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 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 outermost 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 `0.125 , 1.5 "and" 1.0 W//m//^(@)C` 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.