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Homework Help: Thermodynamics - please guide me through

  1. May 11, 2009 #1
    Can someone please guide me through these problems..

    (1) An inventor claims to have constructed an engine that has an efficiency of 72.0% when operated between the boiling and freezing points of water. Calculate the difference between the percentage difference between the claimed efficiency and the actual efficiency. (ans: 45.2%)

    (2) An ideal refrigerator, with coefficient of performance 4.3 extracts heat from the cold chamber at the rate of 258J/cycle.
    (i) How much work per cycle is required to operate the refrigerator? (ans: 60.0 J)

    (ii) How much heat is added to the room environment each cycle? (ans: 318.0 J)
  2. jcsd
  3. May 11, 2009 #2


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    Homework Helper

    So find the maximum efficiency of the engine (meaning it uses a carnot cycle) and check the difference.

    What is the definition of the coefficient of performance?

    What is the first law of thermodynamics for a closed cycle?
  4. Apr 8, 2010 #3
    The efficiency of the engine is given by

    [tex]e= \frac{W_{eng}}{|Q_h|}=\frac{|Q_h|-|Q_c|}{|Q_h|}= 1-\frac{|Q_c|}{|Q_h|}[/tex]

    We are only told that the they claim the engine has efficiency 72.0%


    But we don't have any other information. So what do we need to do?


    The coefficient of preformance is given by

    [tex]COP=\frac{|Q_c|}{W}[/tex] (cooling mode)

    4.3= 258/W

    W=60 :biggrin:

    It think for cyclic process, the internal energy is equal to 0. Because Q=-W

    But here Q=-W=-60

    But this is wrong, the correct answer has to be 318.0 J. :rolleyes:
  5. Apr 8, 2010 #4
    (A) The maximum efficiency of a thermal engine operating between Ta and Tb>Ta is 1-Ta/Tb (see Carnot efficiency at http://en.wikipedia.org/wiki/Thermal_efficiency)

    (C) The internal energy is decreased at each cycle by 258J, you didn't consider that...
  6. Apr 9, 2010 #5
    Thank you so much, I've got it! :smile:
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