Suppose we have a system (isolated) which has two partitions. Initially the one partition has temperature $T_1$ and the other has temperature $T_2$ with $T_2 \neq T_1$. We know that in the new equilibrium state the temperature on both partitions must be the same.
Can the second law of thermodynamisc predict that? The Second Law of Thermodynamics it is usually stated as:
For an isolated system, $ΔS \geq 0$.
From the above statetement it is not clear why the entropy must increase. I mean if the temperature stays the same $T_1$ and $T_2$ for the two partitions then the entropy won't change and second law isn't violated. Is there something more fundamental that says that the entropy must increase?
Should the second law be formulated different? I gave that example because this is how the second law is introduced in introductory courses.
After reading the answers and the comments I think that I should add the following. We can show using Lagrange multipliers that at equilibrium the temperatures on the partitions (unconstrained) must be the same. We also know that eventually the system will reach an equilibrium state. Now we can deduce that the temperature must be the same. But from second lone alone we can't.
If we consider the initial state before making the partition permeable to heat and the final state where the temperatures are equal and we measure the change in entropy we will find that is greater than zero. But just from the requirement $ΔS \geq 0$ we can't find/predict the final state. So what is so special about $ΔS \geq 0$ if it can't predict the new state?