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Changes in entropy of an isolated system

  1. Nov 24, 2013 #1

    CAF123

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    1. The problem statement, all variables and given/known data
    Consider two fixed volume bricks of mass m1=2kg and m2=1kg with initial temperatures T1=400K and T2=100K. They are enclosed in a system thermally isolated from the surroundings and are made from a material with a heat capacity cv = 1kJ/kg/K.

    A) In Process 1, the bricks are brought into thermal contact. What is their final temperature once they reached thermal equilibrium with each other?
    B)In Process 2, the initial conditions are the same, but the maximum amount of work is extracted from the system (e.g via a heat engine). What is the final temperature in this case?
    C)Find the change in entropy of processes 1 and 2.
    D)Find the amount of work extracted in Process 2.

    2. Relevant equations
    Q = cvmΔT
    ΔS ≥ 0 for an isolated system.

    3. The attempt at a solution
    A) is fine, the final temperature should be between 100 and 400K. I get 300K which is reasonable.

    B)If the max amount of work is extracted, then the heat gained by brick 2, Q2= Q1 - W. I don't see how to proceed here unless I write an explicit expression for W (but that comes in D)). I did find an expression though, but when I use it, it yields a negative temperature.

    C)Entropy lost by brick 1 = entropy gained by brick 2 since system thermally isolated. Hence net change in entropy in process A is zero.

    D)For a thermally isolated system, ΔS ≥ 0. The maximum amount of work is extracted from a Carnot engine, which is reversible, so the equality applies. Therefore, ΔS2-ΔS1 = 0 = Q2/T2-Q1/T1=0 = (Q1-W)/T2 - Q1/T1=0. Rearrange for W gives Q1(1-T2/T1). Is that correct?

    Many thanks for any guidance.
     
  2. jcsd
  3. Nov 24, 2013 #2
    In B), max energy is extracted when the process is reversible, i.e., total change in entropy is zero. You need to write the equations for the entropy change of each brick, equate their sum with zero, and find the final temperature from that.

    In C), obviously process 1 is not reversible, so it cannot have the total change in entropy at zero. The method from B) could be adapted here.

    In D), you are not using the correct equations. By definition, ## T = dQ / dS ##. Note the differentials.
     
  4. Nov 24, 2013 #3

    CAF123

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    Hi voko,
    Why is that so?

    ΔS2 = Q2/(Tf-T2), where Q2 is the heat that enters brick 2. Similarly, ΔS1 = -Q1/(Tf-T1), where -Q1 is the heat lost by brick 1. I can also write Q2 = cvm2(Tf-T2) and Q1=cvm1(Tf-T1)
     
  5. Nov 24, 2013 #4
    This is what Carnot's theorem is about.

    Same mistake as before. You are using finite quantities, while you should be using differentials.
     
  6. Nov 25, 2013 #5

    CAF123

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    So like ##dS_1 = dq_1/dT_1## and ##dS_2 = -dq_2/dT_2##. Then sum and equate to zero for a reversible process => ##dq_1/dT_1 = dq_2/dT_2 = (c_v m_1 dT_1)/dT_1 = (c_v m_2 dT_2)/dT_2##. It looks like I can cancel dT's, but this cannot be right because I end up with a contradiction.
     
  7. Nov 25, 2013 #6
    Not sure why, but you are making the whole thng a lot more complex than it needs to be by inventing equations :) The correct formula is ## T = dQ/dS ##, as I said in #2. Notice the placement of ##d##'s and the minus signs.
     
  8. Nov 25, 2013 #7

    CAF123

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    I think I see what you mean. T here is the temperature of the external source of heat. So, the entropy change of brick 2, ##\Delta S_2 = c_v m_2 \ln \left(T_f/T_2\right)## and for brick 1, ##\Delta S_1 = -c_v m_1 \ln(T_f/T_1)## after integration. When I sum these and equate to zero, I end up with ##T_f = 1600K.## Is that right? How can I tell?
     
  9. Nov 25, 2013 #8
    You are still imaginative about your equations :) Where does that minus sign come from?

    This is manifestly wrong. The temperature is higher than the bricks had originally.
     
  10. Nov 25, 2013 #9

    CAF123

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    ##dS_2 = dQ_2/T = c_vm_2dT/T \Rightarrow \Delta S_2 = c_v m_2 \ln (T_f/T_2)##. Similarly for brick 1, and the negative is there for brick 1 because it's entropy decreases since heat flows from it to brick 2.
     
  11. Nov 25, 2013 #10
    Thermodynamic temperature is defined by ## T = dQ / dS ##. The definition is not dependent on the hand-waving of the kind "the heat flows from X to Y". ## Q ## is the heat of the body in question, ## S ## is the entropy of the body as well. The definition is independent of the state of other bodies, and, as a useful definition, it had better be.
     
  12. Nov 25, 2013 #11

    D H

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    You should immediately know that that is wrong, without asking! The only way the final temperature can be greater than that of the hotter brick is if work is performed on the bricks. Here we're using the temperature difference to extract work from the bricks.


    No! You are over complicating things and making up equations. Don't do that! If ##T_f < T_i##, then ##\ln ( T_f/T_i) ## will be negative. There's no reason for you to add a negative sign. All that adding that negative sign accomplished for you was to invalidate the expression.
     
  13. Nov 25, 2013 #12

    CAF123

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    The fact that the entropy of brick 1 decreases is already encoded in the reasons you gave. With this, I end up with ##T_f \approx 252 K##. This seems more reasonable, yes? Energy is extracted from the system, so it is sensible for it to be lower than the 300K obtained in A).
     
  14. Nov 25, 2013 #13
    Hi CAF123. What DH and Voko are saying is "lose the minus sign in the expression for ΔS1." Then you'll be OK.
     
  15. Nov 25, 2013 #14

    CAF123

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    Hi Chestermiller,
    That's what I did and I end up with about 252K. I do not see, however, why what I did qualified as 'making up equations'.
     
  16. Nov 25, 2013 #15

    D H

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    Very reasonable, and also correct. Very good.
     
  17. Nov 25, 2013 #16

    CAF123

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    Thank you D H. For maximum amount of work to be extracted, the engine is operating at Carnot efficiency, and this is a reversible system. For reversible processes, the change in entropy of the system is zero. Hence the change in entropy of process B is 0.

    In process A, I calculated that the change in entropy was about 0.8 kJ K-1, since ΔS2 = ln(3) kJ K-1 and ΔS1 = -0.288 kJ K-1. For an isolated system, ΔSsystem + ΔSsurr ≥ 0. The second term is zero, since the system is isolated, so the change in entropy of the universe is due to the changes in entropy of the system.

    The work extracted, by the first law is W = Q1-Q2. Q1 is the heat flowing from brick 1, which (I think) I may write as Q1 = -cvm1(Tf-T1) and Q2 = cvm2(Tf-T2). This gives me W approx. 448kJ. Is it okay?
     
  18. Nov 25, 2013 #17
    I don't obtain these results. For ΔS1, you forgot to multiply by the mass. For the work, I get Q1+Q2, with Q1=296 kJ and Q2=-152 kJ, where these Q's apply to the engine, and not to the masses.
     
  19. Nov 25, 2013 #18

    CAF123

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    I also made a negative error in my expression for Q2 when I calculated the temperature difference, I get W = 144J now with this correction.
     
  20. Nov 25, 2013 #19
    Yes. That matches what I got.

    Chet
     
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