In a refrigerator, heat from inside at `277 K` is transferred to a room at `300K`. What is the cofficient of performance of the refrigerator. How many joule of heat will be deliverd to the room for each joule of electric energy consumed ideally?

Answer

`12.04 ; 13.04J`

Explanation

Here, `T_(2)= 277K`
`T_(1)= 300K, COP=? Q_(1)=?`
`W=1J`
`COP= (Q_(2))/W=(T_(2))/(T_(1)-T_(2))`
`COP=(273)/(300-277)= (273)/(23)=12,04`
Again, as `COP= (Q_(2))/W=(Q_(2))/(1J)= 12.4`
`Q_(2)= 12.04J`
`Q_(1)=Q_(2)+W= 12.04+1= 13.04J`