Problem — Hot bath

Problem

A bathtub contains $150$ kg of water at $12\ ^{\circ}$C. How much energy, in MJ and in kWh, is needed to bring it to $38\ ^{\circ}$C? Data: $c_{\text{water}} = 4186$ J/(kg·K).

Solution

Heat required. With $m = 150$ kg and a rise $\Delta T = 38 - 12 = 26$ K: $Q = m \, c \, \Delta T = 150 \cdot 4186 \cdot 26 \approx 1.63 \cdot 10^{7}\ \mathrm{J} \approx 16.3\ \mathrm{MJ}$ Conversion to kWh. Since $1$ kWh $= 3.6 \cdot 10^{6}$ J: $Q = \frac{1.63 \cdot 10^{7}}{3.6 \cdot 10^{6}} \approx 4.53\ \mathrm{kWh}$ Water’s high specific heat explains why heating a full bathtub takes quite a lot of energy. $\ev{Q \approx 16.3\ \mathrm{MJ} \approx 4.53\ \mathrm{kWh}}$

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Topics: Termologia Concepts: Calore specifico e capacità termica