A Carnot engine having an efficiency of eta = (1 / 10) as a heat engine is used as a refrigerators. If the work done on the system is 10 J, what is the amount of energy absorbed from the reservoir at lowest temperature ?

A Carnot engine having an efficiency of eta = (1 / 10) as a heat engin

1.	A Carnot engine having an efficiency of eta = (1 / 10) as a heat engine is used as a refrigerators. If the work done on the system is 10 J, what is the amount of energy absorbed from the reservoir at lowest temperature ?
Answer» Solution :As HEAT ENGINE, LET Q the heat energy ABSORBED from the SOURCE and `Q_(2)` be the energy rejected to the sink. Then `Q_(1)=W+Q_(2)=10+Q_(2)"…(1)"` `"Given :"eta=1-(Q_(2))/(Q_(1))=(1)/(10)` `(Q_(2))/(Q_(1))=1-(1)/(10)=(9)/(10)` `Q_(1)=(10)/(9)Q_(2)"...(2)"` From equation (1) and (2), `10+Q_(2)=(10)/(9)` `10=((10)/(9)-1)Q_(2)` `therefore""Q_(2)=90J`