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# Extracting Work from Heat: Modeling Power Plants and Refrigerators with Entropy Change Constraints | *Six Ideas That Shaped Physics* Chapter T9 Notes

Samson Zhang
Dec 7, 2021Last updated Jan 17, 20226 min read

If hot things -- like the ocean or the atmosphere -- have energy, can we take this energy and use it to do work? Can a ship sailing on the sea simply suck a bit of heat out of the water to power itself?

At first this might seem perfectly reasonable. We're not trying to do work for free and create perpetual motion, just take energy that's lying around and convert it to work.

(figure something)

Sucking heat energy out of the ocean and converting it to work doesn't violate energy conservation: energy from one place is simply ending up somewhere else. But remember the second law of thermodynamics: the entropy of an isolated system can only ever increase, never decrease. Cooling the ocean lowers its entropy, while using that energy for work doesn't automatically increase energy anywhere else, so the entropy of the ocean-work system is lowered overall, violating the second law of thermodynamics.

This doesn't mean that it's impossible to extract work energy from heat -- that would be ridiculous. Power plants work by using heat to turn turbine blades to generate electricity.

Rather, the second law simply adds a constraint on the efficiency of such a process, because we must consider as part of the system some subsystem that increases in entropy at least as much as the entropy of the hot subsystem decreases when energy is extracted from it for work.

Modeling heat engines and generators

A good way to generally model such processes is to consider two reservoirs: a hot one with temperature

T_H $T_H$
and a cold one with temperature
T_C $T_C$
. If connected with each other, heat energy will naturally flow from the hot reservoir to the cold reservoir. If undisturbed, the hot reservoir will lose
Q_H $Q_H$
of energy and the cold reservoir will gain
| Q_C | |Q_C| = |Q_H| $| Q_C | |Q_C| = |Q_H|$
of energy over some period of time.

But consider that the magnitude of entropy change of the two reservoirs is unequal. Since they are reservoirs, their temperatures will remain relatively unchanged when energy is taken or added to them, so their entropy change is given by the equation

\Delta S = \frac{\Delta Q}{T} $\Delta S = \frac{\Delta Q}{T}$
. Since
T_H > T_C $T_H > T_C$
, the hot reservoir will lose less entropy than the cold reservoir gains.

Now consider that instead of letting all the energy

Q_H $Q_H$
flow to the cold reservoir, we can take some energy
W $W$
about of it so
|Q_C| = |Q_H| - W $|Q_C| = |Q_H| - W$
. As long as the entropy of the cold reservoir still increases by more than or as much as the entropy of the hot reservoir decreases, the second law of thermodynamics is not violated.

(figure something)

The maximum value of

W $W$
where this is true can be calculated easily:

\Delta S_C \geq \Delta S_H \implies \frac{Q_C}{T_C} \geq \frac{Q_H}{T_H} \implies \frac{Q_H - W}{T_C} \geq \frac{Q_H}{T_H} \implies T_H(Q_H - W) \geq T_C Q_H \implies W \leq Q_H \frac{T_H - T_C}{T_H}
$\Delta S_C \geq \Delta S_H \implies \frac{Q_C}{T_C} \geq \frac{Q_H}{T_H} \implies \frac{Q_H - W}{T_C} \geq \frac{Q_H}{T_H} \implies T_H(Q_H - W) \geq T_C Q_H \implies W \leq Q_H \frac{T_H - T_C}{T_H}$

We can also calculate the efficiency of such an energy extractor

e = \frac{W}{Q_H} $e = \frac{W}{Q_H}$
:

e = \frac{W}{Q_H} \leq \frac{T_H - T_C}{T_H}
$e = \frac{W}{Q_H} \leq \frac{T_H - T_C}{T_H}$

Some implications of this equation are immediately clear. Unless

T_C = 0 $T_C = 0$
-- an impossible reservoir that stays at absolute zero -- the efficiency of a heat engine is always less than one. In other words, as we stated at the beginning, it's impossible to directly extract heat and turn it into work energy.

The other implication is that maximum efficiency increases when the temperature difference

T_H - T_C $T_H - T_C$
increases. That's why power plants -- coal, nuclear, whatever else -- run as hot as possible, using some fuel to maintain the temperature of the hot reservoir as energy is extracted from it and dumping waste heat necessary to not violate the second law into the cold reservoir of the atmosphere.

Modeling refrigerators

A very similar model explains how refrigerators and air conditioners can cool some space without violating the second law of thermodynamics.

We can model a refrigerator or air conditioner once again as a cold and hot reservoir. Instead of transferring heat from hot to cold, we want to transfer heat out of the cold reservoir (the inside of a refrigerator or a cold room) and put it in the hot reservoir (the outside of a refrigerator or the atmosphere).

To make this transfer possible, we'll add work

W $W$
to
Q_C $Q_C$
so that the total heat flow
Q_H = Q_C + W $Q_H = Q_C + W$
into the hot reservoir makes its entropy increase by more than the entropy of the cold reservoir decreases. Once again we can balance out the entropy changes:

\Delta S_H \geq \Delta S_C \implies \frac{Q_H}{T_H} \geq \frac{Q_C}{T_C} \implies \frac{Q_C + W}{T_H} \geq \frac{Q_C}{T_C} \implies T_C(Q_C + W) \geq T_H Q_C \implies W \geq Q_C \frac{T_H - T_C}{T_C}
$\Delta S_H \geq \Delta S_C \implies \frac{Q_H}{T_H} \geq \frac{Q_C}{T_C} \implies \frac{Q_C + W}{T_H} \geq \frac{Q_C}{T_C} \implies T_C(Q_C + W) \geq T_H Q_C \implies W \geq Q_C \frac{T_H - T_C}{T_C}$

The efficiency we care about here, for refrigerators more commonly referred to as the coefficient of performance or COP, is the ratio of heat extracted from the cold reservoir

Q_C $Q_C$
to the work put in
W $W$
:

COP = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C}
$COP = \frac{Q_C}{W} = \frac{T_C}{T_H - T_C}$

As this value shows, unlike the efficiency of a heat engine, the coefficient of performance of a refrigerator can far exceed 1. The COP also decreases as the temperature gap increases, which makes sense -- it takes more energy to maintain a larger temperature difference.

Heat pumps: using refrigerators as more efficient heaters

Since a refrigerator is able to transfer more heat from a cold to hot reservoir than the amount of energy put in, it has another interesting use case: as a more efficient heater.

Consider a warm house in the winter. The inside of the house can be modeled as a hot reservoir, and the cold air outside as a cold reservoir. Now consider a refrigerator-like process that sucks heat from the cold air and dumps it, using a bit of energy, into the hot inside of the house.

With a normal heater, stored energy in some form (say from a wall socket or the burning of natural gas) is converted into heat with some loss along the way, so less heat energy is added to the house than spent to create and add it.

But the inverted refrigerator, or heat pump, has

COP > 1 $COP > 1$
. That means that, for every unit of energy spent heating the house, more than that amount of heat energy will be deposited.

If everyone used heat pumps instead of traditional heating systems, a huge amount of energy could be saved, as well as natural gas specifically, 20% of which goes towards residential usage in the U.S. But Moore's textbook notes that "heat pumps are quite a bit more expensive (and somewhat less reliable) than standard furnaces" and so they are "less economically feasible than they should be." In class, Prof. Whitaker also noted the bigger picture problem: even if heat pump technology were improved to the point of widespread adoption, it -- and other energy usage and emissions reducing technologies -- wouldn't be enough to make a dent against climate change. "It's too late," he said. Solutions like carbon capture are the only way forward now.

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## Physics 70

Notes for Pomona class with Prof. Whitaker, fall 2021