- #1
Erik Horwath
- 7
- 0
a firebox is at 750K and the ambient temp is 300K. The efficiency of a Carnot engine doing 150J of work as it transports energy between these constant temperature baths is 60%. The Carnot engine must take in energy 150/.60=250J from the hot reservoir and must put out 100J of energy by heat into the environment. To follow Carnot's reasoning, suppose that xome other heat engine S could have efficiency 70%. a)Find the energy input and wasted energy output of engine S as it does 150J of work. b)Let engine S operate as in part a) and run the Carnot engine in reverse. Find the total energy the firebox puts out as both engines operate together, and the total energy transferred to the environment. Show that the Clausius statemetn of the second law of thermodynamics is violated.
Part a) is obvious. They show exactly how to do it within the problem. But the wording in part b) is confusing to me. If the engine is operating in reverse, then it is a heat pump, not a heat engine. Therefore the work under consideration must be input work, not output work. So when it says "let engine S operate as in part a)" does that mean 150J of work in done on the system rather than by the systm? Further, the expression for the coefficient of performance of a heat pump is different from that for the effiecency of a heat engine. The former is e=Wnet/Qin and is always less than unity and the latter is cop=Qout/Win and is generally >1. I don't understand what the problem is asking exactly. Thanks for your help!
Part a) is obvious. They show exactly how to do it within the problem. But the wording in part b) is confusing to me. If the engine is operating in reverse, then it is a heat pump, not a heat engine. Therefore the work under consideration must be input work, not output work. So when it says "let engine S operate as in part a)" does that mean 150J of work in done on the system rather than by the systm? Further, the expression for the coefficient of performance of a heat pump is different from that for the effiecency of a heat engine. The former is e=Wnet/Qin and is always less than unity and the latter is cop=Qout/Win and is generally >1. I don't understand what the problem is asking exactly. Thanks for your help!