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Heat Q flows spontaneously from a reservoir at 383 K into a reservoir

  1. Sep 19, 2010 #1
    Heat Q flows spontaneously from a reservoir at 383 K into a reservoir that has a lower temperature T. Because of the spontaneous flow, thirty percent of Q is rendered unavailable for work when a Carnot engine operates between the reservoir at temperature T and a reservoir at 234 K. Find the temperature T.

    I have absolutely no idea. Please help me. I am getting desperate. I have to pull up my failing physics grade or else i will pretty much be disowned by my family.
     
  2. jcsd
  3. Sep 19, 2010 #2

    rock.freak667

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    Given this, what would the efficiency of the Carnot Engine?

    When you get that, what is the expression for efficiency of a Carnot Engine?
     
  4. Sep 19, 2010 #3
    the efficiency i guess is 2/3.
    e=1-(Tc/Th)
     
  5. Sep 19, 2010 #4
    so what next?
     
  6. Sep 19, 2010 #5

    rock.freak667

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    How did you get 2/3 ?

    So when you get 'e' Th=T, and you were given Tc, can you get T?
     
  7. Sep 19, 2010 #6
    i got 2/3 because it says that 1/3 is unavailable therefore 2/3 is available
     
  8. Sep 19, 2010 #7

    rock.freak667

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    That thinking is correct but thirty percent is not the same as 1/3.
     
  9. Sep 19, 2010 #8
    oh no! im lost again what is the th?
     
  10. Sep 19, 2010 #9
    so the efficiency is actually .7
     
  11. Sep 19, 2010 #10

    rock.freak667

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    Right so, what is T?
     
  12. Sep 19, 2010 #11
    there are two Ts in the problem. do i use the higher one for Th?
     
  13. Sep 19, 2010 #12

    rock.freak667

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    You are only considering the two temperatures associated with the Carnot engine, the high temperature T and the lower temperature 234 K.
     
  14. Sep 19, 2010 #13
    so i use 234K?
     
  15. Sep 19, 2010 #14

    rock.freak667

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    Yes.
     
  16. Sep 19, 2010 #15
    so let me get this straight .7(234)=T where T is the answer im looking for? this seems too simple
     
  17. Sep 19, 2010 #16

    rock.freak667

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    e=1- (Tc/T), so you '0.7' would be '0.3'.

    Nothing is being done to the heat when it is moving from 384K to T, so I don't think you'd need to worry about that for now.
     
  18. Sep 19, 2010 #17
    so .3=1-(234/t)
     
  19. Sep 19, 2010 #18

    rock.freak667

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    No, 0.7 = 1 -(Tc/T)

    when you rearrange you'd get the '0.3'.
     
  20. Sep 19, 2010 #19
    alright i think i almost got this. so the T i am trying to solve for is equal to Tc/.3
     
  21. Sep 19, 2010 #20

    rock.freak667

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    Yes and you know th vaue of Tc.
     
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