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

The situation is shown in figure
The thermal resistances of the wood, the cement and the brick layer are
`R_(W)=(1)/(K)(x)/(A)` . `=(1)/(0.125Wm^(-1)`^(@)C^(-1))(2.5xx10^(-1)m)/(137m^(2))` . `=(0.20)/(137)`^(@)CW^(-1)` . `R_(C)=(1)/(1.0Wm^(-1)`^(@)C^(-1))(1.0xx10^(-2)m)/(137m^(2))` . `=(0.0067)/(137)`^(@)CW^(-1)` . `R_(B)=(1)/(1.0Wm^(-1)`^(@)C^(-1))(25.0xx10^(-2)m)/(137m^(2))` . `=(0.25)/(137)`^(@)CW^(-1)` . As the layers are connected in series, the equivalent thermal resistance is
`R=R_(W)+R_(C)+R_(S)` . `=(0.20+0.0067+0.25)/(137)`^(@)CW^(-1)` . `=3.33xx10^(-3)`^(@)CW^(-1)` . The heat curent is
`i=(theta_(1)-theta_(2))/(R)` . `=(20^(@)C-(-10^(@)C))/(3.33xx10^(-3)`^(@)CW^(-1))~~9000W` . The heater must supply 9000W to compensate the outflow of heat.