The change of entropy is defined $$\Delta S = \int \frac{dQ_\mathrm{rev}}{T}.$$ If a system is isolated the heat transfer between the system and the surroundings is zero ($dQ = 0$), thus $\Delta S = 0$.
However, it is commonly stated that the entropy of an isolated system can increase. How is this possible, given the above definition of entropy?
Answer
The definition of entropy $$dS = \frac{\delta Q}{T}$$ only applies for reversible processes. For every irreversible process, $$dS > \frac{\delta Q}{T}.$$ Therefore, if the sytem is isolated ($\delta Q = 0$), and an irreversible process occurs, $dS > 0$.
Simple irreversible processes include friction, mixing, and heat transfer accross a finite temperature difference.
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