A refrigerator has to transfer an average of `506J` of heat per second from. Temp. `- 20^(@)C "to" 25^(@)C`. Caculate the average power consumed, assuming no energy losses in the process.

Explanation

Here, `Q_(2)=506J`,
`T_(2)= -20^(@)C= (-20+273)K= 253K`
`T_(1)= 25^(@)C= (25+273)K= 298K`
As `(Q_(1))/(Q_(2))= (T_(2))/(T_(2))`
`:. Q_(1)=(T_(1))/(T_(2))Q_(2)= (298)/(253)xx506= 596J`
Average power consumed,
`W=Q_(1)-Q_(2)= 596-506= 90 "watt" `