A Carnot engine absorbs 6 10^5 cal at 227^ C . The work done per cycle by the engine, if its sink is maintained at 127^ C is

Step-by-step solution

Here, Q ₁=6 10^5 cal aligned & T ₁=227^ C =227+273=500 ~K & ~T ₂=127^ C =127+273=400 ~K aligned Work done/cycle, W = ? As Q ₂ Q ₁ = T ₂ ~T ₁ or Q ₂= T ₂ ~T ₁ Q ₁= 400 500 6 10^5=4.8 10^5 cal aligned As, & W = Q ₁- Q ₂=6 10^5-4.8 10^5 ~W & =1.2 10^5 cal =1.2 10^5 4.2 ~J ~W & =5.04 10^5 ~J aligned