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`.

Options

A. 70 cal
B. 50 cal
C. 79 cal
D. 63 cal

Answer

C

Explanation

`T_(1)=300K" "T_(2)=273K" "thereforeT_(1)-T_(2)=27K`
To freeze 10 gram of water at `0^(@)C`, the quantity of heat that must be transferred (removed) from water to the surrounding is
`Q_(2)="mass"xx"latent heat"=10xx80=800cal`
and `beta=Q_(2)/W=T_(2)/(T_(1)-T_(2))`
`thereforeW=(Q_(2)(T_(1)-T_(2)))/T_(2)=(800xx27)/273=79cal`