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Thermodynamics (please state working)

  1. May 11, 2009 #1
    Answers are there- but explanation of working out questions would be great.

    A model Stirling engine uses n=7.16*10^-3 mol of gas (assumed to be ideal) as a working substance. It operates between a high temperature reservoir at T(H)=93.0 degrees Celsius and a low temperature reservoir at T(C)=23.0 degrees Celius. The volume of its working substance doubles during each expansion stroke. It runs at a rate of 0.8 cycles per second. Assume the engine is ideal.

    (i) how much heat is transferred to the gas from the high temperature reservoir during each cycle? (ans - 15.1J)

    I've already worked out that the work done by the engine per cycle is -2.89J; and the power of the engine is 2.31W (if that helps)
     
  2. jcsd
  3. May 11, 2009 #2
    Well, think of it this way... The heat being transferred from the high temp reservoir only happens during the high temp process. They aren't asking for the total heat transferred for the whole cycle.
     
  4. Apr 6, 2010 #3
    So, do we need to use the formula [tex]\Delta E_{int}=Q+W[/tex] to find the heat transferred to the gas from the high temperature reservoir during each cycle?

    The problem is I can't use this formula because the internal energy [tex]\Delta E_{int}[/tex], is not given to us! So how can I find the amount of energy Q?

    Edit: I also know that [tex]W_{eng}=Q_{net}=| Q_h | - | Q_c |[/tex] but I still don't know how to obtain Qh and Qc...
     
    Last edited: Apr 6, 2010
  5. Apr 6, 2010 #4
    The heat transferred during the high temp part is just an isothermal expansion. So what is the work done during an isothermal expansion? Since the temperature is constant, the internal energy will not change, so then you will have:

    [tex]Q=-W[/tex]

    And you will have your heat transfer.
     
  6. Apr 6, 2010 #5
    I'm a little bit confused here :confused:

    The question asks "how much heat is transferred to the gas from the high temperature reservoir during each cycle?". So the question asks for the value of Q, right?

    Yes, I know that in an isothermal process the energy transfer Q must be equal to the negative of the work done on the gas [tex]Q=-W[/tex]. Since we already know that the work done by the engine per cycle is -2.89J, Q will be

    Q=-(-2.89)=2.89

    But this is wrong since the correct answer has to be 15.1J. :confused:
     
  7. Apr 6, 2010 #6
    The whole cycle isn't one giant isothermal expansion, so you can't say the internal energy is constant through out the whole cycle. Only when heat is being transferred from the high temp reservoir do we care about work and heat flow. That means we only want the work done by the system during this specific part of the cycle.

    So the work done by the engine for the full cycle won't help us. You will need to compute the work done by the system just during this isothermal process when it is in contact with the high temp reservoir.
     
  8. Apr 7, 2010 #7
    I see. The internal energy is not constant but I know it equals zero, since it is a cyclic process (i.e. starts and ends at the same state).

    So we want to find the value for Qh (energy taken in from a hot reservoir). But I don't know how to obtain it from the formula:

    [tex]W=|Q_h|-|Q_c|[/tex]

    What value do I substitute for Qc?

    Furthermore I don't know how to find the work done by the engine during the isothermal process. In my previous post I used the formula W=-Q but apparently it is wrong...
     
  9. Apr 7, 2010 #8
    [tex]Q_h[/tex] is just the work done during the isothermal process. And your previous post was wrong because you had the wrong work. Again, the work during the isothermal process is not the work done for the whole cycle. Find only the work done during the isothermal process and you will have the answer.
     
  10. Apr 7, 2010 #9
    What formula should I use??

    I'm frustrated because I tried the following to find the work done during the isothermal process only, but again my answer is wrong:

    [tex]W=nRT ln \left( \frac{V_i}{V_f} \right)[/tex]

    [tex](7.16 \times 10^{-3})(8.314) (93) ln \left( \frac{1}{2} \right)=-3.83[/tex]

    [tex]Q=-W=3.83[/tex]
     
  11. Apr 7, 2010 #10
    You have the wrong temperature. You should use Kelvin and not Celsius.
     
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