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Carnot Engine Question

  1. Oct 7, 2013 #1
    1. The problem statement, all variables and given/known data

    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?

    2. Relevant equations

    A) e = (T2-T1)/T2
    ΔU=Q-W

    3. 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?
     
  2. jcsd
  3. Oct 7, 2013 #2
    Everything is fine, nothing to worry.....
    Regards
     
  4. Oct 7, 2013 #3

    Simon Bridge

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    Well done. Sometimes physics is easy - with that reasoning it sounds like you have understood the material.
     
  5. Oct 7, 2013 #4
    Thanks for the confirmation.
     
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