Homework Set 7: Solutions
Due: Thursday, October 21, 2010
Chapter 11
(Q2) 8 points
Could a simple machine such as a lever, a pulley system, or a hydraulic jack be considered
a heat engine? Explain.
A heat engine is a device that converts thermal energy into work as heat flows. Simple machines, such as
levers or pulley systems do not convert thermal energy into work. Therefore, they can not be considered
heat engines.
(Q6) 12 points
Which law of thermodynamics requires the work output of the engine to equal the difference
in the quantities of heat taken in and released by the engine? Explain.
The first law of thermodynamics states that the net work done by the heat engine is equal to the difference
in the quantities of heat taken in and released by the engine. This follows because, for a complete cycle,
the change in internal energy ∆U is zero so, from the first law of thermodynamics: ∆U = Q − W = 0.
(Q8) 12 points
Is it possible for a heat engine to operate as shown in the following diagram? Explain, using
the laws of thermodynamics.
There are two ways to argue this problem. If you interpret the thickness of the arrows as representing
the amount of energy being transferred, then no, it is not possible for a heat engine to operate as shown
in the diagram. That is because the diagram violates the first law of thermodynamics: more energy is
leaving the engine (as work and as heat transferred to the cold-temperature reservoir), then is leaving
entering the engine (as heat transferred from the high-temperature reservoir).
In class, however, our diagrams looked a little different than those done in the book. If, instead, you
ignored the thickness of the arrows, then yes, a heat engine could operate as shown in the diagram. That
is because it is describing a system where energy is being extracted from a high-temperature reservoir
and moving to a low-temperature reservoir with some energy (work) being extracted along the way.
(Q14) 14 points
If we want to increase the efficiency of a Carnot engine, would it be more effective to raise
the temperature of the high-temperature reservoir by 50◦ or lower the temperature of the
low-temperature reservoir by 50◦ C? Explain.
The efficiency of a Carnot engine is given by
ε=
TH − TC
.
TH
If we were to raise the temperature of the high-temperature reservoir by some amount, ∆T (in this
problem, ∆T = 50◦ ), then
TH − TC + ∆T
εH =
.
TH + ∆T
If, instead, we were to lower the temperature of the low-temperature reservoir by the same amount, then
εC =
TH − TC + ∆T
.
TH
Comparing these two expressions, we see that εC > εH . Therefore, it is more effective to lower the
temperature of the low-temperature reservoir by some amount, then it is to raise the temperature of the
high-temperature reservoir by the same amount.
(Q18) 14 points
Is it possible to cool a closed room by leaving the door of a refrigerator open in the room?
Explain.
No, it is not possible to cool a closed room by leaving the door of a refrigerator open. A refrigerator is
a heat pump: it transfers heat from one location (the ’source’) to another location (the ’sink’ or ’heat
sink’), using work. The heat transferred from the source is QC , the heat transferred to the sink is QH ,
and the work is W . In class, we found that by the first law of thermodynamics, QC + W = QH . In other
words, the amount of heat transferred to the sink is always greater than the amount of heat transferred
from the source. If we open the refrigerator door, the refrigerator is now transferring heat QC from the
closed room. In the process, it transfers heat QH back to the closed room. Because QH > QC , the
refrigerator is transferring more heat into the room than it is transferring heat out of the room. As a
result, it can not be used to cool down the closed room.
(Q24) 10 points
A hot cup of coffee is allowed to cool down, thus warming its surroundings. Does the
entropy of the universe increase in this process? Explain.
This process is irreversible. That is, while it is certainly possible that the hot cup of coffee will cool down
while its surroundings will heat up, it is not possible that the reverse will happen (the hot cup of coffee
heats up while its cool surroundings cool down). As a result, by the second law of thermodynamics, the
entropy of the universe must increase in this process.
(E2) 14 points
A heat engine with an efficiency of 25% does 400 J of work in each cycle. How much heat
must be supplied from the high-temperature source in each cycle?
The efficiency of an ideal heat engine is given by
ε=1−
QC
QH
where, by the first law of thermodynamics
QH = QC + W.
Therefore, by substitution, we find that
W
W
=
.
ε=1− 1−
QH
QH
Solving for QH we find that
QH =
(E6) 8 points
W
400 J
=
= 1600 J.
ε
0.25
A Carnot engine takes in heat at a temperature of 650 K and releases heat to a reservoir
at a temperature of 350 K. What is its efficiency?
The efficiency of this engine is
ε=1−
(E10) 8 points
350 K
TC
=1−
= 0.462 or 46.2%.
TH
650 K
In each cycle of its operation, a refrigerator removes 18 J of heat from the inside of the
refrigerator and releases 30 J of heat into the room. How much work per cycle is required
to operate this refrigerator?
By the first law of thermodynamics
W = QH − QC = 30 J − 18 J = 12 J.