EasyPhysicsHigh School (11–12)Heat & Thermodynamics › Heat Engines, the Second Law and EntropyNumerical

0.35 kg of ice at 0 °C melts, and the water then warms to 20 °C.

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$0.35\ \text{kg}$ of ice at $0\ ^\circ\text{C}$ melts, and the water then warms to $20\ ^\circ\text{C}$ . Find the change of entropy in each stage and in all. (Latent heat of fusion $3.34 \times 10^{5}\ \text{J/kg}$ , $c = 4186\ \text{J kg}^{-1}\text{K}^{-1}$ .)

Need a reminder? The key ideaShow

Heat $Q$ taken in slowly at a steady temperature $T$ raises the entropy by $\Delta S = \dfrac{Q}{T}$ . Warming from $T_1$ to $T_2$ : $\Delta S = mc\ln\dfrac{T_2}{T_1}$ . A gas at a steady temperature: $\Delta S = nR\ln\dfrac{V_2}{V_1}$ . Entropy depends only on the state, and the total for an isolated system never falls.

The worked solution

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