Efficiency of a perfect engine working between –3°C and 27°C (i.e.,T2=270K and T1=300K) ηengine=1−
T2
T1
=1−
270K
300K
=0.1 Since efficiency of the refrigerator (ηref) is 50% of ηengine ∴ηref=0.5ηengine=0.5 If Q1 is the heat transferred per second at higher temperature by doing work W, then
ηref=
W
Q1
orQ1=
W
ηref
=
1kJ
0.05
=20kJ
(as W=1kW×1s=1kJ) Since ηref Is 0.05,heat removed from the refrigerator per second, i.e.,