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

Energy required, `W=?, m=2kg`,
`T_(1)= 25^(@)C=(25+273)K= 298K`
`T_(2)= -15^(@)C= (-15+273)K= 258K`
specific heat of water, `c=1cal//g`
`=4.2xx10^(3)j//kg`
Latent heat of ice, `L=80cal//g`
`=3.36xx10^(5) J//kg`
Amount of heat required to be removed from `2kg` of water at `25^(@)C` to form ice at `0^(@)C`
`Q_(2)= mc(theta_(2)-theta_(1))+mL`
`=2xx4.2xx10^(3)(25-0)+2xx3.36xx10^(5)`
`=(2.1+6.72)10^(5)= 8.82xx10^(5)J`.
Heat rejected to the surroundings
`Q_(1)= (T_(1))/(T_(2))xxQ_(2)= (298)/(258)xx8.82xx10^(5)`
`=10.15xx10^(5)J`
`:.` Energy to be supplied for the purpose
`W=Q_(1)-Q_(2)=(10.15-8.82)10^(5)J`
`=1.33xx10^(5)J=(1.33xx10^(5))/(3600)Wh`
`36.96Wh`