Question

An electric heater is used in a room of total wall area $137m^2$ to maintain a temperature of ${20}^{\circ }C$ inside it, when the outside temperature is - ${10}^{\circ }C$.

The walls have three different layers of materials. The innermost layer is of wood of thickness 2.5 cm, the middle layer is of cement of thickness 1.0 cm and the outer most 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 the ceiling. The thermal conductivities of wood, cement and brick are

$0.125W/m^{\circ }C$, $1.5W/m^{\circ }C$ and $1.0W/m^{\circ }C$ respectively.

• A. 900W.
• B. 19000W.
• C. 9000W.
• D. 1000W.

Answer 1

The correct option is C 9000W.


The thermal resistance of the wood, the cement and the brick layers are,
$R_W=\frac{1}{K}\frac{x}{A}$
$=\frac{1}{0.125W/m^{\circ }C}\frac{2.5\times {10}^{-2}m}{137m^2}$
$={\frac{0.20}{137}}^{\circ }C/W$
$R_C=\frac{1}{1.5W/m^{\circ }C}\frac{1.0\times {10}^{-2}m}{137m^2}$
$={\frac{0.0067}{137}}^{\circ }C/W$
$and{R}_B=\frac{1}{1.0W/m^{\circ }C}\frac{2.5\times {10}^{-2}m}{137m^2}$
$={\frac{0.25}{137}}^{\circ }C/W$
As the layers are connected in series, the equivalent thermal resistance is,
$R=R_W+R_C+R_B$
$={\frac{0.20+0.0067+0.25}{137}}^{\circ }C/W$
$=3.33\times 10{-3}^{\circ }C/W$
The heat current is
$i=\frac{{\theta }_1-{\theta }_2}{R}$
$=\frac{20rC-(-10rC)}{3.33\times {10}^{-3}rC/W}=9000W$
The heater must supply 9000W to compensate the outflow of heat.