Physics 100 Energy in Today’s World Homework Ch. 4b
Prof. Menningen
p. 1 of 3
1.
One possible end to the universe is for it to reach thermal equilibrium, that is, it would have
a uniform temperature. Why could this be called the "heat death" of the universe?
a. The uniform temperature would be absolute zero.
b. The entropy of the universe would be zero.
c. The total energy in the universe would be zero.
d. No thermal energy can flow without a difference in temperature.
2.
A hurricane is a kind of a heat engine. True False
3.
Explain how the following simplified statements of the first and second laws of
thermodynamics are consistent with the versions given in the text.
(a) First Law: You cannot get ahead.
Energy is conserved; that is, you cannot get more energy out of a
thermodynamic system than you put in.
(b) Second Law: You cannot even break even.
Heat engines cannot be 100% efficient, so that some heat is always
exhausted whenever you try to extract mechanical energy from the
thermal energy stored in a hot reservoir.
4.
Why is a condenser needed in a steam-powered electric plant?
a. The condenser increases the efficiency by maximizing both the
temperature difference TH − TC and the energy difference ΔE of the steam.
b. The condenser increases the efficiency by maximizing the amount of heat QC
that is ejected into the environment.
c. The condenser increases the efficiency by maximizing the increase in entropy
of the steam.
d. The condenser increases the efficiency by decreasing the specific heat of the
recycled water.
5.
How would you explain to a group of citizens why waste heat must be released in the
operation of a steam-generating plant?
a. A power plant operates by using a heat flow, so heat must flow to a low
temperature sink.
b. While the plant isn't required to release heat, it must do so because of
federal regulations.
c. The more heat the plant releases, the more efficient it will be.
d. The second law of thermodynamics states that entropy must always decrease.
6.
What happens to the efficiency of an ideal heat engine as its input temperature is increased
while its exhaust temperature is held fixed?
a. The efficiency goes up.
b. The efficiency goes down.
c. The efficiency stays the same.
Physics 100 Energy in Today’s World Homework Ch. 4b
7.
Prof. Menningen
p. 2 of 3
If only conductive heat loss was significant for a house, then by what percentage would
you lower the heat loss if the temperature was reduced inside from 72°F to 66°F, when the
outside temperature is 24°F?
 Q t new ATnew R Tnew Tin, new  Tout 66F  24F




 0.875
 Q t old ATold R Told Tin, old  Tout 72F  24F
 Q t new  0.875  Q t old , so you've reduced the heat loss rate by
1  0.875  0.125  12.5%
8.
A heat engine requires an input of 9.70 kJ per minute to produce 2.50 kJ of work per
minute.
(a) How much heat must the engine exhaust each minute?
QH  QC  W
QC  QH  W  9.70 kJ  2.50 kJ  7.20 kJ
(b) What is the efficiency of this engine?
e
9.
W 2.50 kJ

 0.258  25.8%
QH 9.70 kJ
What is the exhaust temperature of an ideal heat engine that has an efficiency of 50% and
an input temperature of 455°C?
emax  1 
TC
TH

TC
 1  emax
TH
TC  TH 1  emax    455  273 K 1  0.50   364 K
 364 K  273  91C
Physics 100 Energy in Today’s World Homework Ch. 4b
Prof. Menningen
p. 3 of 3
10. An engineer has designed a machine to produce electricity by using the difference in the
temperature of ocean water at different depths. If the surface temperature is 24.0°C and the
temperature at 50 m below the surface is 12.5°C, what is the maximum efficiency that this
machine could produce?
emax  1 
TC
12.5  273 K
1
 0.0387  3.87%
TH
24.0  273 K
11. A typical coal-fired electric power plant has an efficiency of 36.0%, while a nuclear power
plant is 31.0% efficient.
(a) How many joules of thermal energy are required by each plant to generate one
joule of electrical energy? How much heat does each plant exhaust as waste?
e
W
QH
 QH 
W
e
and QC  QH  W
Coal Plant
W 1.0 J
QH  
 2.78 J
e 0.360
QC  QH  W  2.78  1.00 J
QC  1.78 J
Nuclear Plant
W 1.0 J
QH  
 3.23 J
e 0.310
QC  QH  W  3.23  1.00 J
QC  2.23 J
(b) What percent increase in waste heat results from the lower efficiency of the
nuclear power plant?
QC, nuclear  QC, coal
2.23  1.78 J
 100 
 100  25.3%
QC, coal
1.78 J
Note that a 5% decrease in plant efficiency (31% vs. 36%)
results in a 25% increase in wasted energy! Nuclear plants
are less efficient than coal-fired plants because their boilers
must be run at a lower temperature for safety reasons.