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Carnot Engine Question: Efficiency & Work/Heat Rejected
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[QUOTE="1question, post: 4529313, member: 466045"] [h2]Homework Statement [/h2] A Carnot heat engine absorbs heat form a high-temp reservoir at 1200K and concerts some of it into useful work and rejects the rest to a low-temp reservoir at 400K. A) What is the efficiency of the engine? b) If the engine absorbs 2400J per cycle from the high-temp reservoir at 1200K, what is the amount of work done, and how much heat is rejected to the low-temp reservoir at 400K? [h2]Homework Equations[/h2] A) e = (T[SUB]2[/SUB]-T[SUB]1[/SUB])/T[SUB]2[/SUB] ΔU=Q-W [h2]The Attempt at a Solution[/h2] For part a, I got (1200-400)/1200= 0.6666... For part b, my logic is that Total energy = work + heat rejected, based on ΔU formula above. Therefore, work done should be 2400J*0.66 = 1600 J and heat rejected should be the rest, aka 2400*0.33 = 800J. These answers seems to easy, especially for part b. Am I doing something wrong? [/QUOTE]
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Carnot Engine Question: Efficiency & Work/Heat Rejected
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