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Carnot engine efficiency problem

  1. Jan 6, 2017 #1
    1. The problem statement, all variables and given/known data
    A Carnot engine absorbed 1.0 kJ of heat at 300 K, and exhausted 400 J of heat at the end of the cycle. What is the temperature at the end of the cycle?

    2. Relevant equations
    The efficiency of a Carnot engine is given by the formula
    Efficiency = 1 – Qc/Qh
    = 1 – Tc/Th

    3. The attempt at a solution
    Efficiency = 1 – 300J/1000J
    = 0.7 = 70%
    0.7 = 1 – Tc/Th = 1 – Tc/300K
    Tc = 90K
    Is it correct? I am trying to self study the topic Carnot cycle for a report
     
  2. jcsd
  3. Jan 6, 2017 #2

    TSny

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    Gold Member

    Welcome to PF!
    Did you mean to use 300 J for QC? The problem statement gives 400 J.

    Otherwise, I think your calculation is correct. However, to specify the temperature at the end of the cycle, you need to know which point of the cycle is taken to be the "end of the cycle".
     
  4. Jan 7, 2017 #3
    Oh i see i misread the problem so it should be
    Efficiency = 1 – 400J/1000J = 0.6
    Then
    0.6 = 1 – Tc/Th = 1 – Tc/300K
    Tc = 120K
     
  5. Jan 7, 2017 #4

    TSny

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    OK. I'm still not clear on which point of the cycle is the endpoint. So I'm not sure if TC is the temperature at the endpoint. But your work for TC looks good.
     
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