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Homework Help: Carnot heat engine

  1. Jan 26, 2010 #1
    Given :
    QH=W+QC and Power= E/t or W/t

    A heat engine is attached to a 200W lamp. If we power the lamp for 1 hour and extract 700 btu from the heat source, how much heat is rejected?

    Attempt:

    200w(3.412btu/1w) = 682.4 btu now, 700 btu-682.4 btu =17.6 btu rejected??
     
    Last edited: Jan 26, 2010
  2. jcsd
  3. Jan 26, 2010 #2

    ideasrule

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  4. Jan 27, 2010 #3
    what if i was after time?


    Given :
    QH=W+QC and Power= E/t or W/t

    A heat engine is attached to a 200W lamp. we can extract 700 btu from the heat source and engine is 50% efficient, how long can we power it??

    Attempt:

    200w(3.412btu/1w) =682.4 /2 btu =341.2 btu and 700 btu-682.4 btu =17.6 btu
    Total 358.8btu
    lost from there??
     
    Last edited: Jan 27, 2010
  5. Jan 27, 2010 #4

    Andrew Mason

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    You are converting watts (Joules/sec) into BTU/hr (1 w = 3.412 btu/hr). Therefore, the 200w lamp consumes 682.4 btu/hr.

    If the Qh is 700 btu at 50% efficiency, this means that you can extract 350 btu of work. Since the light consumes 682.4 btu in one hour, how long can you keep the light going with 350 btu of energy?

    AM
     
  6. Feb 6, 2010 #5
    less then 2hrs.
     
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