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Change in Entropy, double integral?

  1. Mar 9, 2015 #1
    1. The problem statement, all variables and given/known data
    A thermally conducting, uniform and homogeneous bar of length L, cross section A, density p
    and specific heat at constant pressure cp is brought to a nonuniform temperature distribution by contact at one end with a hot reservoir at a temperature TH and at the other end with a cold reservoir at
    a temperature TC. The bar is removed from the reservoirs, thermally insulated and kept at constant pressure. Show the change in entropy is

    ΔS = Cp ( 1 + ln(Tf) + (Tc/(Th-Tc))lnTc - (Th/(Th-Tc))lnTh )

    3. The attempt at a solution

    I am sorry in advance for my very poor english, but I am studying entropy more in-depth, however I cannot seem to understand where the double integral arises in this problem and if it can be avoided. I have done the problem, and my set-up is identical except I used a single integral only.

    http://www.scribd.com/doc/12398371/Problem-Solution-Thermodynamics#scribd

    The solution is here on page 50, or it might be page 61,62 for some of you.

    Plain and simple, why do I need to integrate over the rod's length as well (dx) ?
     
  2. jcsd
  3. Mar 9, 2015 #2
    This problem statement is very confusing. Are you trying to calculate the change in entropy of the bar? But, in the initial state, the temperature profile is non-uniform, so it is not a thermodynamic equilibrium state. Is the assumption being made that it is possible to calculate the entropy change from a non-equilibrium state to an equilibrium state by assuming that, at each location of the bar, thermodynamic equilibrium exists approximately? Also, what is Tf? Also, I can't get access to the book. Also, please show us what you have done so far.

    Chet
     
  4. Mar 9, 2015 #3

    haruspex

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    If the whole bar were changing from one uniform temperature to another, you would integrate to find the entropy change, yes? Since in this case you effectively have many short bars, length dx each, doing that, you need to integrate wrt x as well.
     
  5. Mar 9, 2015 #4
    Hi haruspex,

    What are your thoughts about the validity of determining the change in entropy between a non-equilibrium thermodynamic state and an equilibrium thermodynamic state?

    Chet
     
  6. Mar 9, 2015 #5

    haruspex

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    Is it not valid to treat it as a number of independent short elements, each going from one given temperature to another?
     
  7. Mar 10, 2015 #6
    There is no question that the initial state of the system is not in a thermodynamic equilibrium state. So, in doing this, we need to consciously make the assumption/approximation of "local thermodynamic equilibrium." As a person trained in chemical engineering, we were never taught that there was even any kind of approximation involved, and we routinely applied this approach to determine state functions such as internal energy, enthalpy, and entropy locally. And then we routinely integrated these spatially to get the overall values of the functions for the system. All transport processes involving heat transfer, momentum transfer, and mass transfer are tacitly based on making this approximation. So, personally, I have no qualms whatsoever about applying the assumption of local thermodynamic equilibrium to a system.

    However, after arriving at Physics Forums, I was taken to task by many physicists who informed me that it is not valid to make this approximation, and that, if a total system is not in thermodynamic equilibrium, all bets are off. So, according to them, all that I had learned about momentum transport, heat transport, and mass transport could not be applied to real world systems, and that, if I did so, I would likely be getting the wrong answer. So all those those processes and equipment worldwide that have been designed and operated successfully for well over a century based on this assumption should not be operating successfully, even though they are. (Can you tell that this is a pet peeve of mine?)

    I just wanted to see what you, as someone I hold in high regard, thought about all this.

    Chet
     
  8. Mar 10, 2015 #7

    haruspex

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    Chet, I would defer to you on such matters any day, but I must say I find this objection incomprehensible. It seems to imply that entropy is not additive.
    Did your admonishers provide any references?
     
  9. Mar 10, 2015 #8
    Basically, their argument was that the thermodynamic state functions are only applicable to a system at thermodynamic equilibrium. I think their point was that the distribution of quantum mechanical energy states is different for a system that is not at thermodynamic equilibrium from one that is. So, if there is a temperature gradient in the system, for example, even locally, the distribution of energy states will be perturbed. The local thermodynamic equilibrium approximation assumes that this effect is negligible. As I said, I have no problems with this approximation, because experience has shown that, in practice, it is valid.

    Chet
     
  10. Mar 10, 2015 #9

    haruspex

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    Hmmm... A standard entropy question would involve transfer of ##\Delta Q## between two bodies at different temperatures. The system is therefore not in thermodynamic equilibrium. I just don't get how this is different from a temperature gradient within a body.
     
  11. Mar 10, 2015 #10
    There is a big difference. The determination of the change in entropy of a system between two thermodynamic equilibrium states involves the integral of dq/T over a reversible path between the two states. This will not necessarily be the actual path that the body experiences (unless that process was reversible to begin with).

    When two bodies at different temperatures are allowed to irreversibly equilibrate with one another, the entropy change of each body has to be calculated separately by evaluating the integral of dq/T over a reversible path between the initial and final equilibrium states of that body. Neither of these paths will not be the same as the actual path for the equilibration process.

    Chet
     
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