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Entropy Change in a Carnot Engine

  1. May 19, 2009 #1
    1. The problem statement, all variables and given/known data
    A Carnot engine using a gas operates between 2 reservoirs of temperatures 1000K and 280K. The work output per cycle is 1kJ.
    a) What is the efficiency?
    b) What is the entropy change of the gas per cycle due to the heat transfer at the hot reservoir?

    2. Relevant equations

    [tex] \eta_{carnot} = 1 - \frac{Q_{1}}{Q_{2}} = 1 - \frac{T_{1}}{T_{2}} [/tex]

    [tex] \Delta S = \int \frac{dQ}{T} [/tex]

    [tex] \Delta U = W_{on} + Q_{in} [/tex]


    3. The attempt at a solution
    a) Seems easy enough, I got 72%.
    b) I think you just use the entropy formula above, which goes to S = Q_in / T_2 (never used Latex before, and wasn't working here - don't have time to work out why!). If internal energy, U, is zero then heat in equals negative work done on the gas (ie. -1kJ) due to the 1st law. In that case I get the change in entropy to be 1.. which just doesn't seem right. Can anyone comment/help?

    Thanks!
     
  2. jcsd
  3. May 19, 2009 #2

    Mapes

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    I'm curious about what answer would "seem right"... OK, does heat in really equal work out? Isn't some energy also rejected to the cold reservoir?
     
  4. May 19, 2009 #3
    I guess I'm not sure what answer would 'seem right'!

    Perhaps some energy does go to the cold reservoir, but I don't know how you would do that. This is a past exam question on the 'easy'/bookwork section of the paper designed to get everyone some marks, so the questions are generally quite simple, so I get the feeling that needing to take the cold reservoir into account would make the question too complicated :confused:
     
  5. May 19, 2009 #4

    Mapes

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    What are [itex]Q_1[/itex] and [itex]Q_2[/itex] in the equation you wrote?
     
  6. May 19, 2009 #5
    I used the temperatures to calculate the efficiency, just wrote down the Q version of the formula because I thought it might be relevant, but am not sure how you can use it considering you know neither of the values. Obviously you can say Q_1/Q_2 = T_1/T_2 = 0.28 though.
     
  7. May 19, 2009 #6

    Andrew Mason

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    Part b) is a rather odd question. The change in entropy of the gas in one cycle is 0 since it returns to its original state after a complete cycle. That occurs whether or not it is a Carnot cycle. So the question must be asking: what is the change in entropy resulting from the absorption of heat from the hot reservoir only, without regard to the rest of the cycle.

    Since a Carnot cycle uses 1. isothermal heat absorption followed by 2. adiabatic expansion, then 3. isothermal heat release and finally 4. adiabatic compression, you are asked only for the change of entropy in 1 (as it is the only part in which heat flowed into the gas). To find that, all you have to know is Qh. You get that from the efficiency and the work output that is given.

    AM
     
  8. May 19, 2009 #7
    What is your h representing? Also though I have Q_1 and Q_2, so which would it be and how would I get it from the efficiency since surely I only know the ratio of the two Q values?

    Thanks both of you for your help so far, by the way.
     
  9. May 19, 2009 #8

    Mapes

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    Definitely.

    You also know the difference between the two values. Of the energy coming in, what isn't sent to the cold reservoir must be output as work.
     
  10. May 19, 2009 #9

    Andrew Mason

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    The efficiency is W/Qh. So you can easily find Qh given efficiency and W. With Qh and T you can work out the entropy. (I am using Qh as the heat flow from the hot reservoir).

    AM
     
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