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Unformatted text preview: 10. This problem is similar to Sample Problem 212. 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 twoblock 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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 Fall '08
 SPRUNGER
 Physics, Mass

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