The temperature inside a refrigator is `t_2^@`C and the room temperature is `t_1^@C`. The amount of heat delivered to the room for each joule of electrical energy consumed ideally will be

Explanation

The area of the roof
`=4mxx 4m=16m^(2)` . The thickness `x=10cm=0.01m` . (a) The thermal resistance of the roof is
R_(1)=(1)/(K)(x)/(A)=(1)/(1.26Wm^(-1)`^(@)C^(-1))(0.01m)/(16m^(2)` . `=4.96xx10_(-3)`^(@)CW_(-1)` . The heat current is
`=(DeltaQ)/(Deltat)=(theta_(1)-theta_(2))/R_(1)=(46^(@)C-32^(@)C)/(4.96xx_(-3)`^(@)CW^(-1)` . `=2822W` . (b) The thermal resistance of the brick layer is
`R_(2)=(1)/(K)(x)/(A)=(1)/(0.65Wm^(-1)`^(@)C^(-1))(7.5xx10^(-2)m)/(16m^(2)`. `=7.2xx10^(-3)`^(@)CW^(-2)` . The equivalent thermal resistance is `R=R_(1)+R_(2)=(4.96+7.2)xx(10)^(-3)`^(@)CW^(-1)`. `=1.216xx10^(-2)`^(@)CW^(-1)` . The heat current is
`(DeltaQ)/(Deltat)=(theta_(1)-theta_(2))/(R)=(46^(@)C-32^(@)C)/(1.216xx10_(-2)`^(@)CW^(-1)` . `=1152W` .