A refrigerator is used to mainatain eatables kept inside it at `9^(@)C`. If the room temperature is `35^(@)C`, then calculate the coefficient to performance. Calculate the amount of heat rejected to the surrounding.

Explanation

Given : `T_(2)=9+273=282K`
`T_(1)=36+273=309K`
Coefficient of performance `prop=(T_(2))/(T_(1)-T_(2))`
i.e., `alpha=(282)/(309=282)=10.44" ""but "(1)/(alpha)=(Q_(1))/(Q_(2))-1`
hence `(Q_(1))/(Q_(2))-(1+alpha)/(alpha)=(1+10.44)/(10.44)" "i.e., (Q_(1))/(Q_(2))=(11.44)/(10.44)=1.096`
i. e., heat rejected to the surroundings is `1.096Q_(2)J`