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Heat Engines and the Carnot Cycle

  1. Mar 18, 2008 #1
    [SOLVED] Heat Engines and the Carnot Cycle

    1. The problem statement, all variables and given/known data

    The operating temperatures for a Carnot engine are Tc and Th= Tc + 55K. The efficiency of the engine is 15%.

    How do i find Tc?
     
  2. jcsd
  3. Mar 18, 2008 #2

    nicksauce

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    [tex]e = 1 - \frac{Tc}{Th} = 1 - \frac{Tc}{Tc+55}[/tex]

    One equation one unknown should not be a problem...
     
  4. Mar 18, 2008 #3
    Thats all mastering physics gives me. A problem like that is in the book but the efficiency of the engine is 11% instead. They get 450 K for Tc
     
  5. Mar 18, 2008 #4

    dynamicsolo

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    Keep in mind that while efficiencies are often quoted as percentages, they must be used in equations as fractions. Set e = 0.15 and you should be able to solve the Carnot equation for Tc.
     
  6. Mar 18, 2008 #5
    Do i set e= 0.15 to the equation above? and if i do how do i go about getting the answer?
     
  7. Mar 18, 2008 #6

    dynamicsolo

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    That's correct.

    You will have to re-arrange the equation algebraically. Starting from

    [tex]0.15 = 1 - \frac{Tc}{Tc+55}[/tex] , the first step would be

    [tex]\frac{Tc}{Tc+55} = 1 - 0.15 = 0.85[/tex]
     
  8. Mar 18, 2008 #7
    thanks for the help
     
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