The cofficient of performance of a refrigerator is 5. If the temperature inside freezer is `-20^(@)C`, the temperature of the surroundings to which it rejects heat is :

Options

A. `21^(@)C`
B. `31^(@)C`
C. `41^(@)C`
D. `11^(@)C`

Answer

B

Explanation

Here, temperature of freezer, `T_(c)= -20^(@)C= -20+273= 253K`
Temperature of surroundings, `T(A)=?`
cofficeint of performance `beta= 5`
coeficient of performance is

`beta= (Q_(2))/W= (Q_(2))/(Q_(1)-Q_(2))=(T_(C))/(T_(H)-T_(C))=5`
or `T_(C)= 5T_(H)-5T_(C) or 6T_(C)= 5T_(H)` or `T_(H)= 6/5T_(C)= 6/5xx253= 303.6K`
`=30.6^(@)C~~31^(@)C`