 A system undergoes a power cycle while receiving 1,000 kilojoules by heat transfer from hot combustion gases at a temperature of 500 Kelvin and discharging 600 kilojoules by heat transfer to the atmosphere at 300 Kelvin. Taking the combustion gases and atmosphere as the hot and cold bodies respectively, determined for the cycle, the network developed in kilojoules and the thermal efficiency. So drawing this in the same cycle diagram we used for our definitions of thermal efficiency and coefficient of performance, I have a power cycle receiving heat at a magnitude of 1,000 kilojoules and rejecting heat at 600 kilojoules. Then with a simplification of our energy balance, I can say that E in must equal E out and for energy entering I only have Q in and for energy exiting I have a network out which remember is work out minus work in and heat transfer out. So if 1,000 kilojoules enters 600 kilojoules leaves as heat and the rest leaves as network out, then the difference between 1,000 and 600 must be how much work left as network out. Then for thermal efficiency I'm taking the network out divided by the heat transfer entering which is going to be 400 kilojoules divided by 1,000 kilojoules which is 0.4 or 40 percent. And the assumptions we had to make to get that far were steady state analysis that the system we analyzed was a closed system.