# Can adiabatic heat exchange between different temperatures ever be reversible?

I've been told, A volume 2X of Temperature 0.5*(T1+T2) is always at higher entropy than thermally insulated volumes, 'X' at T1, 'X' at T2 put together.

Let's take Sys1 (A volume 2X of Temperature 0.5*(T1+T2)) with entropy S1. Let's take Sys2 (thermally insulated volumes, 'X' at T1, 'X' at T2 put together) with entropy S2.

We know S2>S1. It means that in any reversible pathway I go from Sys1 to Sys2, I have the weighted average of Q's flowing into the system to be positive. It says heat must flow into the system. I understand we can't give heat to surroundings because that'd result in equilibrium temperature being less than 0.5*(T1+T2). But Why is zero heat not OK?

Let's take an adiabatic two-chambered vessel with an infinitesimal transmittive layer separating two chambers. Why is this not reversible?

• A reversible path between state 1 and state 2 does not have to be adiabatic. Commented Jan 9, 2022 at 12:44

Entropy flows during heat transfer, but entropy is not conserved; it is also generated from nothing whenever energy flows down a gradient in an intensive property (such as pressure, temperature, chemical potential, electric field, stress, surface tension...). That's why your two-chambered vessel coming to thermal equilibrium isn't reversible.

During (real) heat transfer, entropy flows and is generated simultaneously.

The increase in entropy of a system coming to an equilibrium does not require heating from an external force, and surrounding the system with adiabatic walls will not suppress this entropy increase.

For two identical bodies (with temperature-independent material properties) at temperatures $$T_1$$ and $$T_2>T_1$$ brought into contact, the final temperature is $$\frac{T_1+T_2}{2}$$, as you note. In addition, $$2C\ln\frac{T_2}{T_1}$$ entropy is generated, where $$C$$ is the heat capacity of each body.

There is only one way to conceptualize reversible heat transfer between bodies at different temperatures: use them to run a reversible heat engine, which will provide us with work. One example is the Carnot cycle, which has the distinction of being the only reversible heat engine that requires only two reservoirs (the hot and cold reservoirs).

After this reversible heat transfer is completed, the two bodies are at the shared temperature $$\sqrt{T_1T_2}$$, and the entropy increase in the rest of the universe is zero. We obtain energy of $$2C\left(\frac{T_1+T_2}{2}-\sqrt{T_1T_2}\right)$$ in the form of work.

If we wish to bring the system to the same state as if the bodies were simply pushed together, we could dissipate that work into thermal energy (by running an electrical current through a resistor, for instance). This would raise the temperature to $$\frac{T_1+T_2}{2}$$ and generate $$2C\ln\frac{T_2}{T_1}$$ in entropy.

• So reversible heat transfer without heat loss is possible but it always dissipates heat in form of work and we get equilibrium temperature always lesser than the arithmetic average? So heat transfer with dW=dQ=0 is not possible, is this due to the second law?
– Diza
Commented Jan 9, 2022 at 20:09
• So 'reversibe' heat transfer with dW=dQ=0 is not possible, is this due to the second law?
– Diza
Commented Jan 9, 2022 at 20:40
• That's correct. The reason is that heating a cooler body always increases its entropy more than cooling a hotter body decreases its entropy (from $dS=q/T$). So a balanced energy flow creates an unbalanced entropy flow, and a balanced entropy flow requires some energy to be drawn off as work. Heat transfer alone is reversible only if the temperatures are identical—and in this case heat transfer wouldn't occur spontaneously anyway. This is consistent with the maxim that real spontaneous processes are always irreversible. Commented Jan 9, 2022 at 21:03
• Thanks, @Chemomechanics, especially for the link.
– Diza
Commented Jan 10, 2022 at 4:27
• Any time! I can't emphasize enough how important the last section of the link is in developing your scientific intuition. Given some energy and a refractory target, the layperson will say: let's heat it. The thermodynamicist will say: let's find a cool reservoir, run a heat engine, and heat it more. Commented Jan 10, 2022 at 5:53