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Carnot Engine Question

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Homework Statement



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?

Homework Equations



A) e = (T2-T1)/T2
ΔU=Q-W

The Attempt at a Solution



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?
 

Answers and Replies

  • #2
Everything is fine, nothing to worry.....
Regards
 
  • #3
Simon Bridge
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Well done. Sometimes physics is easy - with that reasoning it sounds like you have understood the material.
 
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  • #4
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Well done. Sometimes physics is easy - with that reasoning it sounds like you have understood the material.
Thanks for the confirmation.
 

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