What is the entropy change when 4 kg of water at 0 $\circ$ C is frozen into 4 kg of ice at 0 $\circ$ C?

Question

Quizplus · Accepted Answer

The entropy change during the freezing process is given by: 
ΔS = ΔH/T 
where ΔH is the enthalpy of fusion, which is 334 kJ/kg for water, and T is the freezing temperature, which is 0°C. Therefore, 
ΔS = (334 kJ/kg)/(273 K) = 1.225 kJ/K 
Since we have 4 kg of water, the total entropy change is: 
ΔS = 4 kg×1.225 kJ/K/kg = 4.9 kJ/K 
Thus, the best choice is answer C.

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