Homework Answers
A) Since the lake is very large, it's temperature will not change. Only the ice will melt. The energy involved in this is:
Q = mL = (1.2)(334) = 400.8 kJ
Temperature of ice T = 0 degree C = (0+273) K = 273 K
Change in entropy of ice = Q/T = 400.8/273 = 1.4681 kJ/k or 1468.3 J/K (ans)
B)
Temperature of system T = 10 degree C = (10+273) K = 283 K
so, Change in entropy of the system = Q/T = 400.8/283 = 1.4163 kJ/k or 1416.3 J/K (ans)
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