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Unformatted text preview: te QH . SOLUTION
a. Solving Equation 15.15 for TH gives
TH = TC
1–e = 378 K
= 1260 K
1–0.700 b. Solving Equation 15.14 for QH gives
T
QH = QC H
T
C 1260 K 4 = (5230 J) = 1.74 × 10 J 378 K 54. REASONING We seek the ratio TC, f TC, i of the final Kelvin temperature of the cold
reservoir to the initial Kelvin temperature of the cold reservoir. The Kelvin temperature TH
of the hot reservoir is constant, and it is related to the temperature of the cold reservoir by
T
the Carnot efficiency: eCarnot = 1 − C (Equation 15.15). We will solve Equation 15.15 for
TH
the temperature TC of the cold reservoir, and use the resulting expression to determine the
ratio TC, f TC, i . Because the temperature of the cold reservoir increases, we expect the ratio
to have a value greater than 1. SOLUTION Solving eCarnot = 1 −
TC yields TC
TH (Equation 15.15) for the coldreservoirtemperature Chapter 15 Problems TC
TH = 1 − eCarnot TC = TH (1 − eCarnot ) or 797 (1) Applying Equation (1) to both the initial temperature and the final temperature, we find that ( TC,i = TH 1 − eCarnot,i ) and ( TC,f = TH 1 − eCarnot,f ) (2) Taking the ratio of Equations (2), we obtain
TC,f
TC, i = (
) = 1 − eCarnot, f = 1 − 0.70 = 1.2
TH (1 − eCarnot,i ) 1 − eCarnot, i 1 − 0.75 TH 1 − eCarnot,f 55. REASONING The smallest possible temperature of the hot reservoir would occur when the
engine is a Carnot engine, since it has the greatest efficiency of any engine operating
between the same hot and cold reservoirs. The efficiency eCarnot of a Carnot engine is (see
Equation 15.15) eCarnot = 1 − (TC/TH), where TC and TH are the Kelvin temperatures of its
cold and hot reservoirs. Solving this equation for TH gives TH = TC/(1 − eCarnot). We are
given TC, but do not know eCarnot. However, the efficiency is defined as the magnitude W
of the work done by the engine divided by the magnitude QH of the input heat from the hot
reservoir, so eCarnot = W / QH (Equation 15.11). Furthermore, the conservation of energy
requires that the magnitude QH of the input heat equals the sum of the magnitude W of
the work done by the engine and the magnitude QC of the heat it rejects to the cold
reservoir, QH = W + QC . By combining these relations, we will be able to find the
temperature of the hot reservoir of the Carnot engine.
SOLUTION From the Reasoning section, the temperature of the hot reservoir is
TH = TC/(1 − eCarnot). Writing the efficiency of the engine as eCarnot = W / QH , the
expression for the temperature becomes TH = TC 1 − eCarnot = TC
W
1−
QH From the conservation of energy, we have that QH = W + QC . Substituting this expression
for QH into the one above for TH gives 798 THERMODYNAMICS TH = TC
W
1−
W =
1− TC
W = W + QC 285 K
= 1090 K
18 500 J
1−
18 500 J + 6550 J 56. REASONING According to Equation 15.11, the efficiency e of a heat engine is given by
e = W / QH , where W is the magnitude of the work and QH is the magnitude of the input
heat. Thus, the magnitude of the work is W = e QH .
The efficiency eCarnot of a Carnot engine is given by Equation 15.15 as eCarnot = 1 – TC/TH,
where TC and TH are, respectively, the Kelvin temperatures of the cold and hot reservoirs.
This expression can be used to determine TC.
SOLUTION Using Equation 15.11, we find the work delivered by each engine as follows:
Engine A W = e QH = ( 0.60 )(1200 J ) = 720 J Engine B W = e QH = ( 0.80 )(1200 J ) = 960 J Equation 15.15, which gives the efficiency eCarnot of a Carnot engine, can be solved for the
temperature TC of the cold reservoir: eCarnot = 1 − TC
TH or TC = (1 − eCarnot ) TH Applying this result to each engine gives
Engine A TC = (1 − 0.60 )( 650 K ) = 260 K Engine B TC = (1 − 0.80 )( 650 K ) = 130 K 57. REASONING AND SOLUTION The efficiency of the engine is e = 1 − (TC/TH) so
(i) Increase TH by 40 K; e = 1 − [(350 K)/(690 K)] = 0.493
(ii) Decrease TC by 40 K; e = 1 − [(310 K)/(650 K)] = 0.523 Chapter 15 Problems 799 The greatest improvement is made by lowering the temperature of the cold reservoir. 58. REASONING AND SOLUTION The amount of work delivered by the engines can be
determined from Equation 15.12, QH = W + QC . Solving for W for each engine gives:
W1 = QH1 − QC1 and W2 = QH2 − QC2 The total work delivered by the two engines is ( )( W = W1 + W2 = QH1 − QC1 + QH2 − QC2 ) But we know that QH2 = QC1 , so that ( )( ) W = QH1 − QC1 + QC1 − QC2 = QH1 − QC2 (1) Since these are Carnot engines,
QC1
QH1 = TC1 ⇒ TH1 QC1 = QH1 TC1
TH1 670 K 3
= ( 4800 J ) = 3.61× 10 J 890 K Similarly, noting that QH2 = QC1 and that TH2 = TC1, we have
QC2 = QH2 TC2
TH2 = QC1 TC2
TC1 ( ) 420 K 3
= 3.61×103 J = 2.26 × 10 J 670 K Substituting into Equation (1) gives
W = 4800 J − 2.26 × 103 J = 2.5 × 103 J 59. SSM REASONING The maximum efficiency e at which the power plant can operate is
given by Equation 15.15, e = 1 − ( TC / TH ) . The power output is given; it can be used to
find the magnitude W of the work output for a 24 hour period. With the efficiency and W
known, Equation 15.11, e = W / QH , can...
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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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