Ex. 2

Two identical blocks of iron, one at 100$^{\rm o}$C and the other at 0$^{\rm o}$C, are brought into thermal contact. what is the maximum work that can be extracted from the hot block in the absence of other heat sinks?

Remember, we can't just extract heat from the hot block and turn it into work; the entropy of the block would decrease without any compensating increase elsewhere. We need to add at least enough heat to the cold block so that its entropy increases by as much as that of the hot block decreases. (Work can always be used to do things which don't increase the entropy of the universe, such as lifting a weight.) Once the two blocks are at the same temperature, no further work can be extracted.

At that point the entropy change of the two blocks together, from the previous example, is

$$\Delta S=c \ln\left({T_f^2\over T_HT_C}\right)\ge0$$

The lowest final temperature is that for which $\Delta S=0$, ie $T_f=\sqrt{T_HT_C}=319$ K $=46^{\rm o}$C. (Any lower and $\Delta S$ would go negative, which isn't allowed.) So although the total heat loss of the hot block is $Q_H=c(T_H-T_f)$, the work extracted is only the difference between this and the heat gained by the cold block, $Q_C=c(T_f-T_C)$, namely $W=c(T_H+T_f-2T_f)$ In this case, the efficiency is

$$\eta={W\over Q_H}={(T_H+T_f-2T_f)\over(T_C-T_f)}=0.144.$$

Only 14% of the heat lost by the hot block was available to do work! Note that the maximum efficiency was obtained from a reversible process.