An electric heater is used in a room: Heat Conduction

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Problem: An electric heater is used in a room of total wall area 137 m2 to maintain a temperature of +20°C inside it, when the outside temperature is -10°C. The walls have three different layers. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outermost layer is of brick of thickness 25.0 cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are 0.125, 1.5 and 1.0 W/(m-°C), respectively.  (IIT JEE 1986)

Answer: The power of the electric heater is 9000 W.

Solution: The thickness of the wood, the cement, and the brick layers are $x_1=2.5$cm, $x_2=1$cm, and $x_3=25$cm and thermal conductivities of these layers are $K_1=0.125$W/(m-°C), $K_2=1.5$W/(m-°C), and $K_3=1$W/(m-°C) (see figure). Total wall area is $A=137$m2, inside temperature is 20°C, and outside temperature is -10°C. Let $T_1$ and $T_2$ be the temperatures at the interface of wood-cemnet and cement-brick layers, respectively.

An electric heater is used in a room

The rates of heat flow through the layers are given by \begin{align} \frac{\mathrm{d}{Q_1}}{\mathrm{d}t}&=\frac{K_1A(20-T_1)}{x_1}, \nonumber\\ \frac{\mathrm{d}{Q_2}}{\mathrm{d}t}&=\frac{K_2A(T_1-T_2)}{x_2}, \nonumber\\ \frac{\mathrm{d}{Q_3}}{\mathrm{d}t}&=\frac{K_3A(T_2+10)}{x_3}. \nonumber \end{align} In the steady state, the rate of heat flow is equal in all the layers i.e., \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}=\frac{\mathrm{d}Q_1}{\mathrm{d}t}=\frac{\mathrm{d}Q_2}{\mathrm{d}t}=\frac{\mathrm{d}Q_3}{\mathrm{d}t}, \end{align} which gives \begin{align} \label{pra:eqn:4} \frac{K_1(20-T_1)}{x_1}&=\frac{K_2(T_1-T_2)}{x_2},\\ \label{pra:eqn:5} \frac{K_2(T_1-T_2)}{x_2}&=\frac{K_3(T_2+10)}{x_3}. \end{align} Solve these equations for $T_1$ and $T_2$ and substitute the values to get the rate of heat loss through the wall as \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}&=\frac{20-(-10)}{\frac{x_1}{K_1A}+\frac{x_2}{K_2A}+\frac{x_3}{K_3A}} \nonumber\\ &=\frac{30}{\frac{0.025}{0.125(137)}+\frac{0.01}{(1.5)(137)}+\frac{0.25}{(1)(137)}}\nonumber\\ &=9000\,\mathrm{W}. \nonumber \end{align} To maintain the room temperature at 20°C, the rate of heat produced in the room should be equal to the rate of heat loss through the walls. Thus, the power of the electric heater should be 9000 W.

Alternate Solution: The thermal resistances of the three layers are $R_1={x_1}/{(K_1A)}$, $R_2={x_2}/{(K_2A)}$ and $R_3={x_3}/{(K_3A)}$. These thermal resistances are connected in series giving the effective thermal resistance of the wall as \begin{align} {R_\text{eff}}&=R_1+R_2+R_3\\ &=\frac{x_1}{K_1A}+\frac{x_2}{K_2A}+\frac{x_3}{K_3A}. \nonumber \end{align} The rate of heat flow through a wall having temperature difference $\Delta T=20-(-10)=30$°C and thermal resistance $R_\text{eff}$ is given by \begin{align} \frac{\mathrm{d}Q}{\mathrm{d}t}&=\frac{\Delta T}{R_\text{eff}}\\ &=\frac{30}{\frac{x_1}{K_1A}+\frac{x_2}{K_2A}+\frac{x_3}{K_3A}}\\ &=9000\,\mathrm{W}. \nonumber \end{align}

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