A refrigerator freezes 5 kg of water at `0^(@)C` into ice at `0^(@)C` in a time internal of 20 minutes. Assume that the room temperature is `20^(@)C`. Calculate the minimum power needed to accomplish it.

Explanation

Amount of heat required to convert water into ice at 0°C,
`Q_(2) = mL =(5kg) xx (80 kj cal//kg) = 400 cal`
Now `T_(1) = 20^(@)C = 273 + 20 = 293 K,`
`T_(2) = 0^(@)C = 273 + 0 = 273 K`
we know `(Q_(2))/(W) =(T_(2))/(T_(1) - T_(2))`
`:.` W (work done) `= Q_(2) xx (T_(1) - T_(2))/(T_(2))`
`= 400 xx (293 - 273)/(273) = 400 xx (20)/(273) = 29.3` kcal
`= 29.3 xx 4.2 xx 10^(3) j` or `W = 123 xx 10^(3) j`
time, t = 20 min `= 20 xx 60 = 1200` s.
`:.` Power needed, `P = (w)/(t) = (123 xx 10^(3))/(1200)`
= 102.5 W