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$\int \dfrac{\delta Q}{T}$ can't be used to calculate entropy of an irreversible process. If you happen to know heat supplied and temperature at which it is supplied for just an instant. Can you then at least write change in entropy at that instant for an irreversible process as $$\dfrac{\text{Heat supplied}}{ \text{Temperature at which transfer occurs. }}$$

I have seen in lot of books where they give the above formula without talking about reversible or irreversible process. Just confused.

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  • $\begingroup$ See section 4 of "Thermodynamics in Terms of Gradient Vectors" av8n.com/physics/thermo-forms.htm#sec-extirpate . The first and second laws of thermodynamics are both expressed by equalities by giving due recognition to the irreversible production of entropy. $\endgroup$
    – Quillo
    Mar 21, 2023 at 0:37

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You can always say that $\Delta S \ge \frac {Q}{T}$ irrespective of whether $Q$ is transferred reversibly or not, but for equality to hold you need a reversible process. If the process is irreversible then $\Delta S \gt \frac {Q}{T}$. In fact, the difference $\Delta S - \frac {Q}{T}$ is characteristic of the dissipation taking place in the system. These classic publications explain how this is done:

(1) Tolman & Fine:"On the Irreversible Production of Entropy", REVIEWS OF MODERN PHYSICS VOLUME 20, NUMBER 1 JANUARY, 1948

(2) Bridgman: "The Thermodynamics of Plastic Deformation and Generalized Entropy", REVIEWS OF MODERN PHYSICS VOLUME 22. NUMBER 1 JANUARY, 1950.

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