How much energy in watt hour may be required to convert `2kg` of water into ice at `0^(@)C`, assuming that the refrigerator is ideal? Take room temp. `=25^(@)C`, which is also the initial temp. of water and temp. of freezer is `-15^(@)C`.

Explanation

Here `T_(1)=273+25=298K` and
`T_(2)=273-15=258K`
Specific heat of water
`C=4.2xx10^(3)J//kgK`
Latent heat of fusion of ice
`L=3.36xx10^(5)J//kg`
The amount of that required to transform water of `25^(@)C` intio ice of `0^(@)C`
`Q_(2)=mC DeltaT=mL`
`=2xx4.2xx10^(3)xx(25-0)+2xx3.36xx10^(5)`
`=2.1xx10^(5)+6.72xx10^(5)`
`=8.82xx10^(5)J`
Heat rejected to the surroundings
We have `(Q_(1))/(Q_(2))=(T_(1))/(T_(2))`
`:.Q_(1)=Q_(2)((T_(1))/(T_(2)))`
`=8.82xx10^(5)(298/258)=10.19xx10^(5)J`
Energy supplied to convert water into ice
`W=Q_(1)-Q_(2)`
`=10.15xx10^(5)-8.82xx10^(5)`
`=1.33xx10^(5)J`
`=(1.33xx10^(5))/3600=36.96Wh`