Entropy change is defined as $\Delta S = \int \frac{\delta Q_{rev}}{T} $, where $Q_{rev}$ is heat transferred through a reversible process. I’m a little confused how this works, so I have a couple of related questions which I hope will clarify what’s going on.
If a reversible process is defined as $\Delta S = 0$ then how can this integral be non-zero for a reversible process? Or is it that just $\Delta S_{Univ} = 0 = \Delta S_{sys} + \Delta S_{surr}$ and therefore $\Delta S_{sys} = -\Delta S_{surr} \iff T_{sys} = T_{surr} $, in which case
How would one calculate it for an irreversible process, in particular, some heat, Q, transferred across a finite temperature difference? I would try (considering isothermal transfer between infinite temperature reservoirs) $$ \Delta S_{Univ} = Q \big( \frac{1}{T_{sys}} - \frac{1}{T_{surr}} \big) $$ But this is no longer reversible, so I need to do something else?
Since entropy is a state function, is the entropy change through any process that is irreversible, but between equilibrium states, equivalent to $\int \frac{\delta Q}{T}$ through a reversible process?
Is this equivalent to pretending I’m doing 2 reversible transfers to some fictitious bodies, the first taking Q reversibly from the system to the fictitious body, both at $T_{sys}$, and then magically dropping the fictitious body to $T_{surr}$ and transferring the heat Q to the surroundings? If so, why use $\delta Q$ rather than $dQ$? My understanding was that $\delta Q$ was for path-dependent processes, but there is only one fixed (adiabatic) path for reversible transfer?
Sorry for yet another entropy question, thanks in advance!