For the refrigerator of this problem, T H = 96 F = 309 K and T L = 70 F = 294 K, so K = (294 K)/(309 K 294 K) = 19.6.

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Darcy Hampton
5 years ago
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1 39. A Carnot refrigerator working between a hot reservoir at temperature H and a cold reservoir at temperature L has a coefficient of performance K that is given by K L = H L For the refrigerator of this problem, H = 96 F = 309 K and L = 70 F = 94 K, so K = (94 K)/(309 K 94 K) = he coefficient of performance is the energy Q L drawn from the cold reservoir as heat divided by the work done: K = Q L / W. hus, Q L = K W = (19.6)(1.0 J) = 0 J..

2 18. (a) It is possible to motivate, starting from Eq. 0-3, the notion that heat may be found from the integral (or area under the curve ) of a curve in a S diagram, such as this one. Either from calculus, or from geometry (area of a trapezoid), it is straightforward to find the result for a straight-line path in the S diagram: Q straight i + = f S which could, in fact, be directly motivated from Eq. 0-3 (but it is important to bear in mind that this is rigorously true only for a process that forms a straight line in a graph that plots versus S). his leads to for the energy absorbed as heat by the gas. (b) Using able 19-3 and Eq , we find Q = (300 K) (15 J/K) = J 3 3 Eint = n R =.0 mol8.31 J/mol K00 K 400 K = J. (c) By the first law of thermodynamics, a f W = Q ΔEint = 4.5 kj 5.0 kj = 9.5 kj.

3 46. (a) We denote the configuration with n heads out of N trials as (n; N). We use Eq. 0-0: 50! 14 W 5;50 = = ! 50 5! (b) here are possible choices for each molecule: it can either be in side 1 or in side of the box. If there are a total of N independent molecules, the total number of available states of the N-particle system is N Ntotal = =. With N = 50, we obtain N total = 50 = (c) he percentage of time in question is equal to the probability for the system to be in the central configuration: With N = 100, we obtain (d) W(N/, N) = N!/[(N/)!] = , (e) N total = N =1.7 (f) and p(n/; N) = W(N/, N)/ N total = 8.0%. Similarly, for N = 00, we obtain (g) W(N/, N) = , (h) N total =1.61, (i) and p(n/; N) = 5.7%. 14 5; W p5; % (j) As N increases, the number of available microscopic states increase as N, so there are more states to be occupied, leaving the probability less for the system to remain in its central configuration. hus, the time spent in there decreases with an increase in N.

4 33. (a) Energy is added as heat during the portion of the process from a to b. his portion occurs at constant volume (V b ), so Q in = nc V. he gas is a monatomic ideal gas, so C 3 R/ and the ideal gas law gives V 3 hus, in b a b = (1/nR)(p b V b p a V a ) = (1/nR)(p b p a ) V b. Q p p V. V b and p b are given. We need to find p a. Now p a is the same as p c, and points c and b are connected by an adiabatic process. hus, pcvc pbv b and 53 V b a = c = b = Pa = Pa. Vc 8.00 p p p he energy added as heat is Q in = Pa Pa m = J. (b) Energy leaves the gas as heat during the portion of the process from c to a. his is a constant pressure process, so 5 5 Qout = nc = p V p V = p V V 5 p a a c c a a c = Pa m = J, 5 or Q J. he substitutions V a V c = V a 8.00 V a = 7.00 V a and C R out were made. (c) For a complete cycle, the change in the internal energy is zero and (d) he efficiency is W = Q = J J = J. = W/Q in = ( J)/( J) = 0.64 = 6.4%. p

5 11. (a) he energy that leaves the aluminum as heat has magnitude Q = m a c a ( ai f ), where m a is the mass of the aluminum, c a is the specific heat of aluminum, ai is the initial temperature of the aluminum, and f is the final temperature of the aluminum water system. he energy that enters the water as heat has magnitude Q = m w c w ( f wi ), where m w is the mass of the water, c w is the specific heat of water, and wi is the initial temperature of the water. he two energies are the same in magnitude, since no energy is lost. hus, macaai + mwcw wi maca ai f = mwcw f wi f =. m c + m c a a w w he specific heat of aluminum is 900 J/kgK and the specific heat of water is 4190 J/kgK. hus, f 0.00 kg900 J/kg K100C kg4190 J/kg K0C 0.00 kg900 J/kg K kg4190 J/kg K 57.0C 330 K. (b) Now temperatures must be given in Kelvins: ai = 393 K, wi = 93 K, and f = 330 K. For the aluminum, dq = m a c a d, and the change in entropy is dq f d f 330 K Sa maca maca ln 0.00 kg900 J/kg Kln ai ai 373 K.1 J/K. (c) he entropy change for the water is dq f d f 330K Sw mwcw mwcw ln ( kg)(4190 J kg.k)ln wi wi 93K 4.9J K. (d) he change in the total entropy of the aluminum-water system is S = S a + S w =.1 J/K J/K = +.8 J/K. 837

6 (a) he efficiency is H L H (35 115)K %. (35+73) K We note that a temperature difference has the same value on the Kelvin and Celsius scales. Since the temperatures in the equation must be in Kelvins, the temperature in the denominator is converted to the Kelvin scale. (b) Since the efficiency is given by = W / Q H, the work done is given by 4 4 W Q H 0.36( J) = J.

Chapter 20. and the corresponding change in entropy is 1 absorbed (see Eq. 20-2). S T dq Q T, where Q is the heat

Chapter 20. and the corresponding change in entropy is 1 absorbed (see Eq. 20-2). S T dq Q T, where Q is the heat Chapter 0 1. INK If the expansion of the gas is reversible and isothermal, then there s no change in internal energy. owever, if the process is reversible and adiabatic, then there would be no change in