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In the related questions Can you take a bath on Mars? & How would swimming on Mars feel, given the lower gravity? we learn that bathing and swimming on Mars is much like it is on Earth. Some comments on those questions suggest that if the pool is exposed to Martian atmosphere it would boil away quickly. What would really happen if your swimming pool or fish pond was suddenly exposed to Martian atmosphere?

Some comments suggest the pool magically appears and boils away, but here is a realistic scenario.

A swimming pool (Olympic size) or fish pond is full of water {25–28 °C (77–82 °F)} and covered by an inflatable dome that keeps Earth atmosphere over the pool. A event occurs: which cuts off power to heat and circulate the water, and the dome is torn badly or complete removed. What happens to the pool now?

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  • $\begingroup$ The Martian atmosphere is about as dense as the Earth's atmosphere at 70,000 feet. $\endgroup$ – Howard Miller Aug 18 '16 at 21:23
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    $\begingroup$ Relevant google query: How not to open a pressure cooker. $\endgroup$ – David Hammen Aug 19 '16 at 10:46
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    $\begingroup$ Something else is that depending on how deep the pool is, some of the water will still be under pressure from the weight of the water on top of it. I don't think the entire volume could boil simultaneously, only the surface would be able to. By my crude calculations, if the pool is 2m deep then the bottom of the pool would be under ~6% atm, and the boiling point would be around 35C. The surface would boil vigorously but I don't think there could be a kablammo explosion because the weight of the water would be greater than the vapor pressure for the bottom layers. $\endgroup$ – Blake Walsh Aug 19 '16 at 12:22
  • $\begingroup$ @BlakeWalsh some of the comments on the related questions discussed that also. Do you want to try and include it in a new answer? $\endgroup$ – James Jenkins Aug 19 '16 at 13:07
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The other answers forget latent heat/enthalpy of phase change, and many neglect pressure rise with column of water. :)

Enthalpy of evaporation of liquid water is considerably higher than its latent heat of fusion. It takes way more energy to bring a 100C liquid water into a 100C steam at 1 bar, than is released as same mass of 0C water freezes into 0C ice. The values vary with pressure, but not by a very large margin.

Water pressure in Earth gravity changes by about 1bar per 10 meters. With Mars, that will be a ballpark of 1 bar, or 100KPa per 30 meters - meaning scarce centimeters deep the water is above the triple point pressure. That means the boil-off would be occurring only near the very surface. It would also drain heat very rapidly, freezing water below. It would very quickly form a layer of ice, which would then proceed to evaporate very slowly, draining heat from water below (ice is a lousy heat conductor, so it would take a lot of time to drain the energy needed for sublimation).

Essentially, your pool's thin top-most layer would boil rapidly for a short moment, and a very short distance below the boiling layer a thin layer of ice would form, which then, subsequently, would thicken over time, as water deeper below freezes, and its level would keep dropping slowly (much slower than ice layer thickness growth) as its surface would evaporate.

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  • $\begingroup$ Unless there is a flaw in your answer this seems, like the best. $\endgroup$ – James Jenkins Jan 10 '17 at 18:52
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    $\begingroup$ @JamesJenkins: There could be effect connected with dynamics of boiling/evaporation. Possibly quite a bit of steam would turn into snow during the initial boil-off; it's hard to predict the actual mechanics and how the very surface would appear immediately afterwards; evaporation rate afterwards would also be related to amount of heat provided from other sources - if the pool is shaded from sunlight, it would be much slower. $\endgroup$ – SF. Jan 10 '17 at 18:57
  • $\begingroup$ also, the amount of heat that can be drained from water by freezing it, and from the surrounding ground is just a drop comparing to what is needed to evaporate comparable amounts of water, so other sources of heat would play dominant role in evaporation of the ice. $\endgroup$ – SF. Jan 10 '17 at 18:58
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    $\begingroup$ @TylerH: Observe the graph in Dean McGregor's answer below mine. This depends on pressure primarily - the triple point is at 611 Pa - below it liquid water can't exist. Mars atmosphere is slightly less. As you go down the green area of the graph (liquid) it tapers to a point, the temperature difference between vapor (yellow) and solid (blue) shrinking. As you heat or cool things or pressurize/depressurize, you can move over the field. But each of these black borders should be imagined as a fence/step, skipping which requires extra heat (going or releases extra heat. $\endgroup$ – SF. May 10 '18 at 15:03
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    $\begingroup$ ...and so, as pressure drops, less pressurized water crosses over to the yellow side - but needs heat, which it sucks out of more pressurized water. Eventually, as all the water (steam too) cools to 0C, the water on the surface (least pressurized) can't suck more heat from liquid water below without pushing it over to become ice. And no more water can become steam, until more heat is extracted - forming more ice, until the layer of ice is thick enough that pressure on its bottom is still "in the green". And the ice surface will evaporate as fast as heat is provided - that is not fast at all. $\endgroup$ – SF. May 10 '18 at 15:11
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Here's a phase diagram of water from here.

enter image description here

If you find Mars's atmosphere of 600Pa on the Y-axis and track right you'll see that H20 can't be water on Mars. It can only be ice or vapor.

In your example where you have a dome of Earth like atmosphere protecting your Martian pool that blows away somehow then you'll either have a skating rink or an empty hole but no water. Going by the average Martian temperature you'd have a skating rink but once the temperature exceeds the sublimation point all your H2O would turn to vapor and that'd be the end of it.

Edit:

The other answer asserts that it would first boil and perhaps there'd be some H20 left as ice after some of it boiled off. I suppose this depends on the exact circumstances of losing the dome. The way OP asks the question, it seems that first heat is lost and then the dome goes away. If the water has already cooled to below the sublimation point before the pressure drop occurs then there's no need for it to boil. If everything is working fine and then the dome flies away while the ambient temperature is above boiling temperature then clearly it would boil off. However, if the ambient temperature is below sublimation when the dome is lost I don't think it is clear that the water would boil rather than freeze. The very act of decompressing would cool the water so we can't simply say that the pressure can drop faster than the water cools because the pressure dropping speeds the cooling.

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    $\begingroup$ The other answer made the point that when the dome rips, the water will be at 25 Celsius, so it will boil until it drops to it's freezing point, which seems to be right at the triple point there at 600 Pa and 0 Celsiuis. $\endgroup$ – kim holder Aug 18 '16 at 17:53
  • $\begingroup$ @kimholder well the other answer makes the assumption that the dome blows away while the ambient temperature is above the sublimation point which, based on the mean, isn't likely. Further, until the edit, it didn't explicitly say that the pool couldn't remain as water which wasn't clear to me until I looked up the phase diagram. $\endgroup$ – Dean MacGregor Aug 18 '16 at 20:23
  • $\begingroup$ That's a legitimate point. I interpreted that as meaning the water doesn't continue to be heated. $\endgroup$ – kim holder Aug 18 '16 at 20:55
  • $\begingroup$ The question says "an event happens", not "multiple events happen, separated by some indeterminate amount of time". I would take that to mean that pool heating stopped at more or less same time that the dome lost integrity. $\endgroup$ – David Hammen Aug 19 '16 at 10:24
  • $\begingroup$ @DavidHammen That's a fair point but I addressed that in the last two sentences of my edit. $\endgroup$ – Dean MacGregor Aug 19 '16 at 12:24
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According to Wikipedia:

The atmospheric pressure on the Martian surface averages 600 pascals (0.087 psi; 6.0 mbar), about 0.6% of Earth's mean sea level pressure of 101.3 kilopascals (14.69 psi; 1.013 bar).

This chart seems to indicate that the 77°F (25°C) water would be above its boiling point by more than 75°F (the granularity of the chart does not give more info).

This would result in a (perhaps explosive) boiling of the pool until it had released enough energy to be below its boiling point in the 6°F to -24°F (-14°C to -31°C) range. At this point the remaining water may freeze and sublimate in the future.

But, according to Wikipedia, the surface of Mars is between -226°F (-143°C) and 95F (35°C), so depending on conditions the pool may never cool below its boiling point, and all of it simply evaporates.

@Dean MacGregor correctly points out that at 600Pa water will never be a stable liquid, so the final result (after 'explosive' boiling) will eventually be that all the water evaporates, or a portion boils off, allowing the remainder to cool to ice, which may or may not sublimate. All depending largely on the temperature of this particular part of Mars.

As a demonstration, this is a video of water being subjected to a vacuum at room temperature. On those conditions the water boils, until it cools to freezing.

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Depending on how deep the pool is, some of the water will still be under pressure from the weight of the water on top of it. I don't think the entire volume could boil simultaneously, only the surface would be able to.

A very crude way to calculate the water pressure is to take the depth of the water column, divide it by 10m (on earth, 10m of water equals 1atm), then multiply by 0.38: the strength of martian gravity relative to earth gravity. For example at 1m the pressure would be 0.038atm, and at 2m it would be 0.076atm.

The temperature required for water to boil at these pressure would be be 26C at 1m and 38C at 2m (I found these numbers on this site). If the pool temperature is ~25 degrees then the top 1m of water would immediately start boiling. An Olympic pool is 2-3m deep which would leave 1-2m of water which is under enough pressure to prevent immediate boiling.

In accordance with this analysis it would not boil explosively because there just wouldn't be enough vapor pressure at that temperature to "blow apart" the pool against the force of gravity. The top 1m of water would merely boil vigorously, rapidly cooling down in the process.

As discussed in other answers, the real question is whether the water would boil away to nothing or freeze, I suspect it would freeze and it might be that a liquid layer remains trapped under the ice, which then freezes solid. Over time the ice pool will probably sublimate away.

Disclaimer: I am not a physicist. My analysis may be flawed.

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