The work left in two buckets of water

Every engine in the ceiling on every engine runs between reservoirs: a hot body and a cold one so large that the heat flowing out of the first and into the second changes neither temperature. That assumption is what makes the efficiency a single number, $1 - T_c/T_h$, fixed for as long as the engine runs.

The temperature an engine really takes its heat at relaxed the other half of the picture — heat taken in over a range of temperatures by the working fluid — and found the exact replacement, the entropy-weighted mean. This essay relaxes the reservoirs. Take a kilogram of water at 90 °C and a kilogram at 10 °C, put the best engine that can exist between them, and ask how much work comes out.

Carnot’s formula at the starting temperatures promises 22 per cent of whatever heat is drawn. The answer is barely half that, and the reason turns out to be one line of entropy bookkeeping with a square root at the end of it.

The engine that keeps entropy fixed #

The best engine is a reversible one, which creates no entropy anywhere. It works in small steps: draw a little heat from the hot water, deliver a little work, pass the rest to the cold water, repeat. In each step the hot water’s entropy falls by $C\,dT_1/T_1$ and the cold water’s rises by $C\,dT_2/T_2$, with $C$ the heat capacity of each, and a reversible engine requires the two to cancel:

$$\frac{dT_1}{T_1} + \frac{dT_2}{T_2} = 0.$$

That says the product $T_1T_2$ does not change. Both temperatures slide along a hyperbola until they are equal, and they are equal at $\sqrt{T_1T_2}$ — the geometric mean of the starting temperatures. For water at 363.15 and 283.15 K that is 320.7 K, or 47.5 °C.

The geometric mean is less mysterious than it looks. Taking logarithms turns the fixed product into a fixed sum, $\ln T_1 + \ln T_2$, so the reversible engine is sharing out the logarithm of temperature evenly, where contact shares out the temperature itself. Entropy is heat capacity times the logarithm of temperature, and an engine that creates no entropy can only move it from one body to the other; a common temperature that keeps the total fixed is the one whose logarithm is the average of the two.

The work is then energy bookkeeping. The hot water gave up $C(T_1 - T_f)$ and the cold water received $C(T_f - T_2)$, and the difference went out as work:

$$W = C(T_1 + T_2 - 2\sqrt{T_1T_2}) = C\left(\sqrt{T_1} - \sqrt{T_2}\right)^2.$$

For the two kilograms of water that is 20.8 kJ, from 177.8 kJ drawn from the hot one. The figure checks the result both ways, against the heat balance and against the entropy balance.

The common temperature is a receipt #

The dashed lines in the figure are the same two buckets poured together, or simply placed in contact. No work is taken out, so all the heat the hot water loses goes into the cold water, and the two meet halfway, at the arithmetic mean, 323.1 K.

The two endpoints differ by 2.5 kelvin, and the difference is not a detail. Two kilograms of water at 323.1 K hold more energy than two at 320.7 K by exactly $2C \times 2.5$ K — the 20.8 kJ the engine delivered. The temperature two bodies settle at is a receipt for the work taken out on the way. An arithmetic mean says none was; a geometric mean says the maximum was; anything between says some was, and how much.

Contact also generated entropy that the engine did not. Mixing the water raises the total entropy by $C\ln(T_\text{mix}^2/T_1T_2)$, about 65 J/K here, and multiplying that by the final temperature gives, to within a fraction of a per cent, the same 20.8 kJ: the work an irreversible process forgoes is the entropy it creates times the temperature at which the energy ends up. The rule is general, and it is the most useful form of the second law for an engineer, because it prices every irreversibility in joules.

Three buckets, and an inequality about averages #

The argument does not need two bodies. Put an engine among any number of equal bodies at different temperatures, run it reversibly until they all agree, and the same entropy bookkeeping says the product of all their temperatures stays fixed. They end at the geometric mean of the starting temperatures, and the work delivered is the total heat capacity times the difference between the ordinary average and the geometric mean.

For three kilograms of water at 10, 50 and 90 °C the ordinary average is 50 °C, 323.15 K, and the geometric mean is 321.49 K. The best engine among them delivers 20.8 kJ and leaves all three at the lower of the two.

Written that way, the second law for these bodies is a familiar piece of mathematics: the geometric mean of positive numbers never exceeds their arithmetic mean, and equals it only when the numbers are all the same. That inequality is the statement that work can be extracted from bodies at different temperatures and never from bodies at the same one. It is also why the work is small for small differences. The gap between the two means grows as the square of the spread of the temperatures, which is why the two buckets, 80 kelvin apart, gave up only 20.8 kJ against 178 kJ of heat — and why the three, with the same extremes and one in the middle, give up almost exactly the same 20.8 kJ: to leading order the work depends on the sum of the squared departures from the average, and the middle bucket, sitting on the average, adds heat without adding to that sum.

Mixing without an engine is irreversible for the reason entropy is a count gives. The number of ways of arranging the energy is far larger when it is shared evenly than when it is bunched into the hot bucket, and the mixture never returns to the unmixed state by itself — not because it cannot, but because it has overwhelmingly more ways to stay mixed. The engine delivers work by never letting the energy explore those extra arrangements; contact lets it, and the work is gone.

An efficiency that runs down #

At every moment the engine is a Carnot engine between the current temperatures, and the figure plots the efficiency that allows. It begins at 22 per cent. Each joule drawn cools the hot water and warms the cold, so the next joule is converted at a slightly lower rate, and the efficiency runs down to zero as the two approach each other. The work is the area under that curve, and the figure’s integral reproduces the 20.8 kJ to a part in ten thousand.

The average over the whole run is the work divided by the heat, and it has a closed form that is worth pausing on:

$$\frac{W}{Q_1} = \frac{(\sqrt{T_1} - \sqrt{T_2})^2}{T_1 - \sqrt{T_1T_2}} = 1 - \sqrt{\frac{T_2}{T_1}}.$$

For the two buckets that is 11.70 per cent. The shortfall from 22 per cent is not a failure of the engine, which is perfect; it is that most of the heat can only be drawn after the temperature difference has already been spent.

A square root where Carnot had a ratio #

Plotted against the ratio of the starting temperatures, the finite-body efficiency is always below Carnot’s and the gap is large. At a ratio of 4, where reservoirs would allow three-quarters of the heat to become work, two equal bodies allow half.

The square root has turned up before, in another engine. The engine that has to finish found that an engine between true reservoirs, run for the most power rather than the most work, has an efficiency of $1 - \sqrt{T_c/T_h}$ — the Curzon–Ahlborn result. The formula is identical and the reasons have nothing in common: there, a finite rate of heat flow through finite conductances; here, finite heat capacities and no rates at all. Two different ways of making an ideal engine less ideal land on the same function, which is a coincidence of the square root rather than a hidden connection, and worth knowing so as not to mistake one argument for the other.

The practical case where this matters most is the one with the smallest temperature difference. An ocean thermal energy plant runs between surface water near 25 °C and deep water near 5 °C. Carnot between those temperatures allows 6.7 per cent. But neither the warm water pumped in nor the cold water pumped up is a reservoir; each stream is heated or cooled by the plant, and for equal flows the best the plant can do is the square-root figure, 3.4 per cent — before any of the pumping, the temperature drops across heat exchangers, or the friction. The plants built have produced net power at a few per cent at best, and the reason is visible in the square root.

How much of hot water is work #

The cold body is often the surroundings themselves: the air, a river, the ground. They are large enough to be a reservoir, and only the hot body is finite. The most work that can be extracted from a hot body cooling to the surroundings’ temperature $T_0$ is called its exergy, and for a body of heat capacity $C$ it is

$$C\left[(T - T_0) - T_0\ln\frac{T}{T_0}\right],$$

which is the Carnot factor averaged over the cooling, and the figure checks it that way.

The numbers are sobering for anyone who thinks of a hot-water tank as a store of energy. At 80 °C, with the surroundings at 20 °C, a kilogram of water holds 251 kJ of heat above the surroundings and 22.7 kJ of work: nine per cent. At 60 °C it is six per cent; at the 40 °C of a shower, three. The heat is real and useful for heating. As a source of work it is poor, and it grows poorer the closer the water is to the room.

A cup of tea makes the point at the scale of a kitchen. Three hundred grams at 80 °C in a room at 20 °C holds 75 kJ of heat above the room and 6.8 kJ of work. Left to cool on the table it loses all of both, and the work is the part that could never be recovered afterwards: the tea ends at the room’s temperature having created the entropy that the lost work is the price of.

Cold is a resource by the same accounting, and the numbers are surprising. An engine can run between the room and something colder than it, taking heat from the room and rejecting it into the cold body. A kilogram of ice melting at 0 °C absorbs 334 kJ at a fixed temperature, and relative to a room at 20 °C the work that heat flow can deliver is 334 kJ times $(T_0/T - 1)$, about 24.5 kJ — slightly more than the 22.7 kJ in a kilogram of water at 80 °C. The ice holds its temperature while it melts, as heat that changes no temperature does, so it behaves like a small reservoir rather than a cooling body, and that is what makes a block of ice a better store of work, per kilogram, than a tank of hot water that looks far more energetic.

That is also why heating a house by burning gas is an expensive use of the gas. A flame at well over a thousand kelvin could run an engine at a high efficiency; using it to make water at 60 °C throws almost all of that capacity away. The engine that pays back more than it takes runs the argument the other way: a heat pump spends a small amount of work to move a large amount of heat, and the small amount is roughly the exergy of the warm water it produces, which is why it can deliver three or four joules of heat for each joule of electricity. Combined heat and power stations exist for the same reason: generate electricity from the high-temperature part of the fuel’s heat, and use the low-temperature remainder, whose work content is small, for heating.

The larger the cold body, the closer to the limit #

Between the two cases — two equal buckets and a bucket against the ocean — the cold body’s size decides how much of the hot water’s work can be had. With unequal heat capacities the reversible engine brings both bodies to a weighted geometric mean, pulled towards the larger body’s temperature. A cold body a tenth the size of the hot one warms quickly and the engine stops early; one a thousand times the size hardly warms at all, and the work approaches 39.9 kJ, the exergy of the hot water against a reservoir at 10 °C. Equal bodies get barely more than half of that.

The pattern is the same one that ran through the efficiency curve: a temperature difference is a resource that is spent by using it, and a small cold body spends it on itself.

This is the general shape of every problem in which a temperature difference is the resource. The engine a fluctuation cannot run found that no device, however small or clever, can extract work from a single temperature; the figures here are the quantitative sequel, pricing how much a difference is worth once the difference itself is used up by being used. The price is always the same kind of quantity — a heat capacity times the gap between two averages — and it is why thermal storage, waste-heat recovery and heat pumps are all argued in the same units. Engineers who design heat recovery from exhaust streams, cooling water or industrial waste heat are working in exactly this regime, where the question is less “how efficient is the engine” than “how fast does the stream being cooled or heated change temperature”.

Where the model stops #

The heat capacities are constant. Water’s changes by less than one per cent between 10 and 90 °C, so the figures are good for it; for a body that melts or boils partway, the latent heat arrives at a single temperature and changes no temperature, which makes a phase-change store closer to a reservoir and raises its exergy per joule.

The engine is reversible. It runs infinitely slowly and through ideal heat exchangers. Any real engine between the two buckets delivers less, and the lost work is the entropy it generates times the temperature of the surroundings it ends up in.

Nothing leaks. The buckets are insulated from the room. If the room is at 20 °C, the cold bucket at 10 °C is itself a source of work relative to the room, and the analysis has three bodies rather than two.

And pressure and volume are ignored. Exergy in general includes mechanical and chemical parts — compressed air, a fuel waiting to burn — and only the thermal part is drawn here.

What the pictures cannot show #

None of the figures shows the engine. The results are true of any reversible engine whatever, which is their strength and the reason no mechanism needs drawing; but it means the pictures cannot show how an engine running between two buckets would actually be built, or how slowly it would have to run to approach these numbers.

Nor can they show where the lost work of contact went. When the buckets are simply poured together, 20.8 kJ of work capacity disappears, and nothing visible happens except a temperature two and a half kelvin higher than it could have been. The loss is recorded in the entropy of the mixture, which no thermometer reads, and whose increase is, as the second law with a probability attached puts it, overwhelmingly likely rather than strictly certain for a system of any finite size.

Still open: how much low-temperature heat is worth recovering #

Most of the energy that power stations, factories and data centres reject leaves as heat below 100 °C, and there is a great deal of it. The thermodynamics says exactly how little work it contains, as the figures here compute. What is not settled is how much of that small share can be recovered economically: organic Rankine cycles, thermoelectric generators, and heat pumps that upgrade waste heat for district heating compete on cost as well as efficiency, and their best uses depend on the temperatures and flows of each source and on what the recovered energy displaces.

Whether large-scale recovery of low-grade heat — or the long-proposed ocean thermal plants — can ever be worth building at scale is an engineering and economic question that the second law bounds without answering. It says the prize is a few per cent of the heat. It does not say whether a few per cent of so much heat is worth the equipment.

The habit worth keeping is the one the receipt teaches. When two things come to the same temperature, ask which mean they came to. The arithmetic mean says the temperature difference was wasted; the geometric mean says it was used; and the gap between them is the work, measured in kelvin.