An ideal refrigerator has a freezer at a temperature of `-13^(@)C`. The coefficient of performance of the engine is `5`. The temperature of the air (to which heat is rejected) will be

Options

A. `325^@C`
B. `325 K`
C. `39^@C`
D. `320^@C`

Answer

C

Explanation

`T_2=-13^@C=-13+273=260K`
`eta'=5`
Coefficient of performance
`eta'=(Q_2)/(W)=(Q_2)/(Q_1-Q_2)=(1)/((Q_1)/(Q_2)-1)=(1)/((T_1)/(T_2)-1)`
`(T_1)/(T_2)-1=(1)/(eta')implies(T_1)/(T_2)=1+(1)/(eta')`
`(T_1)/(260)=1+(1)/(5)=(6)/(5)`
`T_1=312K=312-273=39^@C`