Calculate the least amount of work that must be done to freeze one gram of wate at `0^@C` by means of a refrigerator. Temperature of surroundings is `27^@C`. How much heat is passed on the surroundings in this process? Latent heat of fusion `L=80cal//g`.

Answer

A, C

Explanation

`Q_2=mL=1xx80=80cal`
`T_2=0^@C=273K`
and
`T_1=27^@C=300K`
Least amount of work will be needed for carnot's type of cycle.
`Q_2/W=(T_2)/(T_1-T_2)`
`:.` `W=(Q_2(T_1-T_2))/(T_2)`
`=(80(300-273))/(273)`
`=7.91cal`
`Q_1=Q_2+W`
`=(80+7.91)`
`=87.91 cal`