Physics for Scientists and Engineers, 6e
Chapter 22 – Heat Engines, Entropy, and the
Second Law of Thermodynamics
The energy input to an engine is 3.00 times greater
than the work it performs. What is its thermal
efficiency?
1
1.
3.00
2.
1.00
3.
0.333
4.
impossible to determine
2
3
4
25% 25% 25% 25%
5
1
2
3
4
Equation 22.2 gives this result directly.
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?
1
1.
0.333
2.
0.667
3.
1.00
4.
impossible to determine
2
3
4
25% 25% 25% 25%
5
1
2
3
4
The work represents one third of the input
energy. The remainder, two thirds, must be
expelled to the cold reservoir.
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.
25% 25% 25% 25%
1
1.
4.00
2.
2.00
3.
0.500
4.
0.250
2
3
4
5
1
2
3
4
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.
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.
1
1.
A, B, C
2.
A, C, B
3.
B, C, A
4.
B, A, C
5.
C, A, B
6.
C, B, A
2
3
4
5
17% 17% 17% 17% 17% 17%
1
2
3
4
5
6
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.
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?
1
1.
1
2.
2
3.
3
4.
4
2
3
25% 25% 25% 25%
4
5
1
2
3
4
One microstate – all four deuces
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?
1
1.
3
2.
4
3.
5
4.
6
2
3
25% 25% 25% 25%
4
5
1
2
3
4
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.
Which of the following is true for the entropy change
of a system that undergoes a reversible, adiabatic
process?
1
1.
ΔS < 0
2.
ΔS = 0
3.
ΔS > 0
2
3
4
5
33%
1
33%
2
33%
3
Because the process is reversible and
adiabatic, Qr = 0; therefore, ΔS = 0.
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
1
1.
ΔSA > ΔSB
2.
ΔSA = ΔSB
3.
ΔSA < ΔSB
2
3
4
5
33%
1
33%
2
33%
3
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.
The entropy change in an adiabatic process must
be zero because Q = 0.
1
1.
True
2.
False
2
3
4
50%
5
1
50%
2
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.