Chapter 16 Solutions
7.
Picture the Problem: The plot shows the pressure of the gas as a
function of temperature. Note that at T2 = 105C the pressure is
P2 = 93.5 kPa, and at −273.15C the pressure extrapolates linearly
to zero. We also indicate the points corresponding to temperature
of T1 = 50C, and the point corresponding to a pressure of P3 =
115 kPa.
Strategy: We assume that the pressure lies on a straight line. Using the known pressures at T2 = 105C and at −273.15C,
calculate the rate at which pressure increases as a function of temperature. This rate can then be used to calculate the pressure at
any temperature, and the temperature at a specified pressure.
rate 
2. (a) Use the rate to find
the pressure at 50C:
P1   0.2473 kPa/C°  50C   273.15 C    79.9 kPa
3. (b) Solve the rate equation for the
temperature as a function of pressure:
17.
93.5 kPa
 0.2473 kPa/C°
105°C   273.15°C 
Solution: 1. Calculate the rate that
pressure increases with temperature:
P3   rate T3  T0 
T3 
P3
115 kPa
 T0 
 273.15°C  192°C
0.2473 kPa/C°
 rate 
Picture the Problem: The Akashi Kaikyo Bridge in Japan is made of steel. When steel is heated it expands and when it is
cooled it contracts.
Strategy: In this problem we wish to find the change in length of the bridge between a cold winter day and a warm summer
day. Use equation 16-4 to determine the change in length. The coefficient of linear expansion for steel is given in Table 16-1.
Solution: Insert the given
values into equation 16-4:
18.
L   L0 T
 1.2 105 (C)1   3910 m  30.0 C  (5.00 C)   1.6 m
Picture the Problem: An aluminum plate has a hole cut in its center. The plate expands
as it is heated.
Strategy: We want to find the size of the hole after the temperature has increased to
199.0C. The hole will expand at the same rate as the aluminum. Since the diameter of the
hole is a unit of length, use equation 16-4 to calculate the diameter as a function of the
increase in temperature. The coefficient of linear expansion is given in Table 16-1.
Solution: 1. (a) Solve equation 16-4
for the final diameter:
2. Insert the given data:
 d  d   d   d T
d   d   d T  d 1   T 
d   1.178 cm 1+  24 10–6 K 1  199.0 C  23.00 C   1.183 cm
3. (b) Solve equation 16-4 for
the change in temperature:
 d  d   d   d T
d  d
T  T  T0 
d
4. Solve for the final temperature:
T  T0 
d  d
1.176 cm  1.178 cm
 23.00 C 
  48 C
d
 24 10–6 K 1  1.178 cm 
21.
Picture the Problem: A steel gasoline tank is completely filled with gasoline, such that the gasoline and the tank have the same
initial volumes. When the gas and tank are heated, the gas expands more than the tank, causing some of the gas to spill out of
the tank.
Strategy: Since the initial volumes of the gas and tank are equal, the amount that will spill out is the difference in the increase in
volume of the gas and tank, namely: The volume of spilled gasoline Vspill  Vgas  Vtank . Use equation
16-6 to calculate the changes in volume for the gas and tank. The coefficient of volume expansion for steel is 3 times the
coefficient of linear expansion, which is given in Table 16-1.
Solution: 1. Write the volume difference
Vspill  Vgas  Vtank    gasV0  3 tankV0  T    gas  3 tank V0 T
in terms of equation 16-6:
2. Insert the given data:
26.
Vspill   9.5 104  3 1.2 105   C
0.93 L
 3600 s 
2.6 104 Cal/  s  kg   75 kg 8.0 h  
  560 Cal
 h 
Picture the Problem: A lead bullet traveling at 250 m/s has kinetic energy. As the bullet encounters a fence post it slows to a
stop, converting the kinetic energy to heat. Half of the energy heats the bullet resulting in an increase in bullet temperature.
Strategy: Solve equation 16-13 for the change in temperature. Set the heat equal to one half of the initial kinetic energy of the
bullet. The specific heat of lead is given in Table 16-2.
Solution: Set Q in equation 16-13 equal to half
the initial kinetic energy and solve for T:
38.
51 L 25  5.0 C 
Picture the Problem: The metabolic rate is the number of calories expended in bodily functions per second per kilogram.
Strategy: We wish to calculate the total calories expended during a full night’s rest Multiply the metabolic rate by the person’s
mass to calculate the calories expended per second. Multiply this result by 8.0 hours to calculate the calories expended in a full
night’s sleep.
Solution: Multiply together the
metabolic rate, mass, and time:
36.
1
T 
1
K
Q
 2 
mc mc
1
2

1
2
mv 2 
mc
 250 m/s 
v2

 120 K
4c 4 128 J/  kg  K  
2

Picture the Problem: Heat transfers from the hot lead ball to the cool water, causing the lead to cool and the water to heat up.
Eventually the water and lead will come to the same equilibrium temperature.
Strategy: Use equation 16-15 to calculate the equilibrium temperature. The specific heats of water and lead are given in Table
16-2.
Solution: Insert the given data into equation 16-15:
T
mPb cPbTPb  mw cw Tw
mPb cPb  mw cw
0.235 kg 128 J/  kg  K   84.2 C



  0.177 kg  4186 J/  kg  K   21.5 C 

0.235 kg 128 J/  kg  K    0.177 kg  4186 J/  kg  K  
T  23.9 C
41.
Picture the Problem: A hot object is immersed in water in a calorimeter cup. Heat transfers from the hot object to the cold
water and cup, causing the temperature of the object to decrease and the temperature of the water and cup to increase.
Strategy: Since the heat only transfers between the water, cup, and object, we can use conservation of energy to calculate the
heat given off by the object by summing the heats absorbed by the water and cup. Use the heat given off by the object and its
change in temperature to calculate its specific heat.
Solution: 1. Let
Q  0
0  QOb  Qw  QAl
and solve for QOb :
QOb    Qw  QAl     mw cw  mAl cAl  Tw
2. Solve for the specific heat
using equation 16-13:
cOb 

Qob
 m c  mAl cA1 Tw  T 
 w w
mOb T  TOb 
mOb T  TOb 
 0.103 kg 4186 J/  kg  K   0.155 kg 900 J/  kg  K    20  22  C
0.0380 kg  22.0  100  C
cOb  385 J (kg  C)  385 J (kg  K)
3. Look up the specific
heat in Table 16-2:
The object is made of copper.