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

Evidently, the three layers (wood, cement and brick) are arranged in series.
So, the equivalent thermal resistance (R ) of the combination will be
`R=R_1+R_2+R_3`
`(1)/(150m^2)=[(2xx10^-2)/(0.15)(m^(2@)C)/(W)+(1xx10^-2)/(1.75)(m^(2@)C)/(W)+(20xx10^-2)/(1.0)(m^(2@)C)/(W)]`
From, rate of heat loss `(dQ)/(dt)=(30-(-5)^@C)/(2.26xx10^(-3@)C//W)=15.49kW`
Since the heat lost from the inside of the room to the outside through the walls is compensated by the room heater, so the power of the heater is `15.49kW`.