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Carbon Capture... (slrpnk.net)

submitted 11 months ago by [email protected] to c/[email protected]

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[–] NielsBohron 22 points 11 months ago* (last edited 11 months ago) (1 children)

Thermodynamics tells us it takes exactly as much to put the carbon back in as you got out of it by taking it out.

Thermo says it takes at least as much energy to put the carbon back in. If the process is done in a reversible way (reversible in the thermo sense), it would take exactly as much energy. And since real-world spontaneous processes are never reversible, it will always have energy lost.

I know you said down below that energy is lost, but I'm just saying that from a physics POV, there is not a possible way that reactions can ever be done in a reversible way, so it's not like there's even a possible theoretical world where you could approach 100% efficiency.

By definition, you will always pay the heat tax to the second law of thermodynamics.

[–] [email protected] 6 points 11 months ago* (last edited 11 months ago) (1 children)

To be extremely pedantic, operations on physical systems can be performed and perfectly reversed without loss of energy, but you couldn't ever extract anything anywhere along the way - not even direct evidence that it happened. Our models predict that this happens literally all the time in quantum mechanics.

Edit: fun fact: this prediction is actually central to what makes quantum computers work.

[–] NielsBohron 3 points 11 months ago

That's a good point. I hadn't considered that, as I'm a chemist approaching thermo from the stat mech/free energy POV. I was mostly just thinking that a process where ΔG<0 is spontaneous but if ΔG=0, the system is at equilibrium, so nothing happens.

Now that I type it out, even I know that's not exactly accurate as equilibrium is a dynamic state with lots of things happening at the molecular/electronic scale, so I guess I should have added a qualifier of "at the macroscopic scale" to my original post.