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Entropy For Irrevesible Process

  1. May 12, 2013 #1

    morrobay

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    1. The problem statement, all variables and given/known data
    Two 100g Aluminum cubes with T1 = 80° C and T2 = 20°C
    Specific heat of Al is .22 cal/g/°C. The cubes are placed in contact and in short period of time
    a quantity of heat, dQ , is transferred from T1 cube to T2 cube.
    The cubes temperatures now are 75°C and 25°C. After a little math : dQ = 640 Joules
    The total entropy of the two cubes before dQ was: 22.3 J/K and after dQ was 27.5 J/K

    2. Relevant equations
    S2-S1 > ∫12 dQ/T
    ΔS = 5.2 J/K > ∫12 dQ/T This should be evaluated :
    ln dQ/T - ln dQ/T . The question on this is what are the T values here ?
    Is it necessary to do two evaluations, one for hotter object and another for cooler object ?
    ,




    3. The attempt at a solution
    S2-S1 = 5.2J/K
    For hotter object ln 460 J/348K - ln 460J/353K = .01
    For cooler object ln 460J/293K - ln 460J/298K = .02
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. May 12, 2013 #2
    Throughout the heat interchange, the temperature at the interface between the two cubes was 50C = 323 K. For an irreversible processes, if you are evaluating dQ/T for testing the Clausius inequality, you use the temperature at the boundary (interface). What was the integral of dQ/T for the hotter body? What was the integral of dQ/T for the colder body? What was the sum of these? How does that compare with the actual entropy change?
     
  4. May 12, 2013 #3

    morrobay

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    In the attempt of solution above is the evaluation of integral for hotter and cooler object correct ?
    Total entropy change was + 5.2J/K
    dQ/T for 323 K = 640 J/323K = 1.98 J/K

    The inequality certainly holds , my question is whether I evaluated the two integrals correctly
    with temperatures before and after dQ for both objects.
    S2-S1 = 5.2J/K > 1.98 J/K (.03 )
     
    Last edited: May 12, 2013
  5. May 12, 2013 #4
    No. For the hotter body, the dQ/T is -640/323 = -1.98. For the colder body dQ/T is +640/323 = +1.98. So, the total dQ/T for both bodies is zero. This is the value that you are supposed to compare with the sum of the actual entropy changes for the two bodies, which is

    [tex]ΔS=mC_v(\ln (348/353)+\ln (298/293))[/tex]
     
  6. May 13, 2013 #5

    morrobay

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    With m = mass = 200g and Cv= specific heat Al = .22cal/g/C
    ΔS = 200g*.92J/g/C * ln 348/353 + ln 298/293 = .48J/K
    Good news , with math correction total entropy change for both objects S2 - S1 =
    27.54J/K - 27.11J/k = .43J/K This value ( see work below ) is very close to above value
    arrived at independently

    Example for entropy , Q/T, for 100g object at 80 C , specific heat = .92J/g/C before dQ , S1
    100g*80 C * .92J/g/C = 7360 J/353K = 20.84J/K + 6.27J/K for 20c object = 27.11J/K
     
    Last edited: May 13, 2013
  7. May 13, 2013 #6
    ΔShot=92ln(348/353)=-1.05 J/K

    ΔScold=92ln(298/293)=+1.55 J/K

    (dQ/T)hot=-(92)(5)/323=-1.43 J/K

    (dQ/T)cold=+(92)(5)/323=+1.43 J/K

    The inequality must be satisfied not only for the system of two masses as a whole, but also for each mass individually. For the system as a whole,

    ΔShot+ΔScold=-1.05 +1.55=0.50 J/K

    (dQ/T)hot+(dQ/T)cold=-1.43+1.43=0 J/K

    For the hot mass:
    -1.05>-1.43

    For the cold mass:
    1.55>+1.43
     
  8. May 15, 2013 #7

    DrDu

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    As I said in the other forum, you are generally right that T may not even be defined during an irreversible process. I think here you are supposed to assume that the thermal bridge between the two cubes has much lower heat conductance than the aluminium blocks. Then the temperature within the blocks is homogeneous and you can relate it to dQ=C_v dT.
     
  9. May 15, 2013 #8
    Not really. If the interface is infinitely conductive, then, by symmetry, the temperature at the interface would have to be 50 C throughout the heat exchange in this problem. If you could not be sure what the temperature at the interface would be in any irreversible process, then the temperature in Clausius' inequality would be meaningless, except for a reversible process, in which case the equality would apply.

    Chet
     
  10. May 15, 2013 #9

    Andrew Mason

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    I have not seen the Clausius inequality used in this way. The Clausius inequality relates to a cyclical process.

    Since entropy is a state function, when the system returns to its original state there is no change in entropy: ΔSsys = ∫dQrev/T = 0.

    However, if the cycle is not reversible ∫dQ/T ≤ 0 over the entire cycle where dQ is the actual heat flow into/out of the system (ie. this is strictly a path function, not a state function). This is because for a given amount of heat flow into the system, less work is done than in the reversible case so there is more heat flow out of the system to the cold reservoir than in the reversible case. This makes ∫dQ/T < ∫dQrev/T.

    AM
     
    Last edited: May 15, 2013
  11. May 15, 2013 #10
    The form of the Clausius inequality that I am used to is
    [tex]dS≥dQ/T[/tex]
    There have been numerous recent references to this form of the Clausius inequality on PF.

    Chet
     
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