An ordinary refrigerator is thermally equivalent to a box of corkboard 90 mm thick and 5.6`m^2` in inner surface area. When the door is closed, the inside wall is kept, on the average, `22.2^@C` below the temperature of the outside wall. If the motor of the refrigerator runs `15%` of the time while the door is closed, at what rate must heat be taken from the interior whicle the motor is running?
The thermal conductivity of corkboard is `k=0.05 W//mK`.

Explanation

Consider a time interval `Deltat` during which the door is closed. As approximation, take the heat conduction to be steady ober `Deltat`.
Then the rate of heat into the box is
`(DeltaQ)/(Deltat)=kA((DeltaT)/(Deltax))=(0.05)(5.6)((22.2)/(0.090))=69.1W`
To remove this heat, the motor must, since it runs only for a time (0.15) `Deltat`, cause heat to leave at the rate `69.1//0.15=460W`.