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Thermodynamics (Carnot Engines)

  1. Apr 22, 2008 #1
    1. The amount of heat that an engine rejects to the cold reservoir is 3 times the amount of work that it does. What is the efficiency of the engine?

    e = W/ Qh = 1 - Qc/Qh

    For this problem, I said e= w/3w. Thus e would equal 1/3 = 0.333. This however is not correct. What did I do wrong?

    2. A Carnot engine operates between the temperatures 390 K and 300 K. How much heat is discarded for the engine to produce 1900 J of work?

    e = 1 - Tc/Th
    e = 1 - 300K/390K = 0.2307
    e = W / Qh
    thus 0.2307 = 1900J/ Qh
    I solved for Qh and i got the answer to be 8233.3 J

    This isn't correct either, but I'm not sure what I'm doing wrong.

    3. During each cycle a reversible engine absorbs 1550 J of heat from a high-temperature reservoir and performs 1200 J of work. How much heat is exhausted to the low-temperature reservoir during each cycle?

    Qh = W + Qc
    1550 = 1200 + Qc and then I solved for Qc. This isn't correct either.

    Please Help! I am desperate and I really don't understand these problems.
     
  2. jcsd
  3. Apr 22, 2008 #2

    Mapes

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    You're rushing through these and making algebra errors. How did you go from e = 1 - Qc/Qh to e = W/3W? What is the relationship between W, Qc, and Qh, according to conservation of energy?
     
  4. Apr 22, 2008 #3

    Shooting Star

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    Who said e=W/Qc?

    It's given that Qc = 3W => e = 1 - 3W/Qh. Also, e = W/Qh.

    Eliminate Qh from these two eqns and see how much is e.

    Same for the other eqns.
     
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