Entropy is a measure of unavailable energy?

Question

I came across a statement in my text book which said that entropy is a measure of unavailable energy.

What does this statement signify?

Answer 1

When the first law was first formulated it was assumed that the most efficient utilisation of a chemical reaction to produce work was to use all the heat to produce work. The second law was not fully understood and thus at maximum efficiency $-\Delta H$ was expected to represent the maximum amount of work. Many experiments failed to confirm this. We now know that it is the free energy which measures the maximum capacity to do work, $\Delta G=\Delta H-T\Delta S$. Clearly $\Delta G$ and $\Delta H$ are equal only if there is no entropy change between the beginning and end of an isothermal reaction. If this is not the case the work obtainable in a reversible process can be either greater than or less than the heat of reaction.

From the first law the external work performed must be equal to the loss in energy of the system, unless some heat is taken from or given to the surroundings. This is exactly the point first clearly seen by Gibbs. In a reversible isothermal reaction $T\Delta S$ is the heat absorbed from the surroundings and if this is positive the work done will be greater than the heat of reaction.

So 'available/unavailable energy' seems to depend on the particular case. Hopefully some thermo experts will give you a more detailed argument.

edit: It occurs to me that perhaps the textbook was referring to the fact that one can 'convert all the work into heat but not all the heat into work'. In other words not all the random thermal motions of molecules can be made to do work.

Answer 2

Another way of seeing this issue is to write the energy conservation law two ways: $$dU = \delta Q - \delta W= TdS - \sum_{k}Y_kdX_k $$ and this is combined with Clausius's inequality $$\delta Q \le TdS \\ \delta W \le \sum_{k}Y_kdX_k$$ Here $T$ is the temperature at which $\delta Q$ heat transferred to and $\delta W$ is the work done by the system whose internal energy and entropy have thereby changed by $dU$ and $dS$. The $Y_k$ and $X_k$ are the internal intensive and extensive system parameters.

Evidently less work is done by the system for the same amount of change $dU$, $dS$ and $dX_k$ when the process is irreversible (strict inequality) than when it is reversible (equality).

Because of the Clausius inequality $0 \le \delta \mathcal{N} = TdS- \delta Q$ and this, of course, is the same as $\sum_{k}Y_kdX_k - \delta \mathcal{N} = \delta W$. The quantity $\delta \mathcal{N} = T\delta \sigma $ with $\delta \sigma \ge 0$ represent the amount of internally generated "heat" and entropy, resp., by the irreversible process. Because of the internally produced entropy $\delta \sigma $ the total entropy change $dS$ will be greater by this amount than in a reversible process connecting the same states, and the resulting $\delta \mathcal {N}$ is then the amount of deliverable work that is "lost" by the irreversibility.