A carnot engine takes ( 300 ) calories of heat from a source at ( 500 ~K ) and rejects ( 150 ) calories of heat to the sink. The temperature of the sink is

Step-by-step solution

Given, heat taken from source, Q₁=300 calories; temperature of source T₁=500 ~K ; heat rejected to the sink, Q₂=150 calories. Temperature of sink, T₂=? Now, efficiency, = Q₁-Q₂ Q Also, efficiency, = T₁-T₂ T₁ Therefore, Q₁-Q₂ Q₁ = T₁-T₂ T₁ 300-150 300 = 500-T₂ 500 150 300 =1- T₂ 500 1 2 =1- T₂ 500 T₂ 500 =1- 1 2 T₂=500 1 2 =250 ~K Thus, temperature of sink =250 ~K