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I Carnot refrigerator

  1. Sep 18, 2016 #1
    If a Carnot refrigerator requires a work input of only 230 J to extract 346 J of heat from the cold reservoir.
    Doesn’t this discrepancy imply a violation of the law of conservation
    of energy?
     
  2. jcsd
  3. Sep 18, 2016 #2

    Simon Bridge

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    Does it? Please state your reasoning.
     
  4. Sep 18, 2016 #3
    My reasoning to that is it does not violate the 1st law because the heat Qh that will be rejected will be bigger than the extracted Qc.
    what I wondering is how 230J of work results in 346J of heat?
    According to the conservation of energy, 230J of work will be converted,If so, to 230J of heat or any other kind of energy, is that right?
     
  5. Sep 18, 2016 #4

    Simon Bridge

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    Why not look up the carnot engine equations and do some math?
    The work is the extra energy that has to come from outside in order to get the heat to flow "backwards" against the temperature gradient between the two reservoirs ... the gas starts out with 346J, and ends up with 576J. You had to do 230J of work to do that. The simple model does not tell you exactly how this happened.
     
    Last edited: Sep 18, 2016
  6. Sep 18, 2016 #5
    I may not deliver my question right or you may not got it.
    Anyway I did the math and it get out right, but I am asking that the work applied to the refrigerator to extract the heat from the cold staff should not be equal or more than the heat extracted?
     
  7. Sep 18, 2016 #6

    Simon Bridge

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    It's like carrying a 5kg weight up 4.7m of stairs ... that would mean you have to do 230.3J of work.
    If the bottom of the stairs was 7.3m above sea level, maybe the house is on a hill, then the energy of that 5kg weight was 357.7J
    So you just did about 230J of work to extract 357.7J from the bottom of the stairs.
     
  8. Sep 18, 2016 #7
    I understand this, but how is that related?
     
  9. Sep 18, 2016 #8

    CWatters

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    How much gasoline does it take to transport 1000 gallons of gasoline 100 yards down the road?
     
  10. Sep 18, 2016 #9
    We get the volume and the distance, if I am correct about it what's the formula that gives us the work done, so we can find the amount of gasoline by using Q=mLc?
    ??
     
  11. Sep 18, 2016 #10

    CWatters

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    I don't want you to get bogged down in the detail. The point I am making is that that moving energy from one part of a system to another (eg from cold to hot reservoir) is not the same as adding energy to a system. There is no law that says moving energy costs more energy than the energy being moved.

    A lorry might consume less than 1gallon moving 1000gallons that short distance.

    Perhaps worth reading up on "closed systems" as conservation of energy only applies to those. If you don't include a hot reservoir in your system it's not closed.
     
  12. Sep 18, 2016 #11
    This helps a little, but I am still confused!
     
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