Question

# Nonmetric version: (a) How long does a$2.0 \times 10^5 Btu/h$water heater take to raise the temperature of 65 gal of water from$70 ^ { \circ } \mathrm { F }$to$100 ^ { \circ } \mathrm { F }$? Metric version: (b) How long does a 59 kW water heater take to raise the temperature of 246 L of water from$21 ^ { \circ } \mathrm { C }$to$38 ^ { \circ } \mathrm { F }$?

Solution

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a)

We can use expression for amount of heat $Q$ required to change the temperature of a given material and it is proportional to the mass $m$ of the material present and to the temperature change $\Delta T$:

\begin{align*} Q&=mc_{water} \Delta T \\ c_{water}&=1 \, \frac{\text{Btu}}{\text{lb}\cdot \text{\textdegree F}} \tag{specific heat of water from expression 18-15} \\ P&=\frac{Q}{\Delta t} \tag{expression for power} \\ \Delta t&=\frac{Q}{P} \tag{express \Delta t} \\ \Delta t&=\frac{mc_{water} \Delta T}{P} \tag{substitute Q} \\ \rho&=\frac{m}{V} \tag{expression for density} \\ m&=\rho V \tag{express m} \\ 1 \text{ gal}&=231 \text{ in}^3 \tag{from apendix D} \\ 1 \, \frac{\text{kg}}{\text{m}^3}&=3.613 \cdot 10^{-5} \, \frac{\text{lb}}{\text{in}^3} \tag{from apendix D} \\ \rho&=1000\, \frac{\text{kg}}{\text{m}^3}=3.613 \cdot 10^{-2} \, \frac{\text{lb}}{\text{in}^3} \tag{conversion of the density of the water} \\ \Delta t&=\frac{\rho V c_{water} \Delta T}{P} \tag{substitute m} \\ \Delta t&=\frac{3.613 \cdot 10^{-2} \cdot 65 \cdot 231 \cdot 1 \cdot (100-70)}{2 \cdot 10^5} \tag{substitute} \\ \Delta t&=0.08137 \text{ h}=\boxed{4.88 \text{min}} \end{align*}

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