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

`33.2J ; 369.2J`

Explanation

Here, `W=?, m= 1g= 10^(-3) kg`,
`T_(1)= 27^(@)C= (27+273)K= 300K`
`T_(2)= 0^(@)C=(0+273)K= 273K , Q_(1)=?`
Heat extracted from cold reservoir, `Q_(2)=mL= 10^(-3)xx3.36xx10^(5)J`
Coeff. Of performance of refrigerator,
`COP= (Q_(2))/W= (T_(2))/(T_(1)-T_(2))`
`W=(Q_(2)(T_(1)-T_(2)))/(T_(2))=(10^(-3)xx3.36xx10^(5)(300-273))/(273)`
`=33.2J`
Heat passed on the surroundings,
`Q_(1)=Q_(2)+W= 336+33.2= 369.2J`