Heat flow, the third law and residual entropy

Heat flows from hot to cold because that raises total entropy

Intuition

A hot cup left on a cool table cools down. The table warms a tiny bit. Nobody has to be told which way. Whichever direction makes the total number of microstates bigger is the direction it goes.

Bring two isolated systems $x$ and $y$ into thermal contact, keeping the pair isolated from the rest of the universe. The total entropy cannot fall:

$dS_{\text{total}} = dS_x + dS_y \ge 0$

Use $dS = dQ/T$ for each system:

$\frac{dQ_x}{T_x} + \frac{dQ_y}{T_y} \ge 0$

The pair is isolated, so whatever heat $y$ gives up, $x$ takes in: $dQ_y = -dQ_x$. That step is not written on the slide but is what gets the next line:

$dQ_x\left(\frac{1}{T_x} - \frac{1}{T_y}\right) \ge 0$

If $T_x < T_y$, then $1/T_x$ is bigger than $1/T_y$ and the bracket is positive, so $dQ_x$ may be positive. Heat flows into $x$ when $T_x < T_y$. In other words, heat flows spontaneously from hot to cold. This is another way of stating the second law, and it is the “more satisfactory notion of heat” the lecture promised: heat as a flow that always runs downhill in temperature.

Pitfall

The rule needs the pair to be isolated from everything else. A fridge moves heat from cold to hot, but only because something else (a motor) is also changing. That fits the second-law version that says heat cannot pass from colder to hotter “without some other change occurring”.

The third law fixes the zero point of entropy

Entropy approaches a constant value as $T \to 0$. For a perfect crystal of a pure substance, the ground state is the only possible microstate at $T = 0$. One microstate means $W = 1$, so

$S(0) = k\ln 1 = 0$

Think of a tidy library with exactly one right place for every book. Only one arrangement is allowed, so there is no uncertainty to count.

Some systems keep entropy at absolute zero

A system such as a glass can have more than one distinct lowest-energy arrangement. Then $W > 1$ at $T = 0$, and the leftover is called residual entropy.

Carbon monoxide. Each CO molecule can sit in the lattice in one of two orientations. For $N$ molecules there are $2^N$ arrangements, so $S = k\ln 2^N$ and

$S_{\text{residual}} = Nk\ln 2$

Ice. Water ice has residual entropy because of the many ways to arrange the covalent and hydrogen bonds. Around each oxygen, of the four hydrogens, two are close and two are far. The slide shows six such arrangements around one oxygen.

Spin ice. In some magnetic materials the spins cannot all align, which is called frustration. The first spin-ice materials were the pyrochlores dysprosium titanate $\text{Dy}_2\text{Ti}_2\text{O}_7$ and holmium titanate $\text{Ho}_2\text{Ti}_2\text{O}_7$.

Absolute entropies are measured from the T = 0 reference

The third law lets entropies be given relative to $T = 0$. The slide’s graph plots molar entropy $S^\circ_m$ in J K$^{-1}$ mol$^{-1}$ against temperature from 0 to 300 K. It rises through solid III, II and I, steps up at each phase change, and the gas region reaches about 200 at 300 K. The slide does not name the substance. The pattern to remember is that entropy climbs with temperature and jumps at each phase change.

Exam detail

Lecture 15 summary: entropy measures the statistical weight of the configuration and always increases in an isolated system; heat flows from higher temperature to lower; residual entropy at absolute zero is possible in glassy or frustrated systems. Formula to quote: $S_{\text{residual}} = Nk\ln 2$ for CO.

Recall

Why does a perfect crystal have zero entropy at T = 0 but CO does not?

A perfect crystal has one microstate, so $S = k\ln 1 = 0$. CO molecules can each take one of two orientations, which gives $2^N$ microstates and $S_{\text{residual}} = Nk\ln 2$.