- For a reversible path between two states (1 and 2), entropy change of a system is NOT zero. It is
$$\Delta S = \int_a^b \frac{dQ}{T}$$
For reversible path between two states, entropy of the universe (Or any isolated system) is zero.
$$\Delta S + \Delta S_\text{surroundings} = 0$$
So You cannot just take any system and say that entropy change between two states for this system will be zero because it is zero for a reversible process. It is not. So when you say
Surely the total change of entropy is zero.
for reversible process of closed system, it is not true.
Answer to This question might help you here.
- As for the first part of your question, I don't understand what the question is. Could you edit it to be more specific?
Also, You said the following, which is false.
The entropy changes of the system are same for both cases, reversible
and irreversible processes because the first and final states are
unchanged. In this situation I think the surrounding also have the
same first and final states for both reversible and irreversible
processes.
We don't know whether surrounding has same first and final states or not. We only know about the system's first and final states. Think about it this way: In a reversible process, system is going from state A to B, and so is surrounding. Since it is reversible, $ \Delta S_{System} = - \Delta S_{Surrounding} $. So ultimately, $ \Delta S_{Universe} = 0$.
Now for an irreversible process, we know that through this irreversible path, the System goes from A to B. We don't know about surroundings. Now, since system's states are same, $ \Delta S_{System} $ will be same as above case. For the surrounding, you say that states are same as the reversible case. But then, here also $ \Delta S_{Surrounding} $ would be same as before and again $ \Delta S_{Universe} = 0$. But we know that that is not true for irreversible process. Hence, Surrounding's states must be different. So, in irreversible process, while the system goes A to B same as before, the surrounding must go from A to some C. There is no reason to believe that it would go from A to B again.
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The formula you have written applies to a system, with $S_{gen}$ being the entropy generated in the interior of the system, and $\int~dQ/T$ being the entropy flux across the system's boundary. To apply this equation to the example of heat exchange $Q$ between two heat reservoirs at temperatures $T_1$ and $T_2$, with $T_1>T_2$, you must first define your system. If you take the hot reservoir or cold reservoir as your system, then indeed there is no entropy generated within their interior, and only flux term remains (equal to $-Q/T_1$ or $+Q/T_2$ depending on your choice of system). However if you take both of them together as a system, then flux term will become zero (because heat flux is then internal) while entropy generation term will be non-zero (equal to $Q/T_2-Q/T_1$).
For any process to be labeled as irreversible, entropy change not of any particular system but that of the entire universe (system+surroundings) must be considered. Obviously, for the universe flux terms are always zero (there is nothing else to exchange heat with), and any increase in entropy is always $S_{gen}$.
P.S. Any isolated system is a universe unto itself.
Best Answer
In your case, heat conduction is considered to be quasi-static for the subsystems. Let $\Delta Q$ be positive number, energy transferred as heat. Then, change of entropy $S_2$ of the subsystem 2 is $\Delta Q/T_2$, change of entropy $S_1$ of the subsystem 1 is $-\Delta Q/T_1$. For the whole system containing both subsystems, the change of entropy is
$$ \Delta S = \frac{\Delta Q}{T_2} + \frac{- \Delta Q}{T_1}. $$
This is greater than 0, so $\Delta S > 0$. This is consistent with the fact that even quasi-static heat conduction, if it is across finite temp difference, is a non-equilibrium process.
There is single $T$ in that statement. This is meant to be temperature of reservoir the system is connected to.
We can apply it, say, for subsystem 2. The temperature of the reservoir connected is $T_1$ (there are no other systems considered). The statement then reads
$$ \Delta S_2 \geq \frac{\Delta Q}{T_1}. $$ So $\Delta S_2 = \Delta Q/T_2$ and the above non-equality hold at the same time, because different temperatures are to be used.