Unformatted text preview: t. We will apply this expression to each engine and utilize the fact that in each
case the same work is done. We expect to find that engine 2 requires less input heat to do
the same amount of work, because it has the greater efficiency.
SOLUTION Applying Equation 15.11 to each engine gives
W = e1 QH1
14244
4
3 and W = e2 QH2
14 244
4
3 Engine 1 Engine 2 Since the work is the same for each engine, we have
e1 QH1 = e2 QH2 or e QH2 = 1 QH1
e 2 It follows, then, that
e 0.18 QH2 = 1 QH1 = ( 5500 J ) = 3800 J
e 0.26 2 which is less than the input heat for engine 1, as expected. 87. SSM REASONING AND SOLUTION
a. Since the energy that becomes unavailable for doing work is zero for the process, we
have from Equation 15.19, Wunavailable = T0 ∆Suniverse = 0 . Therefore, ∆Suniverse = 0 and
according to the discussion in Section 15.11, the process is reversible . Chapter 15 Problems 821 b. Since the process is reversible, we have (see Section 15.11)
∆S universe = ∆Ssystem + ∆Ssurroundings = 0 Therefore, ∆Ssurroundings = −∆Ssystem = –125 J/K 88. REASONING AND SOLUTION We know that
QC = QH − W = 14 200 J − 800 J = 13 400 J Therefore,
Q 13 400 J TC = TH C = ( 301 K ) = 284 K
Q 14 200 J H 89. REASONING The coefficient of performance of an air conditioner is QC / W , according
to Equation 15.16, where QC is the magnitude of the heat removed from the house and W
is the magnitude of the work required for the removal. In addition, we know that the first
law of thermodynamics (energy conservation) applies, so that W = QH − QC , according to
Equation 15.12. In this equation QH is the magnitude of the heat discarded outside. While
we have no direct information about QC and QH , we do know that the air conditioner is a
Carnot device. This means that Equation 15.14 applies: QC / QH = TC / TH . Thus, the
given temperatures will allow us to calculate the coefficient of performance.
SOLUTION Using Equation 15.16 for the definition of the coefficient of performance and
Equation 15.12 for the fact that W = QH − QC , we have
Coefficient of performance = QC
W = QC
QH − QC = QC / QH
1 − QC / QH Equation 15.14 applies, so that QC / QH = TC / TH . With this substitution, we find 822 THERMODYNAMICS Coefficient of performance = = 90. REASONING QC / QH
1 − QC / QH
TC TH − TC = = TC / TH
1 − TC / TH 297 K
= 21
( 311 K ) − ( 297 K ) The efficiency eCarnot of a Carnot engine is eCarnot = 1 − TC TH
(Equation 15.15), where TC and TH are, respectively, the Kelvin temperatures of the cold
and hot reservoirs. After the changes are made to the temperatures, this same equation still
applies, except that the variables must be labeled to denote the new values. We will use a
“prime” for this purpose. From the original efficiency and the information given about the
changes made to the temperatures, we will be able to obtain the new temperature ratio and,
hence, the new efficiency.
SOLUTION After the reservoir temperatures are changed, the engine has an efficiency that,
according to Equation 15.15, is
T′
′
eCarnot = 1 − C
′
TH where the “prime” denotes the new engine. Using unprimed symbols to denote the original
′
′
engine, we know that TC = 2TC and TH = 4TH . With these substitutions, the efficiency of
the new engine becomes
T′
2T
1T ′
(1)
eCarnot = 1 − C = 1 − C = 1 − C 2T ′
4TH
TH H
To obtain the original ratio TC / TH , we use Equation 15.15:
eCarnot = 1 − TC
TH TC or TH = 1 − eCarnot Substituting this original temperature ratio into Equation (1) gives
1T
′
eCarnot = 1 − C
2T
H 1 = 1 − 2 (1 − eCarnot ) = 1
2 (1 + eCarnot ) = 1 (1 + 0.40 ) = 0.70
2 Chapter 15 Problems 823 91. SSM REASONING AND SOLUTION
a. Starting at point A, the work done during the first (vertical) straightline segment is
W1 = P1∆V1 = P1(0 m3) = 0 J
For the second (horizontal) straightline segment, the work is
W2 = P2∆V2 = 10(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1200 J
For the third (vertical) straightline segment the work is
W3 = P3∆V3 = P3(0 m3) = 0 J
For the fourth (horizontal) straightline segment the work is
W4 = P4∆V4 = 15(1.0 × 104 Pa)6(2.0 × 10–3 m3) = 1800 J
The total work done is
W = W1 + W2 + W3 + W4 = +3.0 × 103 J
b. Since the total work is positive, work is done by the system . 92. REASONING AND SOLUTION
a. The work is the area under the path ACB. There are 48 "squares" under the path, so that ( )( ) W = − 48 2.0 × 104 Pa 2.0 × 10–3 m3 = −1900 J
The minus sign is included because the gas is compressed, so that work is done on it. Since
there is no temperature change between A and B (the line AB is an isotherm) and the gas is
ideal, ∆U = 0, so
Q = ∆U + W = W = −1900 J
b. The negative answer for W means that heat flows out of the gas. 824 THERMODYNAMICS 93. SSM REASONING AND SOLUTION The change in volume is ∆V = – sA, where s is the
distance through which the piston drops and A is the piston area. The minus sign is included
because the volume decreases. Thus, s= – ∆V
A The ideal gas law states that ∆V = nR∆T/P. But...
View
Full
Document
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

Click to edit the document details