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114 學年度臺灣大學轉學生招生考試｜第 67 題

普通物理學(A)

A copper can of negligible heat capacity contains 1.0 kg of water just above the freezing point. A similar can contains 1.0 kg of water just below the boiling point. The two cans are brought into thermal contact. What is the change in entropy of the system? The specific heat capacity of water is 4180 J kg-1 K-1.
  1. +404 J/K
  2. +202 J/K
  3. +101 J/K
  4. The entropy is unchanged
  5. -202 J/K
提示
Equal masses of water mix to a final temperature of 50 °C. The hot water's entropy falls while the cold water's rises; sum ΔS = mc·ln(T_f/T_hot) + mc·ln(T_f/T_cold) with absolute temperatures, and the net must be positive.
正確答案

正解：C

詳解
A. (A) +404 J/K — incorrect. About four times too large; likely from a factor or temperature-conversion error.
B. (B) +202 J/K — incorrect. About twice the correct value, a common error from double-counting or omitting one logarithm's sign.
C. (C) +101 J/K — correct. Final T = 50 °C = 323 K. ΔS = mc·ln(323/273) + mc·ln(323/373) = 4180·[ln(1.183) + ln(0.866)] = 4180·(0.1682 − 0.1439) ≈ +101 J/K. Net positive, as required for an irreversible mixing.
D. (D) The entropy is unchanged — incorrect. Mixing hot and cold water is irreversible, so total entropy must increase, not stay constant.
E. (E) -202 J/K — incorrect. A negative total entropy change would violate the second law for this spontaneous, irreversible process.

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