Thermodynamics does not preclude the possibility of extracting heat from the atmosphere or the ocean water and convert some of it into mechanical work. In principle, is this possible to do it in practice to arrest global warming? Forget the cost.
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asked Aug 27, 2019 at 9:03
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$\begingroup$ Converting heat to mechanical work is just one step - Afterwards in order to achieve a "cooling"-effect, the mechanical work has either to be stored in a way that is not heat (for example by a giant wheel with Zero friction, that will be Spinning faster and faster throughout the centuries....), or it has to be brought off the planet again (for example by shooting particles in to space). $\endgroup$– QuantumwhispAug 27, 2019 at 9:18
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$\begingroup$ Given that the only real way of doing this is to use a cold reservoir which is not on Earth (because the very last thing you want to do is to further heat the cold reservoirs we have on Earth, such as the ice sheets), then something like 'big mirrors in orbit reflecting sunlight away' is probably hugely more practical (or, less impractical). $\endgroup$– user107153Aug 27, 2019 at 10:59
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2 Answers
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Here's why this is a very bad idea if you do it without a cold reservoir which is not on Earth.
Let's assume we can build an ideal Carnot cycle engine. This has an efficiency
$$\eta = 1 - \frac{T_C}{T_H}$$
where $T_C$ is the temperature of our cold reservoir and $T_H$ is the temperature of our hot reservoir. The work done by such an engine is
$$ \begin{aligned} W &= \eta Q_H\\ &= \left(1 - \frac{T_C}{T_H}\right)Q_H \end{aligned} $$
Where $Q_H$ is the energy we are extracting from the hot reservoir.
Further let's assume that we can dump the work it does somewhere that it does not end up as heat in due course: we're using it to fire things into space, or perhaps to build some chemical (for instance to turn carbon dioxide into carbon), and we'll assume we can dump all the work somewhere it doesn't end up as heat since we can't do better than that.
Well, now we can compute $Q_C$, which is the heat we're going to dump into the cold reservoir:
$$ \begin{aligned} Q_C &= Q_H - W\\ &= Q_H - \left(1 - \frac{T_C}{T_H}\right)Q_H\\ &= Q_H\frac{T_C}{T_H} \end{aligned} $$
OK, so finally let's invent some numbers for $T_C$ & $T_H$: We're working between somewhere in the atmosphere or the oceans and somewhere in an ice sheet, so I'll assume that $T_H = 30\mathrm{^\circ C} = 303\,\mathrm{K}$ and $T_C = -30\mathrm{^\circ C} = 243\,\mathrm{K}$. I don't know what the real values of these are, but this is good enough to make the point.
This gives
$$Q_C \approx 0.8 Q_H$$
In other words about 80% of the heat we pull out of the atmosphere & oceans ends up in the cold reservoir. What it will do there is melt ice, and if we do this on a large enough scale to matter it will melt a lot of ice. And this is something we really do not want to happen fast, because there are tens of metres worth of sea-level rise sitting in ice sheets (more than $50\,\mathrm{m}$ in the East Antarctic sheet alone). Significant, rapid, sea-level rise will cause extremely serious problems: coastal cities will be lost (London, New York, &c), entire countries such as Bangladesh & The Netherlands will be lost and so on.
So, if you do this you need to have a cold reservoir which is off the Earth. And if you are doing that, well, giant-mirrors-in-space is hugely simpler and cheaper (and aerosols in the stratosphere is cheaper and simpler again).
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I don't see why not, in theory at least. Although, bear in mind that a heat engine needs both a hot and a cold reservoir, i.e. a temperature differential. But, we have the moon, which is pretty cold. So, perhaps a heat engine could be constructed between the Earth and the moon, which uses the thermal differential to generate electricity and help reduce the heat of the Earth.
But of course there would be many practical engineering challenges, such as how to connect two planetary bodies with a stable structure and how to avoid huge losses in the piping. Also there would be the cost (but you said to ignore that).