# A water drop in vacuum

Let's imagine the following situation:

At an initial moment $t=0$, a large water drop with diameter for example $D=10\ \text{cm}$ is placed in deep space (Say an astronaut is experimenting). Let's the initial temperature of the drop be moderate $T_0=283 K$ and the drop itself is at rest at $t=0$. What will happen with the drop? Maybe it will decompose into smaller droplets while boiling? Or maybe it is going to flash-freeze and the ice shell is forming? Or maybe something else?

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If there is one thing that will not happen, it is flash freezing. Heat transfer in deep space is a major problem. The only way to conveniently transfer heat is radiation which is quite hard for a cool matter. Other than that, I do not know. – Cem Nov 19 '10 at 11:02
It will most likely boil and decompose into smaller droplets because of the huge amount of gas being created in a vacuum. Almost instantly I might add. – Cem Nov 19 '10 at 11:04
Boil isn't quite the right word as boiling requires the average energy of each particle (the macroscopic temperature) to be larger than the escape energy of the liquid, and generally requires an external energy source. – Dom Nov 19 '10 at 12:56
@Dom, this is wrong! Water with 238 K placed in vacuum will boil (and cool thereby) vigorously! – Georg May 11 '11 at 9:46

I think what would happen is that any water molecule with enough energy to escape the surface tension would escape. Because there is no air to provide the water molecule with a way of turning round and going back in, it would permanently leave.

This means that the highest energy molecules would selectively evaporate, lowering the average energy of the remaining water. This process is known as evaporative cooling, where you selectively remove the most energetic molecules (and yes, it's what happens when you blow on a hot drink).

Depending on the size, shape and method of getting it into the vacuum at that temperature (in a jar, take the jar away, blasted out a tube with gas) it may or may not freeze on its way to total evaporation. This is assuming zero pressure in outer space and close enough to zero temperature. If you look at the pressure / temperature phase diagram for water (wikipedia page of generic phase diagram) it depends if the rapid drop in external pressure causes the evaporative cooling to drop the internal temperature fast enough to briefly solidify on it's way to pure gas. Carbon dioxide will go from solid straight to gas at atmospheric pressure, (smoke machines) and, as the pressure is so low, if the water did freeze solid, it wouldn't be for long, as it would continue evaporating in such a strong vacuum.

Edit: Fixed spelling typos.

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The water drop will immideately start to boil because below $0.006~\mathrm{atm}$ it boils even at $273~\mathrm{K}$. However, boiling in vacuum is something special since there is no buoyancy that would force bubbles to leave the droplet. I think that the process will look similar to the one shown in this beautiful video and the droplet will soon end up as a ball of ice foam.

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I'm not to sure it would boil, because the effects of surface tension would raise the internal pressure above zero. Also in order to have boiling, you have to overcome surface tension. Surface tension basically means that the pressure inside a microbubble is greater than the fluid pressure by an amount that is proportional to the inverse of its diameter. So
you need to have seed bubbles of a certain size inorder to get internal bubbles to grow.

I think it would cool by a combination of radiation (Stefan-Boltzmann) and evaporation. It wouldn't flash freeze, but over perhaps a hour or less it would start to freeze. So we would end up with a nearly spherical ice ball. It is easy to compute the cooling rate from radiation, but I'm not sure of the evaporative contribution to the cooling rate.

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