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

Coefficient of performance of a refrigerator,
`beta = Q_(2)/W = T_(2)/(T_(1)-T_(2))` (Where `Q_(2)` is heat removed)
Given: `T_(2) = 4^(@) C = 4+273 = 277 K`
`T_(1) = 30^(@)C = 30 + 273 = 303 K`
`therefore beta = (600 xx 4.2)/(W) = 277/(303-277)`
`rArr W = 236.5` Joule
Power `P = W/t = (236.5 "joule")/(1 "sec") = 236.5` watt.