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, in the middle layer is od 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 conductives of wood, cement and brick are `0.125, 1.5 and 1.0 watt//m// .^(@)C` respectively.

Explanation

Here, `x_(1) = 0.025 m , x_(2) = 0.01 m`,
`x_(3) = 0.25 m`
`K_(1) = 0.125 Wm^(-1) .^(@)C^(-1), K_(2) = 1.5 Wm^(-1) .^(@)C^(-1)`,
`K_(3) = 1.0 Wm^(-1) .^(@)C^(-1)`
Equivalent thermal conductivity of the series combination of three walls is
`K= (x_(1)+x_(2)+x_(3))/((x_1)/(K_1)+(x_2)/(K_2)+ (x_3)/(K_3))=(0.25+0.01+0.25)/((0.025)/(0.125)+(0.01)/(1.5)+(0.25)/(1.0))`
`=(0.285 xx 300)/(137)Wm^(-1) .^(@)C^(-1)`
Rate of flow of heat is
`(Delta Q)/(Delta t) = KA(Delta theta)/((x_(1)+x_(2)+x_(3)))`
`=(0.285 xx 300)/(137) xx 137 xx ([20(-10)])/(0.285)`
`= 9000W`.