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# A thermodynamic sys-tem is taken from state a to state c ISBN: 9780321675460 31

## Solution for problem 42P Chapter 19

University Physics | 13th Edition

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Problem 42P

A thermodynamic sys-tem is taken from state ?a to state ?c in ?Fig. P19.38 along either path ?abc or path ?adc?. Along path ?abc?, the work W done by the system is 450 J. Along path ?adc?, ?W is 120 J. The internal energies of each of the four states shown in the figure are Ua = 150 J, Ub = 240 J, Uc = 680 J, and Ud = 330 J. Calculate the heat flow Q for each of the four processes ?ab?, ?bc?, ?ad?, and ?dc?. In each process, does the system absorb or liberate heat?

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Solution 42P This question is based on the first law of thermodynamics, which is expressed mathematically as Q = U + W , where Q is the amount of heat added to a system, U is the change in internal energy of the system and W is the work done by the system. Along path abc, W) abc = 450 J From path a to b, the volume remains constant, so W) = 0…..(1) ab So, W) = 450 J …..(2) bc Along the path adc, W) = 120 J adc From path d to c, the volume remains constant, so W) dc= 0…..(3) So, W) ad = 120 J …..(4) Now, change in internal energy from a to b U ab= U bU a= 240 J 150 J = 90 J …..(5) Similarly. U bc= U cU b = 680 J 240 J = 440 J …..(6) U dc= 680 J 330 J = 350 J …..(7) U = 330 J 150 J = 180 J …..(8) ad Heat flow calculations: Heat flow for ab, From equations 1 and 5, Q) ab= U ab+ W ab = 90 J + 0 J = 90 J Heat flow for bc, From equations 2 and 6, Q) = U + W = 440 J + 450 J = 890 J bc bc bc Heat flow for dc, From equations 3 and 7, Q) dc= U dc+ W) dc= 350 J + 0 = 350 J Heat flow for ad, From equations (4) and (8), Q) = U + W) = 180 J + 120 J = 300 J ad ad ad Q value is positive for each path. Therefore, the system will absorb heat.

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##### ISBN: 9780321675460

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A thermodynamic sys-tem is taken from state a to state c