
pV
(3.0 atm)Vi
i i

 5.0 atm.
Vf
(0.60)Vi
given: Vi  5.0 104 m3 , from which it follows that Vf  (0.60)Vi  3.0 104 m3.
Solve: After descending, the fish pumps gas into its swim bladder to go from the present volume of
3.0 104 m3 back to its original volume of 5.0 104 m3. The work done is as follows:
 1.013 105 Pa 
4
3
4
3
W  pV  (5.0 atm) 
 (5.0 10 m  3.0 10 m )  100 J
1 atm


Assess: The units work out to joules since 1 Pa  1 N/m2.
We also are
Thermal Properties of Matter
12-29
P12.77. Prepare: The rate of heat transfer of solar energy is Q/t, which is equal to solar power. That is, the
heat absorbed is (solar power)  t which is also equal to Mcwater T according to Equation 12.21. From the given
information, we can then easily find t.
Solve: The area of the garden pond is A  (2.5 m)2  19.635 m2 and the mass of water in the pond is 5.9  103 kg. The
water absorbs all the solar power, which is
(400 W/m 2)(19.635 m2 )  7854 W
This power is used to raise the temperature of the water. That is,
Q  (7854 W)t  Mcwater T  (5900 kg)(4190 J/(kg  K))(10 K)  t  31,476 s  8.7 h
Assess: It is a common experience that a pool of water takes a few hours to warm up. A value of 8.7 h to heat
5900 kg water by 10C is reasonable.
P12.78. Prepare: Since the compression is adiabatic, there is no heat flow and the first law of
thermodynamics says that Eth  W where W is the work done on the gas. So we need to find the change in
thermal energy from Equation 12.30.
Solve: Combining the two equations mentioned above, we have the following:
W  Eth 
3
3
nRT  (0.030 mol)(8.315 J/(mol  K)(50 C  10 C)  15 J
2
2
Assess: We were able to use degrees Celsius in the above calculation since we were subtracting two
temperatures.
P12.79. Prepare: Since this is an adiabatic process, there is no heat flow and so the first law of
thermodynamics says that Eth  W . Now W in this context is work done on the gas. In this problem, the
thermal energy decreases so that Eth and W are negative. A negative value of W means that work is done by
the gas. We can use Equation 12.30 to find the change in thermal energy.
Solve: Combining the two equations mentioned above gives us the following:
W  Eth 
3
3
nRT  (0.15 mol)(8.315 J/(mol  K)(  10 C  20 C)  56 J
2
2
The work done by the gas is 56 J.
Assess: The reduction in temperature is made because the gas is doing work in expanding and the “fuel” it uses
in doing that work is its own thermal energy. As the available thermal energy goes down, the temperature also
goes down.
P12.80. Prepare: We will need the formula for gravitational potential energy, U g  mgh, as well as a
formula which relates a change in temperature to a change in thermal energy. Equation 12.24 gives the heat
needed to change the temperature of an object. In this case, the heat produced by the body’s metabolism is turned
into thermal energy so we write Eth  mcT . We will use the mammalian specific heat from Table 12.4.
12-30 Chapter 12
Solve: (a) Susan’s change in potential energy is given by the following equation:
Ug  mgh  (68 kg)(9.8 m/s2 )(59 m)  39300 J  39000 J
(b) If this increase in potential energy represents 25% of the total energy used by her body, then she has used
four times this much energy, or 157 kJ  160 kJ.
(c) 75% of the energy used is wasted and goes to thermal energy so Eth  (0.75)(157 kJ)  118 kJ . We can
solve the equation Eth  mcT for T to obtain
T 
Eth
118 103 J

 0.51 K
mc
(68 kg)(3400 J/(kg  K))
Assess: This increase is about 0.5 C which can be converted to 0.9 F. This doesn’t seem like much, but on
the other hand, this exercise was less strenuous than it might seem. The energy spent, 160 kJ can be converted
to about 38 dietary Calories. The same amount of energy would be spent in walking about two fifths of a mile.
P12.81. Prepare: There are two interacting systems: the nuclear reactor and the water. The heat
generated by the nuclear reactor is used to raise the water temperature. For the closed reactor –water system,
energy conservation per second requires Q  Qreactor  Qwater  0 J. The heat from the reactor in t  1 s is
Qreactor = –2000 MJ = –2.0  10 9 J and we will use Equation 12.21 for Qwater.
Solve: The heat absorbed by the water is
Qwater  mwater cwater T  mwater (4190 J/(kg  K)(12 K)
 2.0 109 J  mwater (4190 J/(kg  K)(12 K)  0 J  mwater  3.98 10 4 kg
Each second, 3.98  104 kg of water is needed to remove heat from the nuclear reactor. Thus, the water flow per
minute is
3.98 104
kg 60 s

 2.4  106 kg/min
s min
P12.82. Prepare: From Table 11.4, the metabolic power of a 68 kg cyclist is 480 W. We assume that 25% of this
goes to propelling the cyclist and the other 75%, or 360 W, becomes heat which serves to evaporate perspiration.
Equation 12.26 gives the heat needed to evaporate a liquid. From the discussion following Table 12.5, a good value for
the latent heat of vaporization of sweat is Lv  2.4 106 J/kg.
Solve: We solve Equation 12.26 for m and get m  Q / Lv . This is the mass of perspiration which heat Q
could evaporate. We know
Q
 360 W and we can combine this with the preceding equation to obtain the
T
following:
m Q / Lv Q / t
360 J/s



 1.5 104 kg/s  0.54 kg/hr
t
t
Lv
2.4 106 J/kg
Thermal Properties of Matter
12-31
Assess: A kilogram of water has a volume of about one liter, so this is about half a liter per hour. A value on
the order of one liter per hour seems reasonable.
P12.83. Prepare: The entire kinetic energy of the car goes into heating the four brakes. Equation 12.21
applies. Table 12.4 gives the specific heat of iron, c  449 J/(kg  K). We are given the mass of each brake disk,
m  4.0 kg and the mass of the car M  1200 kg. The speed of the car is v  60 mph, which we convert to SI units
(keeping one extra significant figure in the intermediate calculation).
 0.447 m/s 
v  60 mph 
  26.8 m/s
 1 mph 
Solve: Setting the kinetic energy equal to the heat going into the four brakes gives
Q  4mcT 
1
Mv 2
2
Now solve for the temperature increase.
T 
1 Mv 2 1 (1200 kg)(26.8 m/s) 2

 60 K
8 mc 8 (4.0 kg)(449 J/(kg  K))
Assess: This is a significant rise in temperature, but brakes really do heat up quite a bit. The kinetic energy of
the car must go somewhere. One other option is to let the translational kinetic energy of the car be transformed
into rotational kinetic energy of a flywheel, so that it can be recovered after the light turns green.
Careful examination of the units shows they cancel properly.
P12.84. Prepare: We need Equation 12.26 which gives the amount of heat need to evaporate a mass of
liquid. We are given the rate that heat must be absorbed by the water as follows: Q / t  2500 W. We can use
the latent heat of vaporization given in the discussion following Table 12.5, Lv  2.4 106 J/kg.
Solve: Solving Equation 12.26 for the mass gives us: m  Q / Lv and dividing both sides by t gives our
answer
m Q / Lv Q / t
2500 J/s  3600 s 



 3.8 kg/hr
t
t
Lv
2.4  106 J/kg  1 hr 
Assess: This works out to 3.8 L/hr, which seems reasonable. During vigorous exercise, humans sweat a
couple of liters per hour. An elephant at rest has about twice the metabolic rate of a human exercising vigorously
(using the data from Table 11.4), so we would expect a value of around 4 L/hr.
P12.85. Prepare: Use the conservation of energy. First the gravitational potential energy of the water is
converted into kinetic energy, and then, on impact, into thermal energy. We’ll set the change of gravitational
potential energy (which decreases, hence the negative sign) equal to the change in thermal energy (which
increases).
Mgh  Mcw T
Solve: Solve for T .
T 
Mg h  g h (9.8 m/s 2 )(10 m)


 0.023 K
Mcw
cw
4190 J/(kg  K)
The answer is small enough to not be noticeable, which probably jibes with your experience.
Assess: If it is not obvious that the mass of the water should cancel (that is, the temperature rise is the same for
any amount of water dropped from 10 m), keep thinking about it until it makes sense; each little chunk of water
starts with some gravitational potential energy, which changes the thermal energy (and raises the temperature) of
that little chunk. That happens for each chunk, so the temperature goes up the same amount no matter how much
water you drop.
P12.86. Prepare: Changing solid lead at 20C to liquid lead at its melting point (Tm  328C) requires two steps:
raising the temperature to Tm and then melting the solid at Tm to a liquid at Tm. We will use Equations 12.21 and 12.22
and work in SI units. The values of the specific heat and the heat of fusion are given in Tables 12.4 and 12.5.
12-32 Chapter 12
Solve:
The equation for the total heat is
Q  Q1  Q2  1000 J  Mclead(Tf – Ti)  MLf
 1000 J  M(128 J/( kg  K )(328 – 20) K  M(0.25  105 J/kg)  M 
1000 J
 15.5 g
(64,424 J/kg)
The maximum mass of lead you can melt with 1000 J of heat is 15.5 g.
P12.87. Prepare: Heating the material increases its thermal energy. Please refer to Figure P12.87. The material
melts at 300C and undergoes a solid-liquid phase change. The material’s temperature increases from 300C to
1500C. Boiling occurs at 1500C and the material undergoes a liquid-gas phase change. We will use Equations 12.21
and 12.21 to determine the specific heat and the heat of vaporization of the liquid.
Solve: (a) In the liquid phase, the specific heat of the liquid can be obtained as follows:
Q  McT  c 
 20 kJ 
1 Q 
1


  83 J/(kg  K)
M T  0.200 kg  1200 K 
(b) The latent heat of vaporization is
Lv 
Q
40 kJ

 2.0 105 J/kg
M (0.200 kg)
Assess: The values obtained are of the same order of magnitude as in Tables 12.4 and 12.5 for a few materials.
P12.88. Prepare: There are three interacting systems: aluminum, copper, and ethyl alcohol. The aluminum,
copper, and alcohol form a closed system, so Q  QAl  QCu  Qeth  0 J. We will use Equation 12.21 for each of
the three systems in the above equation. The specific heats of the three materials are given in Table 12.4. We must
work in SI units, so have mass in kg and temperature in Kelvins. The mass of the alcohol is Meth  V 
(790 kg/m3)(50  10 –6 m3)  0.0395 kg.
Solve: Expressed in terms of specific heats and using the fact that T  Tf – Ti, the Q  0 J condition is
M AlcAlTAl  M CucCu TCu  M ethceth Teth  0 J
Substituting into this expression,
(0.010 kg)(900 J/(kg  K)(298 K  473 K)  (0.020 kg)(385 J/(kg  K)(298 K  T )
 (0.0395 kg)(2400 J/(kg  K)(298 K  288 K)  1575 J  (7.7 J/K)(298  T )  948 J  0 J
 T  216.6 K  56C
P12.89. Prepare: There are two interacting systems: aluminum and ice. The system comes to thermal
equilibrium in four steps: (1) the ice temperature increases from 10C to 0C, (2) the ice becomes water at 0C,
(3) the water temperature increases from 0C to 20C, and (4) the cup temperature decreases from 70C to 20C.
We will use Equations 12.21 and 12.22. The specific heats and the heat of fusion are given in Tables 12.4 and 12.5.
Solve: The aluminum and ice form a closed system, so Q  Q1  Q2  Q3  Q4  0 J. These quantities are
Q1  M icecice T  (0.100 kg)(2090 J/(kg  K)(10 K)  2090 J
Q2  M ice Lf  (0.100 kg)(3.33 105 J/kg)  33,300 J
Q3  M icecwater T  (0.100 kg)(4190 J/(kg  K)(20 K)  8380 J
Q4  M Al cAl T  M Al (900 J/(kg  K)( 50 K)  (45,000 J/kg) M Al
The Q  0 J equation now becomes
43,770 J – (45,000 J/kg)MAl  0 J
The solution to this is MAl  0.97 kg.
P12.90. Prepare: We have a thermal interaction between the thermometer and the water. Equation 12.21
will be used for both the thermometer and water along with the conservation of energy equation.
Solve: The conservation of energy equation Qthermo  Qwater  0 J is
Thermal Properties of Matter
12-33
Mthermocthermo(Tf – Ti thermo)  Mwatercwater(Tf – Ti water)  0 J
The thermometer slightly cools the water until both have the same final temperature Tf  71.2C. Thus
(0.050 kg)(750 J/(kg  K)(71.2C  20.0C)  (0.200 kg)(4190 J/(kg  K)(71.2C  Ti water )
 1920 J  838 (J/K)(71.2C  Ti water )  0 J  Ti water  73.5C
Assess: The thermometer reads 71.2C for a real temperature of 73.5C. This is reasonable.
P12.91. Prepare: There are two interacting systems: coffee (i.e., water) and ice. Changing the coffee
temperature from 90C to 60C requires four steps: (1) raise the temperature of ice from 20C to 0C, (2) change ice
at 0C to water at 0C, (3) raise the water temperature from 0C to 60C, and (4) lower the coffee temperature
from 90C to 60C. We will use Equations 12.21 and 12.22. The specific heats of water and ice and the heat of
fusion of ice are given in Tables 12.4 and 12.5.
Solve: For the closed coffee-ice system,
Q  Qice  Qcoffee  (Q1  Q2  Q3 )  (Q4 )  0 J
Q1  M icecice T  M ice (2090 J/(kg  K)(20 K)  M ice (41,800 J/kg)
Q2  M ice Lf  M ice (330,000 J/kg)
Q3  M icecwater T  M ice (4190 J/kg)(60 K)  M ice (251,400 J/kg)
Q4  M coffeeccoffee T  (300  106 m3 )(1 000 kg/m3 )(4190 J/(kg  K)( 30 K)  37,000 J
The Q  0 J equation thus becomes
M ice (41,800  330,000  251,400) J/kg  37,710 J  0 J  M ice  0.0605 kg  61g
Assess: 61 g is the mass of approximately one ice cube.
P12.92. Prepare: We will use the first law of thermodynamics to describe this isobaric process. Wgas is
negative because the gas is compressed. Compression transfers energy into the system, that is, work done on the gas
is positive. Also, 100 J of heat energy is transferred out of the gas, that is Q  –60 J.
Solve: The first law of thermodynamics is
Eth  Wgas  Q   pV  Q  –(4.0  105 Pa)(200 – 600)  106 m3 – 100 J  60 J
Thermal energy increases by 60 J.
P12.93. Prepare: Treat helium as an ideal gas. The process in part (a) is isochoric and the process in part
(b) is isobaric. Initially Vi  (0.20 m)3  0.0080 m3  8.0 L and Ti  293 K. Helium has an atomic mass number A 
4, so 3 g of helium is n  M/Mmol  0.75 mole of helium. We will use the ideal gas Equations 12.12 and 12.14.
Solve: (a) We can find the initial pressure from the ideal-gas law as follows:
pi 
nRTi (0.75 mol)(8.31 J/(mol  K)(293 K)

 228 kPa  2.25 atm
Vi
0.0080 m3
Heating the gas will raise its temperature. A constant volume process has Q  nCVT, so
T 
Q
1000J

 107 K
nCV (0.75 mol)(12.5 J/(mol  K)
This raises the final temperature to Tf  Ti  T  400 K. Because the process is isochoric,
pf
p
T
400 K
 i  pf  f pi 
(2.25 atm)  3.1 atm
Tf
Ti
Ti
293 K
(b) The initial conditions are the same as in part (a), but now Q  nCPT. Thus,
T 
Q
1000 J

 64.1 K
nCP (0.75 mol)(20.8 J/(mol  K)
12-34 Chapter 12
Now the final temperature is Tf  Ti  T  357 K. Because the process is isobaric,
Vf Vi
T
357 K
  Vf  f Vi 
(0.0080 m3 )  0.00975 m3  9.75 L
Tf Ti
Ti
293 K
which is to be reported as 9.8 L.
P12.94. Prepare:
Please refer to Figure P12.94. The monatomic gas is an ideal gas, which is subject to
isobaric and isochoric processes. We will use the ideal gas Equation 12.12 and Equations 12.25 and 12.26.
Solve: (a) For this isobaric process, p1  4.0 atm, V1  800  106 m3, p2  4.0 atm, and V2  1600  106 m3. The
temperature T1 of the gas is obtained from the ideal-gas equation as
pV
T1  1 1  390 K
nR
where n  0.10 mol. Also,
T2  T1
 1600  106 m3 
V2
 T1 
 2T1  780 K
6
3 
V1
 800 10 m 
Thus, the heat required for the process 1  2 is
Q  nCP (T2  T1 )  (0.10 mol)(20.8 J/(mol  K)(390 K)  811 J
which is 810 J in two significant figures.
This is heat transferred to the gas.
(b) For the isochoric process, V2  V3  1600  106 m3, p2  4.0 atm, p3  2.0 atm, and T2  780 K. T3 can be obtained
from the ideal gas equation as follows:
 2.0 atm 
p2V2 p3V3

 T3  T2 ( p3/p2 )  (780 K) 
  390 K
T2
T3
 4.0 atm 
The heat required for the process 2  3 is
Q  nCV (T3  T2 )  (0.10 mol)(12.5 J/mol K)(390 K  780 K)   488 J
which is –490 J in two significant figures.
Because of the negative sign, this is the amount of heat removed from the gas.
(c) The change in the thermal energy of the gas is

Eth  (Q12  Q23 )  (W12  W23 )  811J  488J  W12  0J  324 J  pV
 324 J – (4.0  1.013  105 Pa)(1600  106 m3 – 800 106 m3)  0 J
Assess: This result was expected since T3  T1.
P12.95. Prepare: The gas is assumed to be an ideal gas that is subjected to isobaric and isochoric processes.
Please refer to Figure P12.95. We will use SI units, the ideal gas Equation 12.12, and Equations 12.25 and 12.26.
Solve: (a) The initial conditions are p1  3.0 atm  304,000 Pa, V1  100 cm3  1.0  104 m3, and T1  100C  373 K.
The number of moles of gas is
n
p1V1 (304,000 Pa)(1.0 104 m3 )

 9.81103 mol
RT1
(8.31 J/(mol  K)(373 K)
At point 2 we have p2  p1  3.0 atm and V2  300 cm3  3V1. This is an isobaric process, so
V2 V1
V
  T2  2 T1  3(373 K)  1119 K
T2 T1
V1
The gas is heated to raise the temperature from T1 to T2. The amount of heat required is
Q  nCP T  (9.81103 mol)(20.8 J/mol  K)(1119 K  373 K)  152 J
which is to be reported as 150 J.
This amount of heat is added during process 1  2.
Thermal Properties of Matter
12-35
(b) Point 3 returns to T3  100C  373 K. This is an isochoric process, so
Q  nCVT  (9.81103 mol)(12.5 J/mol  K)(373 K  1119 K)  91.5 J
which is to be reported as –92 J.
This amount of heat is removed during process 2  3.
P12.96. Prepare: The heat engine follows a closed cycle. Please refer to Figure P12.96.
Solve: (a) The work done by the gas per cycle is the area inside the closed p-versus-V curve. We get
1
1
Wout  (300 kPa  100 kPa)(600 cm3  200 cm3 )  (200 103 Pa)(400 106 m3 )  40 J
2
2
The heat exhausted is QC  180 J  100 J  280 J. So, the heat extracted from the hot reservoir is QH  280 J 
40 J  320 J.
(b) The thermal efficiency of the engine is

Wout
40J

 0.125
QH 320 J
or 0.13 to two significant figures.
P12.97. Prepare: While the Prepare, Solve, Assess method of problem solving is meant to discourage blind
hunting for an equation and then plugging and chugging without understanding, we do recognize in this case that
the equation relating the rate of thermal energy transfer due to conduction and the thermal conductivity (the thing
we want) is given by Equation 12.32.
Q  kA 
=
T
t  L 
We pause to see what affects Q/t. If the area is larger, then the rate of transfer is greater. If the material is
thicker (larger L), then the rate of transfer is less. If the temperature difference between the two sides is greater,
then the rate of transfer is greater. All of this makes sense, so we proceed.
We are given L  0.050 m, A  1.0 m2, and T  55 K. We also convert the rate of transfer to SI units as follows:
Q
Cal  1000 cal  4.19 J  1 h  1 min 
 25




  29 W
t
h  1 Cal  1 cal  60 min  60 s 
Solve: Since we want the thermal conductivity of goose down, we solve the equation for k.
 Q  L  1
k    
 t  A  T
 0.050 m  1 

  (29 W)  1.0 m2  55 K   0.026 W/(m  K)




Assess: This answer is at the lower end of the range listed in Table 12.7 for fur and feathers, which is as we
expect, since goose down is well known as a good insulator. The one number that may not be quite realistic is the
35C inside the bag; this is 95F, which would be warmer than “cozy.”
P12.98. Prepare: Assume all the insulation comes from the air layer. Assume the value for the thermal
conductivity of air in Table 12.7 is a good approximation for these conditions. Equation 12.31 applies.
Solve: Using Equation 12.31,
Q  kA 
(0.026 W/ (m  K)(1.1 m2 )
   T 
(34C  (20C))  62 W
t  L 
0.025 m
Assess: This is a large amount of power, even though air has the lowest thermal conductivity of all the
materials in Table 12.7.
P12.99. Prepare: We’ll use Equation 12.33 in a ratio for the two plates. First, solve the equation for T.
12-36 Chapter 12
1/ 4
 Q  1  
T =  

 t  e A  
We are given the rates of energy transfer for the two plates. Since they are the same, and since they are both copper,
their emissivities are the same. A2  42A1  16 A1.
Solve:
1/4
1/4
Q
1/4
1
A 
 16 A1 
T1 [( t )( e A1 )]
1/ 4
 Q
 2  
  16  2
T2 [( t )( e1A2 )]1/4  A1 
A
 1 
Plate 1 is twice as hot as plate 2 on the absolute temperature scale.
Assess: Plate 2 has 16 times the area of plate 1, but because plate 1 is twice as hot as plate 2, they radiate energy
at the same rate because of the T 4 in the formula.
P12.100. Prepare: Heat loss by conduction can be calculated with Equation 12.31. Equation 12.33 applies
to heat lost by radiation.
Solve: (a) The dead layer of air separates and insulates you from the air in the room. The conduction through
the air layer is
 (0.026 W/ (m K)(1.8 m 2 )
Q  kA 

T  

t  L 
0.005 m


 (9 K)  84 W

(b) The heat lost through radiation is given by Equation 12.23. Body temperature is T  34  273  307 K. The
temperature of the walls is T  17  273  290 K.
Qnet
 e A(T 4  T0 4 )  (.97)(5.67 108 W / (m 2  K 4 )(1.8 m 2 )((307 K) 4  (290 K)4 )  180 W
t
(c) The heat lost to radiation is greater.
(d) If the person is metabolizing food at a rate of 155 W, he feels chilly because he is producing heat at a rate of
155 W and losing heat at a rate of 84 W  180 W or 260 W.
Assess: Putting some clothes on would decrease the heat lost by radiation, convection, and conduction.
P12.101. Prepare: Volume thermal expansion indicates that we should use Equation 12.2.
V  Vi T
We want to know the change in depth, which is given by d  V/A. Similarly the initial volume in the formula
is given by Vi  Adi. We will assume that the temperatures given (about 17C) are close enough to 20C that we
can use the coefficient of volume expansion for water given in Table 12.3,   210  10–6K–1. We are also given
di 500 m and T  1.00 K.
Solve:
V Vi T  Adi T
d 


  di T  (210 106 K 1 )(500 m)(1.00 K)  0.105 m
A
A
A
The correct choice is A.
Assess: The answer of 11 cm may not seem like much, but it could affect low-lying coastal areas.
The reason the area cancels out is because each square centimeter of water will increase in depth by the same
amount for a given temperature change.
P12.102. Prepare: We can use Equation 12.21. Table 12.4 lists the specific heat of water.
Solve: The mass of the water in the top layer is M  di A. The area of the top level of the oceans is
2
 103 m 
14
A  (3.6  10 km ) 
  3.6 10 m 2
 km 
8
2
The heat required to change the temperature by one degree Celsius is
Q  McT   di AcT  (1000 kg/m3 )(500 m)(3.6 1014 m2 )(4190 J/(kg  K)(1 K)  7.5  1023 J
Thermal Properties of Matter
12-37
To one significant figure, this is 1  1024 J.The correct choice is A.
Assess: This result makes sense. The density and specific heat of water are relatively large.
P12.103. Prepare: Carefully examine Figure 12.20a. At 2C the volume of a mole of water decreases with
increasing temperature.
Solve: Because the graph has a negative slope at 2C, the volume would decrease in going from 2C to 3C.
The correct choice is C.
Assess: It is unusual for a substance to have a negative coefficient of expansion, but the fact that water does at
temperatures close to freezing is biologically important.
P12.104. Prepare: Conduction is the transfer of heat directly through physical material. Radiation is energy
transfer through electromagnetic waves. Evaporation transfers energy through removal of molecules with high
thermal energy.
Solve: Since there is no mixing in the warmer surface water, convection does not happen effectively. The correct
choice is B.
Assess: Note that the water is heated by the light from the sun, so it is heated through radiation.