P21_010 - 10. This problem is similar to Sample Problem...

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Unformatted text preview: 10. This problem is similar to Sample Problem 21-2. The only difference is that we need to find the mass m of each of the blocks. Since the two blocks are identical the final temperature Tf is the average of the initial temperatures: 1 1 Tf = (Ti + Tf ) = (305.5 K + 294.5 K) = 300.0 K . 2 2 Thus from Q = mc∆T we find the mass m: m= Q 215 J = = 0.101 kg . c∆T (386 J/kg · K)(300.0 K − 294.5 K) (a) ∆SL = mc ln Tf TiL = (0.101 kg)(386 J/kg · K) ln 300.0 K 305.5 K = −0.710 J/K . (b) Since the temperature of the reservoir is virtually the same as that of the block, which gives up the same amount of heat as the reservoir absorbs, the change in entropy ∆SL of the reservoir connected to the left block is the opposite of that of the left block: ∆SL = −∆SL = +0.710 J/K. (c) The entropy change for block R is ∆SR = mc ln Tf TiR = (0.101 kg)(386 J/kg · K) ln 300.0 K 294.5 K = +0.723 J/K . (d) Similar to the case in part (b) above, the change in entropy ∆SR of the reservoir connected to the right block is given by ∆SR = −∆SR = −0.723 J/K. (e) The change in entropy for the two-block system is ∆SL + ∆SR = −0.710 J/K + 0.723 J/K = +0.013 J/K. (f) The entropy change for the entire system is given by ∆S = ∆SL + ∆SL + ∆SR + ∆SR = ∆SL − ∆SL + ∆SR − ∆SR = 0, which is expected of a reversible process. ...
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