Question

An electric heater is used in a room of total wall area $137{\text{m}}^2$ to maintain a temperature of ${20}^{\circ }\text{C}$ inside it when the outside temperature is $-{10}^{\circ }\text{C}.$ The walls have three layers of different materials. The innermost layer is of wood of thickness $2.5\text{cm}$, the middle layer is of cement of thickness $1.0\text{cm}$ and the outermost layer is of brick of thickness $2.5\text{cm}$. Find the power of the electic heater. Assume that there is no heat loss through the floor and ceiling. The thermal conductivities of wood, cement and brick are $0.125{\text{W/m}}^{\circ }\text{C}$, $1.5{\text{W/m}}^{\circ }\text{C}$ and $1.0{\text{W/m}}^{\circ }\text{C}$ respectively.

• A. $9\text{W}$
• B. $22.5\text{W}$
• C. $9\text{kW}$
• D. $22.5\text{kW}$

Answer 1

The correct option is C $9\text{kW}$

Thermal resistance offered by the walls in series combination is given by
$R=R_1+R_2+R_3$
i.e $\frac{L_1+L_2+L_3}{K_{eq}A}=\frac{L_1}{K_{1}A}+\frac{L_2}{K_{2}A}+\frac{L_3}{K_{3}A}$


From the diagram, we can write the above equation as,
$\frac{L_1+L_2+L_3}{K_{eq}}=\frac{L_1}{K_1}+\frac{L_2}{K_2}+\frac{L_3}{K_3}$
From the data given in the question, $\frac{(2.5+1+25)\times {10}^{-2}}{K_{eq}}=(\frac{2.5}{0.125}+\frac{1}{1.5}+\frac{25}{1})\times {10}^{-2}$
$\therefore K_{eq}=0.624{\text{W/m}}^{\circ }\text{C}$
The power of the heater $H=K_{eq}A\frac{\Delta T}{\Delta x}$
$=0.624\times 137\left(\frac{30}{28.5\times {10}^{-2}}\right)\approx 9\text{kW}$
Thus, option (c) is the correct answer.