Entropy change during irreversible process

Question

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.

1. 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

2. 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?

3. 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?

4. 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!

Answer 1

The answers to your first and third questions are "yes." Here is a tutorial on how to calculate the change in entropy of the system or the surroundings for an irreversible process: https://www.physicsforums.com/insights/grandpa-chets-entropy-recipe/. The reference also provides worked examples. Note that, in considering entropy changes for both the system and the surroundings from an irreversible process, one needs to first separate the system from the surroundings and then subject each of them separately to a reversible path. These separate reversible paths do not have to match.

It simplifies things a little to think about how entropy of a system can change. There are only two ways: entropy transfer across the boundary of the system by heat flow in and out at the boundary temperature, and entropy generation (due to irreversibility) within the system. In a reversible process, only the first mechanism is present. In both cases, the transfer across the boundary is $\int{dq/T_B}$ where $T_B$ is the boundary temperature.

Regarding questions 2 and 4 (involving ideal constant temperature reservoirs), these are somewhat limiting cases. Here, if you imagine a very thin conductive slab between the two reservoirs (with high temperature at one face of the slab and low temperature at the other face), then for the irreversible process involving the transfer of heat between the reservoirs, all the entropy generation would take place within the intervening slab, while the reservoirs would each separately experience only entropy transfer with the slab. In the limit where the slab becomes infinitesimally thin, we would obtain the system you have envisioned, but all the entropy would still have been generated at the interface, and each reservoir would only experience entropy exchange with the interface.