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Thermal efficiency

  1. Nov 16, 2003 #1
    A heat engine has an efficiency of 25.0% and its heat input is 600J per cycle from the high-temperature reservoir. What is the rate of heat output to the low-temperature reservoir per cycle?

    E=(Qh-Qc)/Qh

    .25=(600J-Qc)/600J
    Qc=450J

    book has the answer as 800J.
    How are they getting 800J and not 450J?
     
  2. jcsd
  3. Nov 16, 2003 #2

    Doc Al

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    Staff: Mentor

    The book's answer makes no sense. Your reasoning is correct. Are you sure you're not mixing up input and output heats? That would explain the book's answer.
     
  4. Nov 16, 2003 #3
    yes, your're right. It looks like I mixed up input and output heats.
    Qc=600J (heat flow into cold res.)

    find heat output from high-temp res. which would be
    Qh, so solving for Qh you get 800J
     
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