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Take a liquid in thermal equilibrium with it's surroundings. Let's say it's water and it happens to be at 20 degrees C.

The molecules of the liquid are obviously at a range of energies, with a mean that results in the 20 degrees measurement. Some have higher energies, some lower. I (perhaps unsurprisingly) want to "extract" the ones with higher energy.

Some of the molecules have enough energy to change state from liquid to gas, separating themselves from the liquid. If i collect the gas and separate it from the system, will i not be left with 1) high temperature water (the collected gas) and 2) correspondingly lower temperature water (the bit left behind). Then i use the temperature gradient to do work, obviously...

I've not violated conservation of energy, but i've certainly felt up entropy in a wholly inappropriate way. Why is this reasoning horribly naive? (and it surely must be, but i can't see why)

 — hennessey, Jan 29 2009

Brownian Ratchet http://en.wikipedia...ki/Brownian_ratchet
[mylodon, Jan 30 2009]

 How will you collect and separate the gas? Remember that when it condenses back into water, it will cool down again (latent 'heat')

We know that water cools as it evaporates - and that it will continue to evaporate until it's all gone, so I think it's not that naive to say that's a 'free' temperature gradient (though you do need some equally free method of replenishing your water once it's all evaporated away to make it worth while)
 — zen_tom, Jan 29 2009

a 'drinky bird' toy works on this principle.
 — xaviergisz, Jan 29 2009

 //I've not violated conservation of energy, but i've certainly felt up entropy in a wholly inappropriate way// [marked-for-tagline]

WTAGTPBAN
 — Custardguts, Jan 29 2009

 Thanks for your response. Since the molecules that become gaseous effectively physically separate themselves from the liquid in doing so, (i.e. "boil" off the surface) collecting them should be simple shouldn't it? Perhaps they could be absorbed into a hydro-phylic substance with a very low specific heat capacity, but i'm trying to avoid practical engineering solutions so that my fundamental reasoning can be attacked (i mean this in a good way)

 Anyway, if they're thermally isolated from the rest of the liquid, my assumption is that they will remain hotter than the liquid from whence they came. They'll have some distribution of energy of course, so, sure, some will condense back into liquid, but they'll still be higher energy on average.

 In removing the molecules at the high energy end of the distribution, we are of course left with liquid with a new and slightly different (lower) distribution of molecular energies. i.e. the liquid cools a bit. It'll still be 'producing' molecules that have enough energy for the state change though, just fewer as time goes by.

So eventually, you'll have your original liquid (at temperature T) separated into 2 halves: one at T+x where x is defined by the boiling point of your liquid and one half at T-x. With regard to replenishing the liquid: perhaps i don't fully understand what you're driving at, but if you have a large reservoir of liquid (at the temperature of the environment) to begin with, you can keep "topping up" the bit you're separating as it evaporates. I guess this would also have the effect of all but negating the cooling effect, and keep the evaporation rate constant.
 — hennessey, Jan 29 2009

Don't look now, but I think that clawed hand on your shoulder belongs to Maxwell's demon.
 — pertinax, Jan 29 2009

...hmmm, i wondered what that feeling was. So in this particular case (using the state change to separate the molecules at different energies) where is my 'reasoning' (i'm flattering myself with that term) wrong? Do the molecules not change state? (surely they do). The brownian ratchet idea was nicely, and unequivocally, dispatched - is it possible to do the same with this?
 — hennessey, Jan 29 2009

I forget the details, but I seem to remember that some evaporation is due to energy exchange at the surface of the liquid - low energy particles are energised with higher energy from above (radiation or from collisions with other material) and so evaporate. Since evaporation only happens at the surface, it's reasonable to assume that it doesn't follow the simple model you describe (i.e. the top percentile of the particles changing state) Otherwise there'd be no difference between evaporation rates between test-tubes and Petri dishes, for example.
 — zen_tom, Jan 29 2009

i know what you mean [zen tom] but the molecules in the body of the liquid (in the test tube say) would only have high energy (sufficient to change sate) for a very short period before hitting other molecules and regressing to the mean. They'd not evaporate; only the ones on the surface would actually break free of the liquid. Hence in the simple model on which i base my reasoning, evaporation rate would indeed be a function of surface area. However, this is no proof that my model is right: Is it fundamentally wrong? is this why the idea is flawed?
 — hennessey, Jan 29 2009

Evaporation complicates this and distracts from the issue. A simpler example is a room, heated by the sun coming through a window. It's much hotter at the top of the room (as is any room - stand on a chair), but remember that the process of extracting work out of this difference is pretty inefficient so it's probably not worth it.
 — hippo, Jan 29 2009

 //will i not be left with 1) high temperature water (the collected gas) and 2) correspondingly lower temperature water (the bit left behind). Then i use the temperature gradient to do work, obviously.// You seem to have left out a convenient bit of equilibrium right there, maybe that's what's confusing you. When water evaporates it subtracts some energy, google "latent heat", when it condenses, it adds some energy. In your experiment (using Gore-tex for e.g) you will be left, for a short period, with steam (water gas) of a certain energy and water (the original pool) of a less energy. As soon as that gas becomes liquid, it will *give off* this energy. If it is close enough to, or in a closed system with, the original pool, your equilibrium will be re-attained. Refrigerators do work to remove the energy of condensation away from the area of evaporation.

I think latent heat is the concept your are missing.
 — 4whom, Jan 29 2009

water that evaporates isn't hot, it's colder than the water it evaporates from. heat is conserved 100%. there is no new temperature differential, no work to be had. I don't know how everyone else missed this flaw in your thesis. Everybody needs a second cup of coffee! Perk up science people!
 — WcW, Jan 29 2009

water that evaporates isn't hot, it's colder than the water it evaporates from. heat is conserved 100%. there is no new temperature differential, no work to be had. I don't know how everyone else missed this flaw in your thesis. Everybody needs a second cup of coffee! Perk up science people!
 — WcW, Jan 29 2009

 //water that evaporates isn't hot, it's colder than the water it evaporates from.//

 Oh boy, here I go. That shit just makes me mad. Like a vacuum that just *sucks* everything into space. It does not happen like that!

 Water that evaporates (steam) is certainly hotter than the water it evaporates from. And the water it evaporates from is certainly cooler, after the fact (Google "the Mpemba effect"). Why else would you sweat? The problem comes when you try and close the system. It is then that the conservation of entropy and its converse (more applicable in this case) the conservation of enthalpy come into play. And it is then that you have to work against the gradient.

The only problem with this idea, (and it is not the physics [WcW]) is using the temperture differential of the gas (converted back to liquid) against the original liquid. When the gas (steam) returns to its liquid phase it *gives up* all the energy potential it had (latent heat). If this happens close enough to the original source (closed system) the system returns to equilibrium, else work needs to be done. This work cannot be done by the original system, else equilibrium. MFD - perpetual motion.
 — 4whom, Jan 29 2009

This problem was famously overcome by Einstein and Szilard. Where the latent heat trapped by evaporation was reclaimed by miscibily. Far from being unworkable, this process is used widely. However, it requires a work input, and is not viewed as self sustaining.
 — 4whom, Jan 29 2009

4whom, evaporation cools both the vapor and the solution being evaporated from. swamp coolers function on this principal. Where does the energy go? INTO THE PHASE CHANGE. NET NEGATIVE TEMPERATURE CHANGE WITH EVAPORATION. COOLING. COLD AIR, COLD WATER. BRRRRR. If the evaporating water "took" the heat with it then we would be double dipping when we condensed. THERMAL->KINETIC is still conservation of energy. The very name, partial pressure, recognizes that equilibrium between gas an liquid phase involves energy and can be made to do work. What we are struggling to pin down her is why does it seem like any open container of water seems to be in a perpetual thermal imbalance with the atmosphere. Mostly this is a mater of false perception and bad education, but it is also a reflection of the fact that we live in a complicated system; on the ground, not in the clouds, so to speak.
 — WcW, Jan 29 2009

Yes [bigsleep] the more energy you move out of a system (work) the colder (or less energetic) it becomes. In addition to simple thermal energy we also have kinetic and chemical energy to reduce and as you get really cold there seem to be other energies that also must be removed before you can reach absolute zero. I suspect that someday there may be a new phase change, the energy of total arrest, that will go with the other phase changes.
 — WcW, Jan 29 2009

A tiny laser basket in a perfect vacuum.
 — WcW, Jan 29 2009

 I think most physics is learned in terms of it's free energy schemes. Please continue to search for them, that is how you learn. I had an idea for a motorcycle that was powered by a generator on the front wheel. I loved the one where scoobie and shaggy were in water floating in an old bath tub and water was squirting in the drain hole like a fire hydrant and they took the shower head and pole pipe, stuck it in the hole and the water flowed out the showerhead and out back. that was great until I found out that the flow could reach no higher than the water level. then I found out about capilary action being able to overcome this but unfortunatly the water won't come out to go downhill and power anything. The demon was mentioned.

 Einstein did work on capilary action as I recall and he was also a theoretical physisist. Playing with ideas in his head because he has no budget for a lab.

The one I haven't figured out is the one where there are molecular ratchets that take bumps from heated molicules and occasionally ratchets up supposedly energy is obtained by cooling the system. wouldn't that be without a heat gradient? It seems to me like the next Maxwell's demon. I imagine that bumps will not be able to ratchet it up without also being able to ratchet the thing down. We probably won't know until we get really good and Nanotech and do it. I wonder how you could prove that you can't do it.
 — MercuryNotMars, Jan 29 2009

I have no doubt that tiny ratchets do exist in the atomic world, even quantum "ratchets", but they all function by moving energy from one form to another. Stealing energy is easy, preventing it from being stolen back is hard. Even if you use tiny quantum thieves the result is that your victim inevitably becomes a cold angry quantum terrorist Eventually it's going to get equal...(ibrium).
 — WcW, Jan 29 2009

 [WcW] Now it seems we are singing from the same psalm,

 //4whom, evaporation cools both the vapor and the solution being evaporated from// This is not true. Or half true. Vapour and steam need to be addressed. Vapour is *not* steam. Yes, vapour will undergo further evaporation into steam, given the large surface area of spheres (or vapour droplets). And this phase change will extract energy. But vapour is no phase change. Changing the phase of water (or whatever) is what extracts energy from a system. i.e liquid to gas (not liquid to suspended droplets). Condensing the gas into a liquid is what adds energy to the system. The gas is never cooler, nor the same temperature, than the liquid, provided we assume a *constant localised pressure*, it is just not fucking possible! Check your phase diagrams, and your coffee!

 So back to the idea: At a constant localised pressure and temperature, some H20 will be inclined to rise. When *selectively* leaving the field (presuming gore-tex, or maxwell's deamon), they take their higher energy with them. This means those left behind have a lesser mean energy. Ergo colder. So far, so good. Evaporative cooling, nothing new.

Now we assume that we have a cooler liquid and a hotter gas (from the post). If we assume we have a cooler liquid, and our gas has condensed to a hotter liquid, we are wrong (check phase diagrams). Basically we will have two seperated liquids at a similar temperature. Anything taken from the inital pool will be given of as condensation heat. MFD - perpetual motion.
 — 4whom, Jan 29 2009

 You are wrong. evaporation, deceptively, is a word with two contradictory meanings. In this case I am talking about the movement of liquid water into the gas phase allowed by the failure of intermolecular forces to retain it in the liquid phase. This process is only checked by the pressure placed on the fluid and the tendency for the process to come to equilibrium. In a system where the "humidity" is below "100%" this tendency to move from liquid to gas phase reduces the temperature of the system by converting thermal energy into kinetic energy.

 There are no droplets in this model, no suspensions, and even if there were it wouldn't make one joule of difference. Give me a sealed system of liquid and gas before equilibrium and given temperature and pressure and molarity I can give you the final temperature, and for water and air that temperature will be colder. No part will ever be any warmer than it starts out.

Gasses don't "carry away" the heat when they evaporate or "give it back" when they condense .The heat (a movement of one sort) in the vibration of the liquid becomes the energy of motion in the gas (to just be a gas requires extra motion). When you force a gas back into the liquid phase this energy simply changes amplitude.
 — WcW, Jan 29 2009

 //vapour will undergo further evaporation into steam// That isn't how I understand it. eVAPORation is what has taken place to get to vapor I think. Either his example doesn't include vapor as in water dropletts or vapor doesn't mean that.

 I think this will be explained with partial pressures. the boiling temperature can maintain a partial pressure of steam at 1atmosphere. lower temperatures maintain lower partial pressures in a closed environment. Your phase diagram will show you the temperature at which no evaporation will take place in a vacuum and water is in ice form at that temp.

My guess is that state changes are a subset of phase changes which would include for example different types of ice formed at different pressures so I don't think anyone is misusing terms by interchanging the two so long as we talk about state changes between solid liquid gas. It couldn't hurt to look it up.
 — MercuryNotMars, Jan 29 2009

//that clawed hand on your shoulder belongs to Maxwell's demon.// Sorry about that; forgot to drop the latch again.
 — MaxwellBuchanan, Jan 29 2009

 You do get energy out of this, but not more energy than was put into the liquid to give its higher energy molecules the speed to escape from the surface.

But that's ok. Puddles turn into rain clouds in this way (solar energy, of course), and then rain clouds fill reservoirs which power hydroelectric generators. So, there's no question that the idea works. An efficient and convenient implementation is the key.
 — colorclocks, Jan 30 2009

Well I think the point is that the atmosphere in many locations is nowhere near 100% humidity. There is a lot of potential energy there, just not in an exothermic form. Endothermic reactions can also do work, this idea should be harnessing that potential.
 — WcW, Jan 30 2009

 I am pretty sure the whole thing is based on no temperature gradient. we will give this experiment a closed environment with constant pressure and really really good insulation and put it in a cave or something. No, this is not going to give any measurable usable energy. This is a thought experiment. yes higher energy molocules do seperate and change state not temperature. to ever take advantage of the heat by bringing it back you must have a heat gradient which you don't have. there will be no dew and no condensation other than high energy particles evaporating and low energy gas condensing.

 //You do get energy out of this, but not more energy than was put into the liquid to give its higher energy molecules the speed to escape from the surface.//

you don't put any energy into this. It is called equillibrium and molicules go back and forth all the time but never in a method you can use to create any energy.
 — MercuryNotMars, Jan 30 2009

 ...trying to figure out how to take advantage of this to no avail; sure if you blow a wind across an open container of water you will blow the evaporated water away thus slowly cooling the remainder... but at the same time the wind will be heating the liquid water back to ambient... resulting in a null exercise.

 Perhaps some experimentation with anhydrous materials ? You'd still have to remove the vapour though to avoid it recondensing and reheating the liquid.

How long would it take a pot of water to turn to ice in an infinite vacuum ?
 — FlyingToaster, Feb 01 2009

 Imagine a jar suspended in a vacuum, half full of dry nitrogen, half full of water. The contents of the jar begin at STP.

 1) As the partial pressure of the gaseous phase of H2O equalizes with the atmosphere of the container the pressure in the container rises.

 2) As this process occurs the temperature of the of the water falls. Also, at the surface, the temperature of the gas falls (participating in the endothermic process).

 3) We reach equilibrium. There is now some water vapor in the gas phase. The pressure in the vessel is higher. The overall temperature in the vessel is lower.

 (some nitrogen enters the water too, but we ignore that)

Since this is an example of one of the founding observations of analytical chemistry that predated the description of the atom and can be found in any good chemistry text book I would hope that it would be widely known. Science allows us understanding that goes beyond the obvious, any person who considers themselves widely versed should know at least the basic principles of inorganic chemistry and physics.
 — WcW, Feb 01 2009

A pot of water in a vacuum will boil itself into an ice cube in less than ten seconds.
 — WcW, Feb 01 2009

But what if you watch it?
 — mylodon, Feb 02 2009

 //A pot of water in a vacuum will boil itself into an ice cube in less than ten seconds.// Holy Fuck! Did I hear right? Is there anyone from temperature here tonight?

 Ignor (I would have said Egor, but Ignor sounds better), go and get me one of these fantastic vacuums. You know... the one that doesn't evapourate water but rather makes it ice.

 A pot of water in space, which is both a vacuum and has a mean temp of 2.735K will be ice. YES! No argument!

I issue a word of caution! Whilst you are generating this vacuum that creates ice at standard temperature and 0 Pressure, please don't get sucked off the face of our planet.
 — 4whom, Feb 04 2009

 //3) We reach equilibrium. There is now some water vapor in the gas phase. The pressure in the vessel is higher. The overall temperature in the vessel is lower.//

 I haven't read all the annos, so I'm sorry if I'm repeating what's already been said, but this is the most important point. At equilibrium, the vapor and liquid will be at the same temperature and you can't get any work out of it.

 //Why is this reasoning horribly naive?//

Because what you've discovered is the enthalpy of vaporization, aka, the latent heat, as 4whom mentioned.
 — ldischler, Feb 04 2009

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