An electric heater is placed inside a room of total wall area `137 m^2` to maintain the temperature inside at `20^@C`. The outside temperature is `-10^@C`. The walls are made of three composite materials. The inner most layer is made of wood of thickness 2.5 cm the middle layer is of cement of thickness 1 cm and the exterior layer is of brick of thickness 2.5 cm. Find the power of electric heater assuming that there is no heat losses through the floor and ceiling. The thermal conductivities of wood, cement and brick are `0.125 W//m.^@C, 1.5W//m.^@C and 1.0 W//m.^@C` 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)=0.624 W//m^(@)C`
The rate of flow of heat is given by
`H=KA(T_1-T_1)/(L)`
`=0.624xx137xx([20-(-10)])/(0.285)`
`=(0.624xx137xx30)/(0.285)=9000W`