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.
The explanation is using an energy argument. That for the normal case of a submerged piece of wood, you can assume that if the wood and a parcel of water above it switch places, then the water (which is heavier/more massive) drops in the gravitational field releasing potential energy. This release is not offset by the rising wood since it is not as massive. This energy imbalance goes into the kinetic energy to move the log (and the water).
This argument says that in the case of the rotating log, none of the water can move downward to release potential energy. Since the water stays in the same place, no energy is released, and there is no energy to move the log.
Why doesn't the object just stay where it is?
Because forces will combine to move it to a position with less energy. You can do a free-body diagram of a ball rolling downhill to see that the net force is down the hill. But you can also simply say that the ball is going to move in a way that lowers its potential energy (which is downward in the gravitational field). Using that argument for the cylinder says that there is no position it is free to move to that lowers the potential energy, so no movement will happen.
Best Answer
Ha! Seems the search phrase "expansion of a gas into vaccum" was key!
I found two sources which agree with each other: H. P. Greenspan and D. S. Butler (1962). On the expansion of a gas into vacuum. Journal of Fluid Mechanics, 13 , pp 101-119 doi:10.1017/S0022112062000543 and Zel'dovich - Physics of shock waves and high-temperature hydrodynamic phenomena.
They give the velocity of the propagating shock/air-vacuum-interface as $u_{max}=\frac{2}{\gamma-1}c_0$. With $\gamma=1.4$ for air and $c_0\approx340\,\mathrm{m/s}$, this gives
$u\approx1700\,\mathrm{m/s}$.
Yeah, it's a solution, and I'm glad about that, but it does not significantly change the fact that Randall still seems to have gotten something wrong by a couple orders of magnitude. At this speed, the shock wave covers the $\approx 5\,\mathrm{cm}$ of vacuum in the glass in about $30\,\mathrm{\mu s}$, which is a factor 60 longer than the $500\,\mathrm{ns}$ he gives.