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Homework Help: The Carnot Engine Problem

  1. Sep 22, 2010 #1
    1. The problem statement, all variables and given/known data
    A Carnot heat engine operates between temperatures Th and Tc.
    (a) If Th = 515 K and Tc = 363 K, what is the efficiency of the engine?
    0.30 Your answer is correct.

    (b) What is the change in its efficiency for each degree of increase in Th above 515 K?
    ? K-1

    (c) What is the change in its efficiency for each degree of change in Tc?
    ? K-1

    (d) Does the answer to part (c) depend on Tc?
    Yes or No?

    Explain.


    2. Relevant equations
    e_c = 1 - T_c/T_h


    3. The attempt at a solution

    (a)
    T_h = 515k
    T_c = 363k
    e_c = 1 - T_c/T_h = 1 -363k/515k = 0.2951456311 = 0.30

    I'm don't know how to start part (b). Do I use the same equation or a different one?
     
  2. jcsd
  3. Sep 22, 2010 #2

    kuruman

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    You want to find the rate of change of the efficiency e with respect to Th. Rate of change is also known as ...? (Hint: It's the "D" word.)
     
  4. Sep 22, 2010 #3
    So do the derivative of e = 1 - T_c/T_h ?
     
  5. Sep 22, 2010 #4

    kuruman

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  6. Sep 22, 2010 #5
    e = 1 - T_c/T_h

    How do you do the derivative of that equation?

    dE = -ln(T_h/T_c) ?
     
  7. Sep 23, 2010 #6

    kuruman

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    Would it help if you said let x = TC and wrote

    e = 1 - x/TH?

    Can you find de/dx ?
     
  8. Sep 23, 2010 #7
    Actually I did not have to do the derivative. It was so simple roflol.

    b) Increase T_h by one and calculate the efficiency

    1 - 363/516= 0.2965

    calculate difference from the original efficiency

    0.2965-0.2951=.0014

    c) Decrease T_c by one and calculate

    1 - 362/515= 0.2971

    Now find the difference

    .2951 - .2971= 0.002

    d) No
     
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