I have here a demonstration of heat transfer direction from my course

Suppose that we have to body (H) and (C) in a isolated system:

$$\Delta U = Q_C + Q_H = 0 \Longleftrightarrow Q_C = -Q_H$$

$$\Delta S_{sys} = \Delta S_{C} + \Delta S_{H} = S_{creation}$$

$$\Delta S_{C} + \Delta S_{H} > 0$$

$$\frac{Q_H}{T_H} + \frac{Q_C}{T_C} = Q_H(\frac{1}{T_H}-\frac{1}{T_C}) > 0$$

First line is first principle, second line is second principle in an isolated system (no exchange term) but in my course it's written that we admit a globally irreversible process while we are ignoring the creation term from each system from equation 3 to 4

How the process can be globally irreversible if the two only process happening are reversible ?

  • $\begingroup$ Your second through fourth equations only apply to an irreversible process. They equal zero for a reversible process. What is it you don’t understand? $\endgroup$
    – Bob D
    Feb 11, 2020 at 20:46

1 Answer 1


In this case, the irreversibility (entropy generation) occurs within the thin heat conductive region (i.e., interface) between the two reservoirs, through which the heat transfer takes place. This interface is assumed to have negligible heat capacity so that, in the end, all the generated entropy is transferred to the combination of the two reservoirs. In short, all the entropy generation takes place at the interface

  • $\begingroup$ Thank you very much ! $\endgroup$
    – Okano
    Feb 11, 2020 at 21:33

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