thermodynamics - Water in vacuum (or space) and temperature in space

1. 
   

   So, water in vacuum will boil first and then freeze. I don't know how the freeze happens. As pressure lowers to zero, what happened to freezing point? (I know heat taken by vapor, and the water cool down, but I don't think it will be that cold, will it? In vacuum, boiling point is so low that water shouldn't need so much heat as it does in normal pressure, which means vapor actually takes more heat away under normal pressure than in vacuum, so water under normal pressure would be cooler? (I'm guessing)

   
   


2. 
   

   And temperature comes from heat generates by motion of molecules (I guess so), so in vacuum, there is no temp?

   
   



3. 
   

   What happen when I heat up a vacuum tube?

   
   


4. 
   

   Does heat need a medium to “travel"?

   
   

Answer

Conventionally, though with justifications, space is said to begin at the Kármán line which is 100km from Earth surface, i.e., still pretty close. The atmospheric pressure at this altitude drops to about 0.032 Pa (wikipedia), which is still a lot more than outer space (less than $10^{-4}$ Pa according to wikipedia)

The phase diagram of water shows that, at this pressure level, water can exist only as a solid or as vapor, depending on temperature, but not as a liquid. The phase transition between solid and gaz at that low pressure takes place near 200°K (around -73°C), which is not that cold.

So, if you drop in space a blob of water at room temperature and pressure it will instanly start to evaporate (boil) and decompress.

Here I am not sure about what happens. There are accounts from astronauts on the web that explain that the water (actually urine) will first vaporise then desublimate into tiny crystals. But no explanation of the actual physical phenomena that drive it.

My own reconstruction of what could happen (before I saw these sites) is the following.

First the loss of pressure propagates very fast in the liquid (speed of sound?) while loss of temperature (heat) propagates slowly (as all beer lovers know from their fridge). So the boiling will essentially take place uniformly in the whole liquid. Phase transition from liquid to gas absorbs heat, and that is what will cool the water very quickly, as it evaporates.

My guess is also that the energy loss will cool the water down to sublimation temperature (solid-gas transition) before it all evaporates, so that some parts of the liquid may be cooled down to freezing before they have time to evaporate. But as boiling takes place everywhere, it actually breaks the remaining water into tiny fragments that cristallize, and possibly also collect some of the vapor to grow.

Anyway, you apparently get snow.

But the cooling is due to evaporation, which is very fast, much more than to radiation which has hardly any time to take place.

Numerical evaluation

We analyze what becomes of available heat to understand whether some water freezes directly. This is a very rough approximation as the figures used are actually somewhat variable with temperature, but I cannot find the actual values for the extreme temperature and pressures being considered.

The specific latent heat of evaporation of water is 2270 kJ/kg. The specific heat of water is 4.2 kJ/kgK Hence, evaporating 1 gram of water can cool 2270/4.2 = 540 grams of water by 1°K, or 5.4 grams by 100°K which is about the difference between room temperature and water (de)sublimation temperature in space. So my hypothesis that there is not enough heat available to vaporize all the water is correct, as only about one sixth of the water can be vaporized with the available heat.

Out of 5.4g of water, 1g will evaporate, though may cool down to just above the sublimation temperature of 200°K, while the remaining 4.4g will be cooled to sublimation temperature without vaporizing, yet. The remaining 4.4g cannot remain liquid, hence, one part freezes, thus freeing some latent het for the other part to vaporize. The ratio between the two part is inversely proportional to the specific latent heat for freezing and vaporizing.

Latent heat for freezing is 334 kJ/kg. The sum of both latent heat is 2270+334=2604 kJ/kg. These figure are very approximate. As a sanity check, the latent heat of sublimation of water is approximately 2850kJ/kg (wikipedia), which show that the figures are probably correct within a 10% approximation.

The ratio divides the remaining 4.4g into approximately 3.8g that freezes and 0.6g that evaporates, making it a total of 1.6g of vaporized water.

So, skipping a quick calculation, we find that about 70% of the water freezes into some kind of snow, while the remaining 30% are vaporized. And it all happens rather quickly.

I was actually uneasy about this account of astronauts stories of water boiling and then desublimating at once, because that would leave us with all the heat to get rid of very quickly. How? Does anyone have a better account?

A last remark is that there always will be some part of the water that gets frozen. I thought initially that very hot water might provide enough heat to vaporize itself completely un low pressure. The critical point of liquid water is at 650°K (with a much higher pressure than you care to create in space: 22MPa), which is only 450° above the sublimation temperature. But the water should be cooled by 540° to provide enough heat to evaporate completely. So the water temperature will drop to the sublimation threshold before enough heat can be supplied to evaporate it completely. This problably a very simplistic analysis, though. I leave the rest to specialists.