# On the predictive character of second law of thermodynamics

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.

Edit

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?

I checked also these links Formulation1 and Formulation2.

The second law states something true and useful to know. You are correct to say that it does not, in and of itself, assert that entropy (of an isolated system) has to increase, and indeed this is the case: the entropy could stay constant. However the behaviour is like that of a ratchet. If, for some reason or other, the entropy increases a bit, then it can't un-increase. The system state has moved inexorably from one surface in state-space to another. So if any internal change disturbs the system, in a general sense---makes it explore some of the available states near to the one it is in---then the ratchet-like effect makes permanent any development towards higher entropy.

In the scenario of the temperatures $$T_1$$ and $$T_2$$ remaining unchanging, you must have surrounded each of the two sides of the system with a passive or active perfect thermal insulator. The Second Law is fine with this.

I don't think you'll find a passive perfect thermal insulator, but you're welcome to try. All the passive materials we know of conduct heat, and even a vacuum can transmit heat through radiation.

I have an active perfect thermal insulator next to my desk: through an enclosure and a solid-state mechanism, my Peltier cooler fridge continuously maintains a colder temperature than its environment. But I have to sacrifice your "isolated" requirement here; the system generates a little entropy and ejects it to the environment.

The equality applies to the change in entropy between two thermodynamic equilibrium states of the system. Once you remove the insulating barrier between the two chambers, the system is no longer in a thermodynamic equilibrium state. If allowed to evolve spontaneously, it will result in both temperatures finally being equal. In that state, the entropy of the system will be higher than the original state when the barrier was present. I don't see any contradiction here.

• I don't think the OP was saying there are any contradictions (if I'm missing something in the post let me know). They are asking why entropy has to increase. $\Delta S=0$ even for two systems not at equilibrium would not violate $\Delta S\geq0$ – BioPhysicist May 28 at 18:27
• @BioPhysicist The inequality only applies to a change between thermodynamic equilibrium states. – Chet Miller May 28 at 18:46
• I'm not the one who needs to know that ;) – BioPhysicist May 28 at 21:40

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 predicts that if the two partitions come into thermal contact with one another there will be energy transfer from the higher temperature partition to the lower temperature partition in the form of heat. If the partitions remain in thermal contact long enough they would typically come to thermal equilibrium with one another.

The Second Law of Thermodynamics it is usually stated as:

For an isolated system, $$ΔS \geq 0$$.

From the above statement 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?

The second law does not say the entropy must increase in an isolated system. The entropy will not increase if the partitions remain insulated from one another since they will each individually remain in internal equilibrium and the equality $$\Delta S=0$$ will apply.

However, if they are brought into thermal contact the heat transfer that occurs would be spontaneous because $$T_{1}\ne T_{2}$$, making the transfer irreversible and generating entropy. The increase in entropy of the lower temperature partition will be greater than the decrease in entropy of the higher temperature partition and the inequality $$\Delta S>0$$ will apply.

Hope this helps.

• However, if they are brought into thermal contact the heat transfer that occurs would be spontaneous because $T1≠T2$, making the transfer irreversible and generating entropy. I don't think the OP is confused on this point. I think they are asking if you can conclude this only from $\Delta S\geq 0$. I think the OP's question can be boiled down to "does the second law tell you when to choose the equality or inequality in $\Delta S\geq0$?" – BioPhysicist May 28 at 21:49
• @BioPhysicist I think I made it clear that the second law tells you to choose the equality if the system is and remains in thermal equilibrium and to chose the inequality if the system is in disequilibrium resulting in heat transfer across a temperature difference. But perhaps that wasn't clear. I suggest you post an answer if you feel you can make it clearer. – Bob D May 28 at 21:55
• Sorry, I wasn't trying to say your post wasn't clear. I think Andrew Steane's answer says what I would say, so I don't think it would be right to post a repeat answer. – BioPhysicist May 29 at 1:24