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Unformatted text preview: , or ten
joules per second, so Q = (10.0 W)t, where t is the ontime for the heater. In addition, we
know that the ideal gas law applies: PV = nRT (Equation 14.1). Since the volume is constant 790 THERMODYNAMICS while the temperature changes by an amount ∆T, the amount by which the pressure changes
is ∆P. This change in pressure is given by the ideal gas law in the form
(∆P)V = nR(∆T).
SOLUTION Using Equation 15.6 and the expression Q = (10.0 W)t for the heat delivered
by the heater, we have
C n ∆T
Q = CV n∆T or (10.0 W ) t = CV n∆T or t = V
10.0 W
Using the ideal gas law in the form (∆P)V = nR(∆T), we can express the change in
temperature as ∆T = (∆P)V/nR. With this substitution for ∆T, the expression for the time
becomes
C n ∆P V
t= V
10.0 W nR bg
bg According to Equation 15.8, CV = 3
2 R for a monatomic ideal gas, so we find ( )( ) 3 5.0 × 104 Pa 1.00 ×10−3 m3
3 ( ∆P ) V
3Rn ( ∆P ) V
t=
=
=
= 7.5 s
2 (10.0 W ) nR 2 (10.0 W )
2 (10.0 W ) 43. REASONING AND SOLUTION Let P, V, and T represent the initial values of pressure,
volume, and temperature. The first process is isochoric, so
Q1 = CVn ∆T1 = (3R/2)n ∆T1
The ideal gas law for this process gives ∆T1 = 2PV/(nR), so Q1 = 3PV.
The second process is isobaric, so
Q2 = CPn ∆T2 = (5R/2)n ∆T2
The ideal gas law for this process gives ∆T2 = 3PV/nR, so Q2 = (15/2) PV. The total heat is
Q = Q1 + Q2 = (21/2) PV.
But at conditions of standard temperature and pressure (see Section 14.2), P = 1.01 × 105 Pa
and V = 22.4 liters = 22.4 × 10–3 m3, so
Q= 21 PV
2 = 21
2 (1.01 × 105 Pa ) ( 22.4 × 10−3 m3 ) = 2.38 × 104 J Chapter 15 Problems 791 44. REASONING
a. The efficiency e of a heat engine is given by e = W
QH (Equation 15.11), where QH is the magnitude of the input heat needed to do an amount of work of magnitude W. The “input
energy” used by the athlete is equal to the magnitude ∆U of the athlete’s internal energy
change, so the efficiency of a “human heat engine” can be expressed as
e= W or ∆U ∆U = W
e (1) b. We will use the first law of thermodynamics ∆U = Q − W (Equation 15.1) to find the
magnitude Q of the heat that the athlete gives off. We note that the internal energy change
∆U is negative, since the athlete spends this energy in order to do work.
SOLUTION
a. Equation (1) gives the magnitude of the internal energy change: ∆U = W 5.1 × 104 J
=
= 4.6 × 105 J
e
0.11 b. From part (a), the athlete’s internal energy change is ∆U = −4.6 × 105 J . The first law of
dynamics ∆U = Q − W (Equation 15.1), therefore, yields the heat Q given off by the athlete:
Q = ∆U + W = −4.6 × 105 J + 5.1 × 104 J = −4.1 × 105 J
The magnitude of the heat given off is, thus, 4.1 × 105 J . 45. SSM REASONING AND SOLUTION The efficiency of a heat engine is defined by
Equation 15.11 as e = W / QH , where W is the magnitude of the work done and QH is
the magnitude of the heat input. The principle of energy conservation requires that
QH = W + QC , where QC is the magnitude of the heat rejected to the cold reservoir
(Equation 15.12). Combining Equations 15.11 and 15.12 gives
e= W
W + QC = 16 600 J
= 0.631
16 600 J + 9700 J 792 THERMODYNAMICS 46. REASONING The magnitude W of the work done by a heat engine is related to its
W
efficiency e by e =
(Equation 15.11), where QH is the magnitude of heat extracted
QH
from the hot reservoir. Solving Equation 15.11 for W, we obtain W = e QH (1) The conservation of energy requires that the magnitude QH of the extracted heat must be
equal to the magnitude W of the work done plus the magnitude QC of the heat rejected by
the engine:
QH = W + QC
(15.12)
Because we know the efficiency e and QC, we can use Equations (1) and (15.12) to
determine the magnitude W of the work the engine does in one second.
SOLUTION Substituting Equation (15.12) into Equation (1) yields ( W = e QH = e W + QC ) (2) Solving Equation (2) for the magnitude W of the work done by the engine, we obtain W = e W + e QC W (1 − e ) = e QC or or W= e QC (1 − e ) (3) The engine rejects QC = 9900 J of heat every second, at an efficiency of e = 0.22, so the
magnitude of the work it does in one second is, by Equation (3),
W= e QC (1 − e ) = 0.22 ( 9900 J )
= 2800 J
1 − 0.22 47. REASONING According to Equation 15.11, the efficiency of a heat engine is 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 . We can apply this result before and after the tuneup to compute the extra work produced.
SOLUTION Using Equation 15.11, we find the work before and after the tuneup as
follows:
WBefore = eBefore QH
and
WAfter = eAfter QH Chapter 15 Problems 793 Subtracting the “before” equation from the “after” equation gives WAfter − WBefore = eAfter QH − eBefore QH = ( eAfter − eBefore ) QH = 0.050 (1300 J ) = 65 J 48. REASONING The efficiency e of a heat engine is given by e = W / QH (Equation 15.11),
where W is the magnitude of th...
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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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