Freezing compartment of a refrigerator is at `0^(@)C` and room temperature is `27.3^(@)C`. Work done by the refrigerator to freeze 1g of water at `0^(@)C` is `(L_("ice") = 80 cal g^(-1))`

Options

A. 336J
B. 33.6J
C. 3.36J
D. 40J

Answer

B

Explanation

Coefficient of performance of refrigerator is
`beta = (T_(2))/(T_(1) - T_(2))`
`T_(2) = 0^(@)C = 0 + 273 = 273K`
`T_(1) = 27.3^(@)C ~~ 300K`
`:. beta = (273)/(300-273) ~~ 10`
Now, if `Q_(2)` is heat extracted and W is work performed them,
`beta = (Q_(2))/(W) or W = (Q_(2))/(beta)`
where, `Q_(2)=` heat extracted from 1g of water to make it ice
`=mL_(1) = 80 cal = 80 xx 4.2J = 336J`
So, W = 336/10 = 33.6J