Assuming that a domestic refrigerator can be regarded as a reversivle engine working between the temperature of melting ice and that of the atmosphere `(170^(@)C)`, calculate the energy which muct be supplied to freeze one kilogram of water already at `0^(@)C`.

Answer

`2.092xx10^(4)J`

Explanation

`T_(2)=0^(@)C= 0+273= 273K`
`T_(1)= 17^(@)C= 17+273= 290K`
Heat required to be taken out to freeze `1 kg` of water already at `0^(@)C` is
`Q_(2)= mL= 1xx80,000 cals. =80000xx4.2J`
Cofficient of performance of refrigerator is
`(Q_(2))/W=(T_(2))/(T_(1)-T_(2))`
`W=(Q_(2)(T_(1)-T_(2)))/(T_(2))=(80000xx4.2(290-273))/(273)`
`W=2.092xx10^(4)J`