A Carnot's engine works as a refrigerator between 250 ~K and 300 ~K . It receives 500 cal heat from the reservoir at the lower temperature. The amount of work done in each cycle to operate the refrigerator is:

Step-by-step solution

Given: Temperature of cold body, T ₂=250 K temperature of hot body; T ₁=300 ~K Heat received, Q₂=500 cal work done, W = ? Efficiency =1- T₂ T₁ = W Q₂+W 1- 250 300 = W Q₂+W ~W = Q₂ 5 = 500 4.2 5 J=420 ~J