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Basic question on temperature

  1. May 14, 2013 #1

    fluidistic

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    Today I've got rid of a lot of questions but one remains so far.
    The definition of temperature I've seen in Callen's book is ##T=\left ( \frac{\partial U}{\partial S} \right ) _{V, \{ n_i \} }## (i=1, 2, etc.) where the entropy is defined only for equilibrium states.
    Therefore if I understand well, T is also defined only for equilibrium states. However in the lab. or in everyday's life (meteorogolists), we all use temperature as if it was defined even for systems that are totally off from any thermodynamic equilibrium. How is that possible? Does it even make sense to talk about temperature for systems that aren't in thermodynamics equilibrium, i.e. systems that have not a well defined entropy?
     
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  3. May 14, 2013 #2

    Andy Resnick

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    You are correct- temperature is only defined in terms of systems in equilibrium. Using *local thermodynamic equilibrium* is why we can extrapolate to kinetic systems. Fully dynamic systems that violate LTE (turbulent flow and shock fronts, for example) do not in general have a well-defined temperature or entropy.
     
  4. May 14, 2013 #3

    fluidistic

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    Thank you very much.
    Does this also mean that U(S,V,n) is also defined only for equilibrium states or at a local viewpoint, only when there's no turbulent flow/shock fronts?
     
  5. May 15, 2013 #4

    Andy Resnick

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    Are you defining 'U' as the internal energy, the total energy, or some other quantity? The internal energy density (that is, the total energy less kinetic and potential) is well-defined always via the Reynolds transport equation- as long as the density is well-defined, the internal energy is then well-defined.
     
  6. May 15, 2013 #5

    fluidistic

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    The book defines it as the internal energy. That's the "U" that appears in the dependence of the entropy, S(U,V,n).
    In principle one could always (if I'm not wrong!) get the dependence U(S,V,n) if one has beforehand the relation S(U,V,n). That's why I think that if S isn't well defined for some states, then U(S,V,n) isn't either.
     
  7. May 15, 2013 #6

    Andy Resnick

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    That's a good point, but it requires that S(U,V,n) is invertible, which is not the case in thermodynamics. For example, while dU is an exact differential (dU = dQ - dW), dQ is not- it is path-dependent, and this is why you often see dU = δQ - δW. Thus the expression dS = δQ/T can be somewhat misleading in the context of thermodynamics (as opposed to thermostatics and thermokinetics), since 'T' may not be well defined.

    So, if 'T' is not well defined, the expression dS = δQ/T means that S is not an exact differential either, so S(U,V,n) cannot be inverted to provide U(S,V,n).
     
  8. May 15, 2013 #7

    fluidistic

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    Hmm I'm confused. My book says
    Sligthly before that, he mentions that S exists only for eq. states. This is why I thought that U cannot be defined if S isn't defined. And so the temperature isn't defined either for non equilibrium states.
     
  9. May 16, 2013 #8

    Jano L.

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    No, this is not right. Energy is more basic notion than entropy - it was introduced in mechanics before entropy was. System has energy even in non-equilibrium states by assumption, since we believe in the law of conservation of energy.

    Only for equilibrium states is the energy function of S,V,N.

    Regarding the original question: the temperature in practice means "degree of hotness", measured by some thermometer. We always get some number, even if thermometer is in touch with body in non-equilibrium state.

    Usually, for water in washing machine or air in weather forecast, the meaning of temperature is quite close to the thermodynamic temperature, since the medium in non-equilibrium state can be understood as a collection of small parts that are almost in equilibrium with spatially varying temperature and density of entropy. Then the relation between energy, entropy and temperature may hold approximately for each small enough constituting part, even if it does not hold for the whole system.
     
  10. May 16, 2013 #9

    Andy Resnick

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    The issue is: when is the entropy function continuous, differentiable, and monotonic? For statics and kinetics, the entropy function meets these criteria. For the fully dynamic case, Jou, Casas-Vazquez, and Lebon's "Extended Irreversible Thermodynamics" has an excellent introductory sequence leading up to non-equilibrium equations of state. This paper discusses the idea of 'uncompensated heat' in nonequilibrium processes:

    http://pre.aps.org/pdf/PRE/v51/i1/p768_1 [Broken]

    And this

    http://www.google.com/url?sa=t&rct=...zkufUlZHf0e0euRAPNqkZcw&bvm=bv.46471029,d.aWc

    Also has a useful introduction. AFAIK, there is no clear definition of entropy for far-from-equilibrium systems, or entropy flux relating to far-from-equilibrium processes.
     
    Last edited by a moderator: May 6, 2017
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