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`.

Options

A. 400 W
B. 500 W
C. 300 W
D. 250 W

Answer

B

Explanation

Consider a time interval `Deltat` while the door is closed.
Then the rate of heat into the box is
`(Q)/(Deltat)=kA((DeltaT)/(Deltax))=(0.05)(6)((22)/(0.09))=73.3W`
To remove heat at this rate, while the motor runs only for a time (0.15) `Deltat`. It must cause heat to leave at the rate,
`H=(73.3)/(0.15)=500W`