Problem — Ice in hot tea

Problem

A glass contains $250$ g of tea at $85\ ^{\circ}$C. An ice cube of $30$ g at $-5\ ^{\circ}$C is added. Neglecting exchanges with the surroundings and with the glass, calculate the equilibrium temperature. Data: $c_{\text{water}} = 4186$ J/(kg·K), $c_{\text{ice}} = 2100$ J/(kg·K), $L_f = 334$ kJ/kg.

Solution

Heat released by the tea. The tea, of mass $0.25$ kg, cools from $85\ ^{\circ}$C down to $T_e$: $Q_{\text{released}} = 0.25 \cdot 4186 \cdot (85 - T_e) = 1046.5 \, (85 - T_e)$ Heat absorbed by the ice (three steps). Warming from $-5\ ^{\circ}$C to $0\ ^{\circ}$C, melting, then warming the melt water from $0\ ^{\circ}$C to $T_e$: $Q_1 = 0.03 \cdot 2100 \cdot 5 = 315\ \mathrm{J}, \quad Q_2 = 0.03 \cdot 334000 = 10020\ \mathrm{J}, \quad Q_3 = 0.03 \cdot 4186 \cdot T_e = 125.58 \, T_e$ Energy balance. Equating heat released and absorbed: $1046.5 \, (85 - T_e) = 315 + 10020 + 125.58 \, T_e$ $88952.5 - 1046.5 \, T_e = 10335 + 125.58 \, T_e$ $78617.5 = 1172.08 \, T_e \;\Rightarrow\; T_e = \frac{78617.5}{1172.08} \approx 67.1\ ^{\circ}\mathrm{C}$ The result is positive and below $100\ ^{\circ}$C, consistent with the assumption that all the ice melts and only liquid water remains at equilibrium. $\ev{T_e \approx 67.1\ ^{\circ}\mathrm{C}}$

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Topics: Termologia Concepts: Calore latente e cambiamenti di stato Skills: Equazione calorimetrica