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Confusion about [itex]T[/itex] in the definition of entropy

  1. Jan 28, 2016 #1
    In the derivation of the Clausius inequality, [itex]T[/itex] is the temperature of the reservoir at that point in the cycle, but in the definition of entropy it becomes the temperature of the system. This seems to work for a Carnot cycle, where the two are the same, but for other processes, such as an object with constant heat capacity [itex]C[/itex] at temperature [itex]T_0[/itex] cooling due to heat exchange with a resevoir at temperature [itex]T < T_0 [/itex], is where I start to get confused.

    In that case we calculate the system's entropy change to be [itex] C \ln (T/T_0)[/itex] and the reservoir's as [itex]C(T_0-T)/T[/itex].

    In fact, I guess we have to come up with a reversible process connecting the two states. What could that be?
     
  2. jcsd
  3. Jan 28, 2016 #2
    To get the entropy change, the first thing you do is to totally forget about the actual process path, and then devise a process path between the initial and final states that is reversible. You then calculate the integral of dq/T for that reversible path.

    For more on your question regarding the correct temperature to use when applying the Clausius Inequality, see this recent thread (particularly the last few responses): https://www.physicsforums.com/threa...reversible-process.797091/page-2#post-5355365

    Also see my Physics Forums Insights article on entropy and the second law: https://www.physicsforums.com/insights/understanding-entropy-2nd-law-thermodynamics/

    Chet
     
  4. Jan 29, 2016 #3
    Hi Chet,

    I did see that thread after I posted this one which answered my question. Thanks for that.

    It has left me wondering though, when bringing two bodies together how do we know what the temperature is immediately after at the boundary? Since temperature is continuous I assume that the final temperature is the obvious choice, but I'm not sure.

    I was also not entirely sure what the reversible path for the heat exchange was.
     
  5. Jan 29, 2016 #4
    It will not always be the final temperature, and, in general, both the heat flux and the temperature at the boundary will be functions of time. To find the temperature at the interface as a function of time, you need to solve a transient (non-steady-state) heat conduction problem. Methods of doing this are covered in Heat Transfer courses, and a great reference for conductive heat transfer is Conduction of Heat in Solids by Carslaw and Jaeger.

    For the particular case we considered in the other thread, the two cubes are identical in terms of both size and properties. In this case, the interface temperature will be the average of the two initial values and will stay at that temperature during then entire irreversible transfer of heat.
    The heat exchange can be carried out reversibly for each body separately by contacting that body with a continuous sequence of constant temperature baths, each of which differs only slightly from the average temperature of that body (at any time). This will result in the same amount of heat being transferred to each body as in the actual process, but each of the bodies will experience the change reversibly. In this case, the temperature at the boundary of each body will not jump initially to the average initial temperature of the two bodies.
     
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