Chapter 17 Solutions
7.
Picture the Problem: Air contained within a car’s tires has a fixed volume and mass. The pressure inside the tires increases as
the temperature increases.
Strategy: Use the ideal gas law to develop a relationship between temperature and pressure. Solve this expression for the final
temperature.
Solution: 1. Arrange equation 17-5 to relate the initial
and final conditions of the gas, and eliminate V and n:
PV
PV
i
 f
nTi nTf
2. Solve for the final temperature:
Tf 

Pi Pf

Ti Tf
Pf Ti  554 kPa  286 K 

 316 K
Pi
501 kPa
10. Picture the Problem: A compressed air tank holds a quantity of air at a fixed pressure, volume, and temperature. This
same quantity of air is released to atmospheric conditions where it occupies a larger volume.
Strategy: Since the number of moles remains constant, it follows from the ideal gas law that the ratio PV T is also
constant. Use this relation to solve for the final volume.
57.
Solution: 1. (a) Set the initial condition equal to final condition:
PV
PV
i i
 f f
Ti
Tf
2. (b) Solve for the final volume:
Vf  Vi
Pi Tf
880 kPa 303 K
  0.500 m3 
 4.63 m3
Pf Ti
101 kPa 285 K
Picture the Problem: Ice is made by extracting heat from water that is initially at 0C.
Strategy: Use equation 17-20 to calculate how much heat must be extracted from the water to convert it to ice.
Q  mLf  0.96 kg  33.5 104 J/kg   3.2 105 J
Solution: Solve equation 17-20 for heat:
58.
Picture the Problem: As heat is added to ice initially at
−15C, the heat first increases the temperature to the
melting point, then melts the ice, and finally raises the
temperature of the melted water to 15C.
Strategy: Set the total heat equal to the sum of the heat
needed to (i) raise the ice to the melting point, (ii) melt
the ice, and (iii) increase the water to the final
temperature. Solve the resulting equation for the mass.
Solution: 1. Sum the heats using
equations 16-13 and 17-20:
Q  Qi  Qii  Qiii
 mcice  T 1  mLf  mcwater  T 2
Q  m cice  T 1  Lf  cwater  T 2 
2. Solve for the mass:
m

Q
cice  T 1  Lf  cwater  T 2
9.5  105 J
 2090 J/  kg  C°   15C°   33.5  104 J/kg   4186 J/  kg  C°   15C° 
m  2.2 kg
59.
Picture the Problem: Heat is added to copper at its melting point to convert it from solid to liquid.
Strategy: Calculate the heat needed to melt copper from equation 17-20.
Q  m Lf  1.75 kg   20.7 104 J/kg   3.62 105 J  362 kJ
Solution: Solve equation 17-20:
70.
Picture the Problem: An ice cube is placed into an aluminum cup filled with water. Heat flows out of the water and aluminum
cup and into the ice until the water, ice, and cup are in thermal equilibrium.
Strategy: Assume that the final temperature is greater than 0C. In this state all of the ice will have melted and the resulting
water will have heated to the final temperature. Use equation 16-13 and equation 17-20 to calculate the heat gained by the ice
Qgained as its temperature rises to the equilibrium temperature. Set that heat equal to the heat lost by the cup and water Qlost as
they cool to the equilibrium temperature and solve for the final temperature.
Solution: 1. (a) Find Qgained :
Qgained  mice Lf  mice cw Tf  0.0 C
2. Find Qlost :
Qlost  mc cAl  23C  Tf   mw cw  23C  Tf    mc cAl  mw cw  23C  Tf 
3. Set the heat absorbed
equal to the heat released:
4. Solve for the
final temperature:
Qgained  Qlost
mice Lf  mice cw Tf   mc cAl  mw cw  23 C  Tf 
Tf 
mice Lf   mc cAl  mw cw  23 C 
cw  mice  mw   mc cAl
   0.035 kg   33.5 104 J/kg  




0.062 kg 900 J/  kg  K    0.110 kg 4186 J/  kg  K    23 C  
  0.22 C

 4186 J/  kg  K    0.035 kg  0.110 kg   0.062 kg 900 J/  kg  K  


5. (b) The equilibrium temperature with silver is less than with aluminum. In fact, not all the ice will melt. Less heat loss is
required to lower the temperature of silver because it has a smaller specific heat.