When $0.50 \mathrm {~kg}$ of water at $0 ^ { \circ } \mathrm { C }$ freezes, what is the change in entropy of the water? The latent heat of fusion of water is $33,400 \mathrm {~J} / \mathrm { kg }$.

Question

Quizplus · Accepted Answer

The change in entropy ($\Delta S$) during the freezing process can be calculated using the formula $\Delta S = -\frac{Q}{T}$, where $Q$ is the heat removed (which is negative for freezing) and $T$ is the temperature in Kelvin. Given $Q = -0.50 \times 33400 \, \text{J}$ (since 0.50 kg of water is freezing and the latent heat of fusion is 33400 J/kg) and $T = 273 \, \text{K}$ (equivalent to $0 ^{\circ} \text{C}$), $\Delta S = -\frac{-0.50 \times 33400}{273} \approx -610 \, \text{J/K}$.

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