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What change in entropy occurs when a 27.9-g ice cube at -12 C is transformed into steam at 115 C?

Add the entropy changes due to: (1) heating the ice to 0C, (2) melting the ice, (3) heatng the water from 0 to 100 C, and (4) Evaporating the liquid water. For (2) and (3), the entropy change is the heat transfered (the heat of fusion or vaporization) divided by the corresponding temperature. For (1) and (3), an integration must be performed, and the entropy change is Mass*C*ln (T2/T1), where T2 is the final temperature and T1 is the initial temperature

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asked by jean
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  2. 0.5×27.9×ln(273÷261)+(80×27.9/273)+ln(388/273)+(540/388)=10.54cal/k

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    posted by Udit

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