Physics Solution Manual for 1100 and 2101

12 for qc we have qc qh w 202 1013 j 73 1012 j

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Unformatted text preview: , or ten joules per second, so Q = (10.0 W)t, where t is the on-time 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 tune-up 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.

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