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m74z00219
2010-Apr-08, 06:02 PM
Hi all, this outta be a fun one.

Imagine we start off on the night side of the moon. Before us is a large cylindrical volume of water, sealed. The contents of the water are maintained at standard temperature and pressure until the lid is suddenly removed. My question is, what happens to the water: does it freeze or does it boil? My guess is that it boils, but is it that simple? Is this question complicated further if we open the lid of the container during lunar daytime?

Thanks all,
M74

John Jaksich
2010-Apr-08, 06:28 PM
Pardon my ignorance---but you are starting at STP----> 760 mm Hg and 273 K ?

Without drawing a phase diagram: all three phases could co-exist if given that for a 1 component Gibbs Phase rule: F= 3 - P where the original (?) Gibbs rule is listed as F = C - P -2

If F= 0 for one component---> water...then possibly we have: F = 3 - P where P is the number of phases. F= 0 implies an equilibrium...although it would be shortlived.

Without further info---> that is all that I can infer(?) for an answer.

So perhaps momentarily---we could have all three phases co-exist---otherwise it is (for me at least) not enough info.

m74z00219
2010-Apr-08, 06:52 PM
Pardon my ignorance---but you are starting at STP----> 760 mm Hg and 273 K ?

Oops, I should have been more specific. I'm accustomed to using NIST's definition (293.15K, 1atm). If we use the IUAPC's definition - as you have - I see how there will be a momentary equailibrium, but only within the bulk of the water - not at the boundary.

Thanks for your input though, I hadn't been aware of Gibb's rule before this.

M74

violentquaker
2010-Apr-09, 05:02 AM
At the warmer STP you describe, we are starting with liquid water when we suddenly remove the lid. The pressure quickly drops to zero, moving at least the water near the top of the cylinder into the gas domain where it quickly (and probably messily) boils. This pulls the temperature down (assuming we are no longer heating it) and the water that hasn't already vaporized will freeze. I would have to look up how fast the remaining ice would sublime in the lunar night temperatures, but it would speed up a lot once the sun rose.

chornedsnorkack
2010-Apr-09, 12:25 PM
At first, the pressure in the water starts dropping from 1013 mbar, and the water naturally expands and accelerates upwards - but not much.

When the pressure inside water falls to 24 mbar (pressure of water vapour at 20 Celsius), boiling becomes possible. But note that boiling is also resisted by surface tension of water. If bubbles cannot form inside water vecause of surface tension, then water will only evaporate from surface.

When water has turned into gas, it expands and escapes to vacuum at a speed comparable to speed of sound, and of molecular motion. Slightly more or less, there are factors depending on the process. Roughly 500 m/s.

2400 Pa is 2400 J per square m. Which means that 1 square m must evaporate 5 kg water per second - a 5 mm layer each second. But evaporating (and accelerating) 1 g of water takes around 540 calories - but cooling 1 g water to 0 C takes just around 20 calories. So before 1 mm water can be evaporated, 27 mm water must be cooled from 20 C to 0 C - and the heat must be able to conduct to surface.

If a thin surface layer of water reaches 0 C, at which point the vapour pressure is about 611 Pa (corresponding to evaporation of 1 mm layer per second), water starts to freeze. Since the heat of ice fusion is about 80 cal/g, boiling 1 g water freezes slightly less than 7 g into ice. The same requirement, of heat conduction from water into ice, still applies.

And then the ice starts to cool. Ice has perhaps half the heat capacity of water.

Now, how long will it go on?

Solar constant is about 1400 W per square m. In other units maybe 330 calories per second per square m. Sufficient to evaporate perhaps 0,5 g per second - which, on escaping at 500 m/s, exerts pressure of roughly 0,25 Pa.

The vapout pressure of ice is 0,261 Pa at -70 Celsius. So ice in full sunlight in vacuum might warm up or cool to roughly -70 Celsius, at which point all the incident sunlight would go to evaporate the ice, at the rate of roughly 0,5 g per second per square m - amounting to 1800 g per hour, or roughly 2 mm layer per hour, 5 cm layer per day (24 hours in full sunlight)... 70 cm during 14 days of lunar day if the sunlight were always perpendicular, but it is not.

If the ice reflected the sunlight, it would evaporate slightly slower and slightly colder than -70 Celsius - but I think water ice is a good absorber for near-IR.

What would the ice be doing at night, though?

neilzero
2010-Apr-09, 01:49 PM
When the lid is removed, the pressure drops from 14.7 PSI to extremely close to zero in about one microsecond. If the container was 90% full, the water vapor near the lid expands and cools the water surface to perhaps 14 degrees c (it was 15 degrees c) Vapor pressure is trying to be several psi, so the water cools due to rapid evaporation, but not boiling. The water also cools by radiating heat into space = infrared photons. The water will freeze to ice in 1 to 100 hours (depending on the container size, shape and water volume, and might evaporate completely before the next sun rise, up to two weeks away. Ice does evaporate, very slowly even at very low temperatures = the ice might cool to minus 200 degrees c.
If the lid was opened with the sun directly overhead (or other angle for a black container without thermal insulation) the solar photons would warm the water faster than the evaporation cools it. Boiling would start (about 82 degrees f, in a vacuum) in about one minute and all the water would be gone in about one hour. If the container is dull black it will heat to 150 degrees c about one hour after the last of the water is gone. Can someone estimate the max temperature of an aluminum colored container in sunlight on the moon's surface at the equator? Neil

m74z00219
2010-Apr-09, 09:44 PM
At the warmer STP you describe, we are starting with liquid water when we suddenly remove the lid. The pressure quickly drops to zero, moving at least the water near the top of the cylinder into the gas domain where it quickly (and probably messily) boils. This pulls the temperature down (assuming we are no longer heating it) and the water that hasn't already vaporized will freeze. I would have to look up how fast the remaining ice would sublime in the lunar night temperatures, but it would speed up a lot once the sun rose.
Wow, thanks VQ, you have painted a very clear picture that I can understand. And yes, it should be assumed that warming of the water ceases at the moment the lid is removed.

The top of the water will rapidly boil. That phase transformation will remove heat from the bulk, thus lower the temperature. So, boiling will continue until the temperature of the bulk liquid has fallen enough to cause freezing. Then, the remaining ice will sublime over time. Cool.



What would the ice be doing at night, though?
First off, thanks for the detailed response. I can understand your thought process, but I will still need to mull over it a bit. I guess we know what happens to the ice at night...nothing!
http://lcross.arc.nasa.gov/

violentquaker
2010-Apr-10, 01:20 AM
Wow, thanks VQ, you have painted a very clear picture that I can understand. And yes, it should be assumed that warming of the water ceases at the moment the lid is removed.

The top of the water will rapidly boil. That phase transformation will remove heat from the bulk, thus lower the temperature. So, boiling will continue until the temperature of the bulk liquid has fallen enough to cause freezing. Then, the remaining ice will sublime over time. Cool.

No problem. A couple of other people have stated that the water will not actually boil, but vaporize due to surface tension. This was a surprise to me but I have no particular reason to think that they are wrong so please interpret my ideas as fallible.

chornedsnorkack
2010-Apr-10, 10:03 AM
The top of the water will rapidly boil. That phase transformation will remove heat from the bulk, thus lower the temperature.
No, it will remove heat from the top. Big difference. The heat has to be conducted to wherever evaporation happens.


First off, thanks for the detailed response. I can understand your thought process, but I will still need to mull over it a bit. I guess we know what happens to the ice at night...nothing!


When ice has cooled to roughly -80 Celsius, the loss of heat through sublimation becomes equal to the heat loss through IR radiation. As the ice further cools, sublimation heat loss decreases exponentially but thermal radiation only with fourth power of temperature.

-81 Celsius is also the Debye temperature of the ice. So the heat capacity starts decreasing drastically. Cooling from 200 K to 180 K releases 16 times the amount of heat as cooling from 100 K to 90 K... and since heat radiation also decreases 16 times, the time needed to cool from 200 K to 100 K is the same as the time needed to cool from 10 K to 5 K.

Cooling will finally slow down when the temperature is below 3 K, as the ice approaches equilibrium with 2,7 K relic radiation.