An electric heater is used in a room of total wall area `137m^(2)` to maintain a temperature of `20^(@)C` inside it, when the outside temperature is `-10^(@)C` .The walls have three different layers of materials. The innermost layer is of wood of thickness 2.5cm, the middle layer is of cement of thickness 1.0cm and the outermost layer is of brick of thickness 25.0cm. Find the power of the electric heater. Assume that there is no heat loss through the floor and the celling. The thermal conductivities of wood, cement and brick are `0.125Wm^(-1)C^(-1)` , `1.5Wm^(-1)C^(-1)` . and `1.0Wm^(-1)C^(-1)` respectively.

Explanation

Equivalent thermal conductivity of the wall
`K=(l_(1)+l_(2)+l_(3))/(l_(1)/K_(1)+l_(2)/K_(2)+l_(3)/K_(3))`
= `(0.025+0.01+0.25)/((0.025/0.125+0.01/1.5+0.25/1.0))`
= `0.285/0.457=0.624W//m-""^(@)C`
The rate of flow of heat is given by `H=KA(T_(1)-T_(2))/L`
= `0.624xx137xx([20-(10)])/0.285=(0.624xx137xx30)/0.285=9000W`