A refrigerator works between `4^(@)C` and `30^(@)C`. It is required to remove `600 calories` of heat every second in order to keep the temperature of the refrigerator space constant.The power required is (Take `1calorie= 4.2 J`)

Options

A. 2.365 W
B. 23.65 W
C. 236.5 W
D. 2365 W

Answer

C

Explanation

Since, `beta= (Q_2)/(W) = (T_2)/(T_1-T_2)`
`implies (600xx4)/(W)=(277)/(303-277) implies W=236.5 J`
Power (P)= `W/t =("236.5 Joule")/("1 sec")= 236.5 W`.