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In Zemansky and Dittman's Heat and Thermodynamics (7th ed., p. 208) change of entropy is computed along a reversible path connecting an initial non-equilibrium state and a final equilibrium state. The example quoted is of a rod heated at one end, with a final equilibrium state having a constant temperature profile.

Details: The rod is initially kept in touch with a high and low temperature reservoir at the two ends, assuming a linear temperature distribution. This is not an equilibrium state. The rod is then removed from the reservoirs, and the final equilibrium state is the average of the temperature of the high and low temperature reservoirs. The authors consider the rod to be composed of thin slices each of which has a (different) initial temperature and the same final temperature. They assume a reversible isobaric process for each slice and find the change in entropy by integrating over one volume element (in contact with a series of reservoirs from initial to final temperature) and then a second integration over the whole volume to get the net change in entropy for the system.

Can we use the same technique for a problem wherein both the initial and final states are not in equilibrium, by replacing the process with a reversible path ?

An earlier query, does not address this issue: Entropy and reversible paths

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  • $\begingroup$ If, in the example, the heated rod is initially all at uniform temperature (prior to applying the heating), then this is an equilibrium state. $\endgroup$ – Chet Miller Nov 24 '19 at 13:48
  • $\begingroup$ What was their answer for the entropy change, and how do you know they did not use a reversible path? $\endgroup$ – Chet Miller Nov 24 '19 at 14:27
  • $\begingroup$ Please find updated details in the question. Thank you. $\endgroup$ – Frost Nov 24 '19 at 17:47
  • $\begingroup$ It depends on whether you are willing to accept the notion that the entropy (and other thermodynamic functions) can be determined for non-equilibrium states of a system by integrating the local entropy per unit volume over the volume of the system, where the local entropy is based on the local temperature and specific volume (for an equilibrium state). I am comfortable with doing this, but some physicists are not. How about you? $\endgroup$ – Chet Miller Nov 24 '19 at 20:03
  • $\begingroup$ Do the authors imply that any non-equilibrium state problem may be replaced with a reversible path (an equilibrium state) problem ? What prevents us from using such a technique for a system which is always in a non-equilibrium state, such as a forced turbulent fluid ? I feel that such a method is valid provided the time-scale of the experiment is much larger than the time-scale of the intrinsic dynamics (system is quasi-static). $\endgroup$ – Frost Nov 25 '19 at 3:18

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