Timeline for Why doesn't the gravitational energy in this system of evaporating and condensing water violate the second law of thermodynamics?
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| Apr 13, 2017 at 12:39 | history | edited | CommunityBot |
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| Feb 10, 2017 at 23:18 | comment | added | Will | @annav Interesting examples ...but I'm going to abandon arguing that point here because I'm afraid I'd just end up hopelessly off-topic and lost in semantics... | |
| Feb 10, 2017 at 11:42 | comment | added | anna v | @WilliamBudd If you do not see that biological processes make order within the skin of a live organism take crystalization. The crystal has high order i.e. small entropy but it was produced in an open system, so crystals are not a violation of the second law. | |
| Feb 10, 2017 at 8:03 | comment | added | Will | @annav This system would generate more electrical energy than the Gibbs free energy contained in the surroundings for its given temperature -- in violation of the 2nd law. Also, living matter does not violate the 2nd law. | |
| Feb 10, 2017 at 7:12 | comment | added | anna v | It is not a closed box i.e. isolated system and the second law is testable only in isolated system. All living matter violates the second law after all, but it is not a closed system. | |
| Feb 9, 2017 at 3:33 | comment | added | Will | From an outside perspective this is a black box that takes heat from its surroundings and converts it into electricity. How is that not a question about (the validity of) the second law of thermodynamics? | |
| Feb 9, 2017 at 3:29 | comment | added | rob♦ | If it's not a closed system, then you have an engineering question about efficiency, not a conceptual question about the second law of thermodynamics. | |
| Feb 9, 2017 at 3:24 | comment | added | Will | There are two separate aspects to do this conundrum. (A) The heat produced by condensation can be transferred to the liquid to be used for evaporation, all within the system. This is not temporary. It is continuous. (B) So where would the energy leaving the system through the waterwheel generator come from? It would enter the system from the surrounding environment as heat. The system would effectively cool the environment over time. Note that I never said this was a closed system. | |
| Feb 9, 2017 at 3:18 | comment | added | rob♦ | No heatsink can overcome the second law of thermodynamics. The convection and condensation is a way to move heat and entropy around within the system; it's temporary. | |
| Feb 9, 2017 at 3:04 | comment | added | Will | (1) Sure, it seems highly likely to me that a ceiling at or near 100 C would probably prevent all condensation. However, I see no reason why it needs to become that hot, provided proper heat transferring means are installed between the ceiling and the colder liquid surface near the bottom (as I mentioned in my question above). Surely a copper heatsink would prevent such a huge temperature gap between its ends? | |
| Feb 9, 2017 at 2:47 | comment | added | rob♦ | (1) Are you really proposing there is no temperature gradient with a hot ceiling that would prevent condensation? I think you're mistaken. Limiting to freezing/boiling isn't reasonable --- if the maximum-entropy endpoint of the system has a layer of ice on the water surface, that's what'll eventually happen. | |
| Feb 9, 2017 at 2:47 | comment | added | rob♦ | Two replies, reverse order. (2) I don't think that appealing to the second law of thermodynamics is circular reasoning. If the system is closed, its total possible entropy is finite, and its entropy tends to increase over time. Eventually it'll fluctuate up to some maximum and stop there. I don't have to calculate those entropies to know that the limits exist. | |
| Feb 9, 2017 at 2:33 | comment | added | Will | Also, your last sentence lacks argumentation. As it stands, it reads as: "This is not perpetual motion, because perpetual motion doesn't exist.", which is circular reasoning (regardless whether the assumption holds or not). | |
| Feb 9, 2017 at 2:31 | comment | added | Will | Thanks for your answer. A temperature difference slows down the cycle, but will not stop it; provided that they remain in-between freezing and boiling point. Vapor pressure describes an approximately logarithmic relationship between temperature and pressure. In/decrementing one parameter does not drop the other to zero. Here's the relevant diagram. | |
| Feb 9, 2017 at 2:08 | history | answered | rob♦ | CC BY-SA 3.0 |
