Unformatted text preview: iven heat to the cold reservoir reversibly. According to the second law of
thermodynamics stated in terms of entropy (see Section 15.11), the reversible engine is the
one for which ∆S universe = 0 J/K , and the irreversible engine that could exist is the one for
which ∆S universe > 0 J/K . The irreversible engine that could not exist is the one for which ∆Suniverse < 0 J/K .
SOLUTION Using Equation 15.18, we write the total entropy change of the universe as the
sum of the entropy changes of the hot (H) and cold (C) reservoirs. ∆S universe = − QH
TH + QC
TC 812 THERMODYNAMICS In this expression, we have used − QH for the heat from the hot reservoir because that
reservoir loses heat. We have used + QC for the heat rejected to the cold reservoir because
that reservoir gains heat. Applying this expression to the three engines gives the following
results:
− QH QC −1650 J 1120 J
Engine I
∆S universe =
+
=
+
= +0.4 J/K
TH
TC
550 K
330 K Engine II ∆Suniverse = Engine III ∆S universe = − QH
TH
− QH
TH + QC + QC TC TC = −1650 J 990 J
+
= 0 J/K
550 K 330 K = −1650 J 660 J
+
= −1.0 J/K
550 K 330 K Since ∆S universe = 0 J/K for Engine II, it is reversible. Since ∆S universe > 0 J/K for Engine I,
it is irreversible and could exist. Since ∆S universe < 0 J/K for Engine III, it is irreversible
and could not exist. 76. REASONING AND SOLUTION Equation 15.19 gives the unavailable work as
Wunavailable = T0 ∆S (1) where T0 = 248 K . We also know that Wunavailable = 0.300 Q . Furthermore, we can apply
Equation 15.18 to the heat lost from the 394K reservoir and the heat gained by the reservoir
at temperature T, with the result that
–Q
Q
∆S =
+
394 K
T With these substitutions for T0, Wunavailable, and ∆S, Equation (1) becomes –Q
Q
0.300 Q = ( 248 K ) + 394 K T or T = 267 K 77. SSM REASONING AND SOLUTION The change in entropy ∆S of a system for a
process in which heat Q enters or leaves the system reversibly at a constant temperature T is
given by Equation 15.18, ∆S = (Q / T ) R . For a phase change, Q = mL , where L is the
latent heat (see Section 12.8). Chapter 15 Problems 813 a. If we imagine a reversible process in which 3.00 kg of ice melts into water at 273 K, the
change in entropy of the water molecules is ( ) (3.00 kg) 3.35 ×105 J/kg mLf Q
∆S = = = 3.68 × 103 J/K
=
273 K T R T R
b. Similarly, if we imagine a reversible process in which 3.00 kg of water changes into
steam at 373 K, the change in entropy of the water molecules is ( ) (3.00 kg) 2.26 ×106 J/kg mLv Q
∆S = = = 1.82 ×104 J/K
=
373 K T R T R
c. Since the change in entropy is greater for the vaporization process than for the fusion
process, the vaporization process creates more disorder in the collection of water
molecules.
78. REASONING According to the discussion on Section 15.11, the change ∆Suniverse in the
entropy of the universe is the sum of the change in entropy ∆SC of the cold reservoir and the
change in entropy ∆SH of the hot reservoir, or ∆Suniverse = ∆SC + ∆SH. The change in entropy
of each reservoir is given by Equation 15.18 as ∆S = (Q/T)R, where Q is the heat removed
from or delivered to the reservoir and T is the Kelvin temperature of the reservoir. In
applying this equation we imagine a process in which the heat is lost by the house and
gained by the outside in a reversible fashion.
SOLUTION Since heat is lost from the hot reservoir (inside the house), the change in
entropy is negative: ∆SH = −QH/TH. Since heat is gained by the cold reservoir (the
outdoors), the change in entropy is positive: ∆SC = +QC/TC. Here we are using the symbols
QH and QC to denote the magnitudes of the heats. The change in the entropy of the universe
is ∆Suniverse = ∆SC + ∆S H = QC
TC − QH 24 500 J 24 500 J
=
−
= 11.6 J/K
TH
258 K
294 K In this calculation we have used the fact that TC = 273 − 15 °C = 258 K and
TH = 273 + 21 °C = 294 K. 814 THERMODYNAMICS 79. REASONING The change ∆Suniverse in entropy of the universe for this process is the sum of the entropy changes for (1) the warm water (∆Swater) as it cools down from its initial
temperature of 85.0 °C to its final temperature Tf , (2) the ice (∆Sice) as it melts at 0 °C, and (3) the ice water (∆Sice water ) as it warms up from 0 °C to the final temperature Tf: ∆Suniverse = ∆Swater + ∆Sice + ∆Sice water. To find the final temperature Tf , we will follow the procedure outlined in Sections 12.7 and
12.8, where we set the heat lost by the warm water as it cools down equal to the heat gained
by the melting ice and the resulting ice water as it warms up. The heat Q that must be
supplied or removed to change the temperature of a substance of mass m by an amount ∆T is
Q = cm∆T (Equation 12.4), where c is the specific heat capacity. The heat that must be
supplied to melt a mass m of a substance is Q = mLf (Equation 12.5), where Lf is the latent
heat of fusion.
SOLUTION
a. We begin by finding the final temperature Tf of the water. Setting the heat lost equal to
the heat gained gives cmwater ( 85.0 °C − Tf ) =
144424444
4
3
Heat lost by water mice Lf + mice c (Tf − 0.0 °C )
1 24
43
144 2444
4
3
Heat gained
by melting ice Heat gained by ice water Solving this relation for the final temperature Tf yields
Tf =...
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This note was uploaded on 04/30/2013 for the course PHYS 1100 and 2 taught by Professor Chastain during the Spring '13 term at LSU.
 Spring '13
 CHASTAIN
 Physics, The Lottery

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