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Unformatted text preview: 17. (a) The ﬁnal mass of ice is (1773 g + 227 g)/2 = 1000 g. This means 773 g of water froze. Energy
in the form of heat left the system in the amount mLF , where m is the mass of the water that
froze and LF is the heat of fusion of water. The process is isothermal, so the change in entropy is
∆S = Q/T = −mLF /T = −(0.773 kg)(333 × 103 J/kg)/(273 K) = −943 J/K.
(b) Now, 773 g of ice is melted. The change in entropy is
∆S = Q
= +943 J/K .
T (c) Yes, they are consistent with the second law of thermodynamics. Over the entire cycle, the change
in entropy of the water-ice system is zero even though part of the cycle is irreversible. However, the
system is not closed. To consider a closed system, we must include whatever exchanges energy with
the ice and water. Suppose it is a constant-temperature heat reservoir during the freezing portion
of the cycle and a Bunsen burner during the melting portion. During freezing the entropy of the
reservoir increases by 943 J/K. As far as the reservoir-water-ice system is concerned, the process is
adiabatic and reversible, so its total entropy does not change. The melting process is irreversible,
so the total entropy of the burner-water-ice system increases. The entropy of the burner either
increases or else decreases by less than 943 J/K. ...
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