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

Question

Quizplus · Accepted Answer

When water freezes, it releases heat to the surroundings, so the change in entropy can be calculated using the formula ΔS = Q/T, where Q is the heat released and T is the temperature at which it is released. Since the water is at 0°C and the ice is also at 0°C, there is no change in temperature and T = 0K. The heat released is given by the heat of fusion of water, which is 334 kJ/kg. Therefore, ΔS = - (334 kJ/kg) / (0K) = - ∞, which is not a physically meaningful answer. However, we know that the entropy change must be negative, since a solid has lower entropy than a liquid. The correct way to calculate the change in entropy is using the equation ΔS = Sice - Swater, where S is the entropy. At 0°C, the entropy of water is 6.023 J/Kg K and the entropy of ice is 1.999 J/Kg K. Therefore, ΔS = (4 kg)(1.999 J/Kg K - 6.023 J/Kg K) = - 17.696 J/K = - 4.9 kJ/K (rounded to one decimal place).

What Is the Entropy Change When 4 Kg of Water