Chapter 22
Heat Engines, Entropy, and
the Second Law of
Thermodynamics
Quick Quiz 22.1
The energy input to an engine is 3.00 times greater than the
work it performs. What is its thermal efficiency?
(a) 3.00
(b) 1.00
(c) 0.333
(d) impossible to determine
Quick Quiz 22.1
Answer: (c). Equation 22.2 gives this result directly.
Quick Quiz 22.2
The energy input to an engine is 3.00 times greater than the
work it performs. What fraction of the energy input is
expelled to the cold reservoir?
(a) 0.333
(b) 0.667
(c) 1.00
(d) impossible to determine
Quick Quiz 22.2
Answer: (b). The work represents one third of the input
energy. The remainder, two thirds, must be expelled to the
cold reservoir.
Quick Quiz 22.3
The energy entering an electric heater by electrical
transmission can be converted to internal energy with an
efficiency of 100%. By what factor does the cost of heating
your home change when you replace your electric heating
system with an electric heat pump that has a COP of 4.00?
Assume that the motor running the heat pump is 100%
efficient.
(a) 4.00
(b) 2.00
(c) 0.500
(d) 0.250
Quick Quiz 22.3
Answer: (d). The COP of 4.00 for the heat pump means that
you are receiving four times as much energy as the energy
entering by electrical transmission. With four times as much
energy per unit of energy from electricity, you need only one
fourth as much electricity.
Quick Quiz 22.4
Three engines operate between reservoirs separated in
temperature by 300 K. The reservoir temperatures are as
follows: Engine A: Th = 1 000 K, Tc = 700 K; Engine B: Th =
800 K, Tc = 500 K; Engine C: Th = 600 K, Tc = 300 K. Rank
the engines in order of theoretically possible efficiency, from
highest to lowest.
(a) A, B, C
(b) A, C, B
(c) B, C, A
(d) B, A, C
(e) C, A, B
(f) C, B, A
Quick Quiz 22.4
Answer: (f) C, B, A. Although all three engines operate over
a 300-K temperature difference, the efficiency depends on
the ratio of temperatures, not the difference.
Quick Quiz 22.5
Suppose that you select four cards at random from a
standard deck of playing cards and end up with a macrostate
of four deuces. How many microstates are associated with
this macrostate?
(a) 1
(b) 2
(c) 3
(d) 4
Quick Quiz 22.5
Answer: (a). One microstate – all four deuces.
Quick Quiz 22.6
Suppose you pick up two cards at random from a standard
deck of playing cards and end up with a macrostate of two
aces. How many microstates are associated with this
macrostate?
(a) 3
(b) 4
(c) 5
(d) 6
Quick Quiz 22.6
Answer: (d). Six microstates – club-diamond, club-heart,
club-spade, diamond-heart, diamond-spade, heart-spade. The
macrostate of two aces is more probable than that of four
deuces in Quick Quiz 22.5 because there are six times as
many microstates for this particular macrostate compared to
the macrostate of four deuces. Thus, in a hand of poker, two
of a kind is less valuable than four of a kind.
Quick Quiz 22.7
Which of the following is true for the entropy change of a
system that undergoes a reversible, adiabatic process?
(a) ΔS < 0
(b) ΔS = 0
(c) ΔS > 0
Quick Quiz 22.7
Answer: (b). Because the process is reversible and adiabatic,
Qr = 0; therefore, ΔS = 0.
Quick Quiz 22.8
An ideal gas is taken from an initial temperature Ti to a
higher final temperature Tf along two different reversible
paths: Path A is at constant pressure; Path B is at constant
volume. The relation between the entropy changes of the gas
for these paths is
(a) ΔSA > ΔSB
(b) ΔSA = ΔSB
(c) ΔSA < ΔSB
Quick Quiz 22.8
Answer: (a). From the first law of thermodynamics, for these
two reversible processes, Qr = ΔEint – W. During the
constant-volume process, W = 0, while the work W is
nonzero and negative during the constant-pressure
expansion. Thus, Qr is larger for the constant-pressure
process, leading to a larger value for the change in entropy.
In terms of entropy as disorder, during the constant-pressure
process, the gas must expand. The increase in volume results
in more ways of locating the molecules of the gas in a
container, resulting in a larger increase in entropy.
Quick Quiz 22.9
The entropy change in an adiabatic process must be zero
because Q = 0.
(a) True
(b) False
Quick Quiz 22.9
Answer: False. The determining factor for the entropy
change is Qr, not Q. If the adiabatic process is not reversible,
the entropy change is not necessarily zero because a
reversible path between the same initial and final states may
involve energy transfer by heat.