Worked exercise — Coffee and steam exchange

Problem

To heat $300$ ml of milk (treat it as water) from $20\ ^{\circ}$C to $65\ ^{\circ}$C, an espresso machine uses steam at $100\ ^{\circ}$C which condenses in the milk. How many grams of steam are needed? Data: $c_{\text{water}} = 4186$ J/(kg·K), $L_v = 2.26 \cdot 10^{6}$ J/kg.

Solution

Energy required by the milk. The milk, of mass $m_L = 0.3$ kg, must go from $20\ ^{\circ}$C to $65\ ^{\circ}$C, a rise of $\Delta T = 45$ K: $Q_L = m_L \, c_{\text{water}} \, \Delta T = 0.3 \cdot 4186 \cdot 45 \approx 5.65 \cdot 10^{4}\ \mathrm{J}$ Energy released by the steam. A mass $m_v$ (in kg) of steam goes through two steps: first it condenses at $100\ ^{\circ}$C releasing the latent heat, then the hot liquid cools from $100\ ^{\circ}$C down to $65\ ^{\circ}$C. $Q_1 = m_v \, L_v = m_v \cdot 2.26 \cdot 10^{6}\ \mathrm{J}$ $Q_2 = m_v \, c_{\text{water}} \cdot 35 = m_v \cdot 1.46 \cdot 10^{5}\ \mathrm{J}$ Balance. The energy released equals that absorbed by the milk: $m_v \left(2.26 \cdot 10^{6} + 1.46 \cdot 10^{5}\right) = 5.65 \cdot 10^{4}$ $m_v = \frac{5.65 \cdot 10^{4}}{2.41 \cdot 10^{6}} \approx 0.0235\ \mathrm{kg}$ The condensation term dominates: almost all the useful energy comes from the latent heat, not from the subsequent cooling. $\ev{m_v \approx 24\ \mathrm{g}}$

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Topics: Termologia Concepts: Calore latente e cambiamenti di stato · Calore specifico e capacità termica Skills: Equazione calorimetrica