Unformatted text preview: e work and QH is the magnitude of the input heat, which
is the 4.1 × 106 J of energy generated by the climber’s metabolic processes.
The work Wnc done in climbing upward is related to the vertical height of the climb via the
workenergy theorem (see Equation 6.8), which is c h Wnc = KE f + PE f − KE 0 + PE 0
14243 14 244
4
3
Final total
mechanical energy Initial total
mechanical energy Here, Wnc is the net work done by nonconservative forces, in this case the work done by the
climber in going upward. Since the climber starts at rest and finishes at rest, the final kinetic
energy KEf and the initial kinetic energy KE0 are zero. As a result, we have
Wnc = W = PEf – PE0, where PEf and PE0 are the final and initial gravitational potential
energies, respectively. Equation 6.5 gives the gravitational potential energy as PE = mgh,
where m is the mass of the climber, g is the magnitude of the acceleration due to gravity,
and h is the vertical height of the climb. Taking the height at her starting point to be zero,
we then have Wnc = W = mgh. SOLUTION Using e = W / QH (Equation 15.11) for the efficiency of a heat engine and
relating the magnitude W of the work to the height h of the climb via the workenergy
theorem as W = mgh, we find that e= W
QH ( ) 2
mgh ( 52 kg ) 9.80 m/s ( 730 m )
=
=
= 0.091
QH
4.1×106 J 49. SSM WWW REASONING The efficiency e of an engine can be expressed as (see Equation 15.13) e = 1 − QC / QH , where QC is the magnitude of the heat delivered to the
cold reservoir and QH is the magnitude of the heat supplied to the engine from the hot
reservoir. Solving this equation for QC gives QC = (1 − e ) QH . We will use this 794 THERMODYNAMICS expression twice, once for the improved engine and once for the original engine. Taking the
ratio of these expressions will give us the answer that we seek.
SOLUTION Taking the ratio of the heat rejected to the cold reservoir by the improved
engine to that for the original engine gives
QC, improved
QC, original = (1 − eimproved ) QH, improved
(1 − eoriginal ) QH, original But the input heat to both engines is the same, so QH, improved = QH, original . Thus, the ratio
becomes
QC, improved
QC, original = 1 − eimproved
1 − eoriginal = 1 − 0.42
= 0.75
1 − 0.23 50. REASONING AND SOLUTION We wish to find an expression for the overall efficiency
e in terms of the efficiencies e1 and e2 . From the problem statement, the overall efficiency
of the twoengine device is
W + W2
(1)
e= 1
QH
where QH is the input heat to engine 1. The efficiency of a heat engine is defined by
Equation 15.11, e = W / QH , so we can write
W1 = e1 QH (2) and
W2 = e2 QH2 Since the heat rejected by engine 1 is used as input heat for the second engine, QH 2 = QC1 ,
and the expression above for W2 can be written as
W2 = e2 QC1 (3) According to Equation 15.12, we have QC1 = QH – W1 , so that Equation (3) becomes ( W2 = e2 Q − W
H 1 ) (4) Chapter 15 Problems 795 Substituting Equations (2) and (4) into Equation (1) gives e= ( e1 QH + e2 QH − W1
QH ) = e1 QH + e2 ( QH − e1 QH )
QH Algebraically canceling the QH 's in the right hand side of the last expression gives the
desired result: e = e1 + e2 − e1e2 51. REASONING We will use the subscript “27” to denote the engine whose efficiency is
27.0% (e27 = 0.270) and the subscript “32” to denote the engine whose efficiency is 32.0%
(e32 = 0.320). In general, the efficiency eCarnot of a Carnot engine depends on the Kelvin
temperatures, TC and TH, of its cold and hot reservoirs through the relation (see
Equation 15.15) eCarnot = 1 − (TC/TH). Solving this equation for the temperature TC, 32 of the engine whose efficiency is e32 gives TC, 32 = (1 − e32)TH, 32. We are given e32, but do not
know the temperature TH, 32. However, we are told that this temperature is the same as that
of the hot reservoir of the engine whose efficiency is e27, so TH, 32 = TH, 27. The temperature
TH, 27 can be determined since we know the efficiency and cold reservoir temperature of this
engine.
SOLUTION The temperature of the cold reservoir for engine whose efficiency is e32 is
TC, 32 = (1 − e32)TH, 32. Since TH, 32 = TH, 27, we have that
TC, 32 = (1 − e32)TH, 27 (1) The efficiency e27 is given by Equation 15.15 as e27 = 1 − (TC, 27/TH, 27). Solving this
equation for the temperature TH, 27 of the hot reservoir and substituting the result into
Equation 1 yields 1 − e32
TC, 32 = 1− e
27 1 − 0.320 TC, 27 = ( 275 K ) = 256 K 1 − 0.270 796 THERMODYNAMICS 52. REASONING AND SOLUTION The efficiency is given by
e = 1 − (TC/TH) = 1 − [(200 K)/(500 K)] = 0.6 = W / QH
The work is
W = e QH = (0.6)(5000 J) = 3000 J 53. SSM REASONING The efficiency e of a Carnot engine is given by Equation 15.15,
e = 1 − (TC / TH ) , where, according to Equation 15.14, QC / QH = TC / TH . Since the efficiency is given along with TC and QC , Equation 15.15 can be used to calculate TH .
Once TH is known, the ratio TC / TH is thus known, and Equation 15.14 can be used to
calcula...
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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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