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

`alpha=5" T_(2)=-20^(@)C`
`=-20+273`
`T_(1)=?`
`alpha=(T_(2))/(T_(1)-T_(2))`
`implies5=(253)/(T_(1)-253)`
`implies5T_(1)-1265=243`
`T_(1)=(1518)/(5)=303.6K`
`=30.6^(@)C~~31^(@)C`
`T_(1)=`temp. of surrounding
`T_(2)`=temp. of source (inside temp.)