I am based far away from the icy storm currently blanketing the US - the 'polar vortex'.

However, I have seen in the TV news footage of reporters throwing boiling water into the air, the water freezing immediately to form snow.

I'm curious as to whether this would also work with cold water. If not, I presume that it is because of the, as yet unexplained, Mpemba effect.

Has anyone in the US tested this?

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    $\begingroup$ Is it a Polar vortex or a jet stream? newscientist.com/article/… $\endgroup$ – Jitter Jan 8 '14 at 15:03
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    $\begingroup$ I assume that there is no such thing as a polar vortex, formally. Hence my use of quote marks. Thank you for enlightening me to the real forces at work here, though! $\endgroup$ – Charon Jan 8 '14 at 15:05
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    $\begingroup$ I think there is a Polar vortex but it's over the pole and much higher up. $\endgroup$ – Jitter Jan 8 '14 at 15:18
  • $\begingroup$ "I assume that there is no such thing as a polar vortex, formally. Hence my use of quote marks. Thank you for enlightening me to the real forces at work here, though!" Um, why would you assume that? Based on what?!? In fact, of course there is such a thing as the polar vortex, as any basic meteorological text book will confirm. $\endgroup$ – Mark Choi Jan 9 '14 at 13:51

At first blush, the Mpemba effect does seem to be in play here. Unfortunately, the Mpemba effect is not very well specified and is generally studied over longer timescales than occur in the "boiling water to snow" phenomenon.

A quick quibble: the boiling water does not become snow, but a cloud of very small ice crystals, so it is more akin to cloud formation than snow formation.

The question is: Why does boiling water make an impressive cloud of ice crystals so rapidly instead of forming a layer of ice and then falling to the ground, and why doesn't cold water form the same impressive cloud?

Most explanations of the phenomenon allude to differences in surface area but neglect to mention how boiling water is able to increase its surface area more than cold water.

This site, while not a peer-reviewed publication, does a good job of discussing various possible contributions.

Let us try to figure out what’s going on, by evaluating various hypothetical contributions. I suspect item (1) is the main contribution, while the others are relatively minor:

  1. In order to make a really impressive fog, there is a huuuuge premium on small particles. Suppose you have a constant amount of water, but you are given the choice of a single 100-micron particle, a thousand 10-micron particles, or a million 1-micron particles. The smaller, more-numerous particles will be vastly more effective at scattering light. They will also stay in suspension better.

    The scenario I have in mind is that a modest percentage (less than 1/3rd, as calculated in section 2) of the water gets turned to vapor, high-density saturated vapor at fairly high temperature. This vapor then cools by contact with the surrounding cold air, becomes supersaturated, and condenses into lots of reeeeally teeny particles, far teenier (and far more numerous) than you could produce just by the violence of throwing the water.

    The other more-than-2/3rds of the water doesn’t disappear. It is still there ... it is just relatively inconspicuous compared to the spectacularly dense cloud.

    We can see why having very hot water is important: the vapor pressure is a steep exponential function of temperature, and we want to push out as much vapor as we can.

  2. There is a very interesting factor we might call the "sizzle" effect or the "rocket" effect. Suppose a drop of hot liquid breaks into two droplets by accident; the question is whether the droplets will coalesce to re-form a larger drop. If the water is hot and the ambient air is very dry, each droplet will be outgassing like crazy. The outward flow of vapor will tend to push the droplets apart. (A single isolated droplet will feel no net force, since the outgassing shoots off in all directions equally. But when two such droplets get near each other, each is repelled by the other.)

    This effect is also exponentially favored by high temperatures.

  3. Hot water has a lower surface tension. Of course this is not the whole story, but it is not negligible either. After all, surface tension is what holds drops together; without surface tension dispersal would be super-easy. OTOH the surface tension does not go to zero even at 100C.

  4. Hot water has lower viscosity. This factor is not overwhelmingly significant; the viscosity of boiling water is not overwhelmingly less than for cold water.

  5. For completeness let’s consider the hypothesis that hot water is useful because you start out with large drops but they become small by evaporation. That’s a nice guess, but the numbers don’t work out. Only about 1/3rd of the water can be lost to evaporation in this way. Sure, some evaporation occurs, making the droplet get smaller, but only by about 13 percent (since the cube root of 2/3rds is 0.87).

  6. It may be worth noting that after the ice cloud has been created, the cloud will be very persistent, much more persistent than the water-droplet fog you’d get under warmer conditions. A big contributing factor is the fact that ice doesn’t stick to ice very well when it’s cold, unlike water droplets that instantly bond due to surface tension effects. This has got nothing to do with the nominal topic of this thread ("how cold does it have to be to make water disappear") but it does address the implicit question of why it is fun to disperse water into the air.

IMHO we now have a pretty decent picture of what is (and isn’t) happening. Remember, the objective is to make an impressive fog. It is exponentially important to have hot water in order to drive a lot of H2O into the vapor phase. It is important to do a decent job of tossing the water, in order to create sufficient surface area for evaporation to occur. It is necessary to have reasonably cold air, to cause recondensation to occur. Extreme cold is not necessary, but doesn’t hurt, and an ice-fog will be more persistent than the other kind of fog.

--- John S. Dunker, 2004, "How to Make Fog"


I have a few things to add:

In videos of the phenomenon, the cloud forms close to the surface of the liquid, leading to a "telescoping" effect as the water moves through the air.

One possible explanation of this is that the latent heat of freezing released by the water closest to the cold air heats the water further from the cold air. This layer of "inner water", since it is already close to boiling, will evaporate, dispersing the water molecules and quickly freezing, exposing and heating another layer, and so on.

It may seem strange that freezing water would heat surrounding water, but remember that the latent heat of freezing must go somewhere, and liquid water is more conductive than air. That, in addition to the rapid expansion and insulation provided by the water vapor (remember the Leidenfrost effect) is what allows it to disperse so quickly without bonding into a mass of ice.

I don't have enough of a background in atmospheric physics to point to relevant literature, but I hope this provides some insight into the subtleties of the phenomenon.

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This is largely due to the high surface area of the water when it is exposed to the cold air. Imagine that when you throw the boiling water into the cold air, you get a many small individual dropletts of water that freeze quickly into snow. They freeze quickly as they have little energy due to their small size.



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    $\begingroup$ Yes, but is that any different to cold water? Why would hot water have a larger surface area than cold water when thrown? $\endgroup$ – Charon Jan 8 '14 at 15:06
  • $\begingroup$ This has little/nothing to do with the polar vortex itself or Mpemba effect. It only has to do with the temperature of the air. $\endgroup$ – kaine Jan 8 '14 at 15:07
  • $\begingroup$ Cold water doesn't have as much surface area, it will form larger droplets. $\endgroup$ – kaine Jan 8 '14 at 15:07
  • $\begingroup$ Why does cold water have a smaller surface area? $\endgroup$ – Charon Jan 8 '14 at 15:09
  • $\begingroup$ Hmmm, OK, I see. So, when the water is thrown in the air, it breaks into many small droplets, each of which releases a significant amount of water vapour. As the air is so cold, the water vapour precipitates out and forms crystals, rather than staying as vapour or condensing into liquid. $\endgroup$ – Charon Jan 8 '14 at 15:16

There is no such thing as the Mpemba effect, at least not as reported. It is a syndrome indicative of poor experimental conditions and rather misguided "physicists" - see the answers to What is the status of Mpemba effect investigations?

Cold water freezes faster than hot, all things being equal. If boiling water helps make snow any better, it's because throwing equal amounts of boiling and cold water into the air probably results in most of the boiling water evaporating, leaving only a little bit left to freeze, while the cold water is more uniformly chilled.

Psychologically, even if both temperatures froze equally well, using boiling water makes the display more impressive.

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Boiling, or hot, water has a lower surface tension and a lower viscosity. Because of this, you get more, smaller droplets. The best xplanation I found is from Fuck yeah Fluiddynamics!:

Several effects are going on here. The first thing to understand is how heat is transferred between objects or fluids of differing temperatures. The rate at which heat is transferred depends on the temperature difference between the air and the water; the larger that temperature difference is the faster heat is transferred. However, as that temperature difference decreases, so does the rate of heat transfer. So even though hot water will initially lose heat very quickly to its surroundings, water that is initially cold will still reach equilibrium with the cold air faster. Therefore, all things being equal, hot water does not freeze faster than cold water, as one might suspect from the video.

The key to the hot water’s fast-freeze here is not just the large temperature difference, though. It’s the fact that the water is being tossed. When the water leaves the pot, it tends to break up into droplets, which quickly increases the surface area exposed to the cold air, and the rate of heat transfer depends on surface area as well! A smaller droplet will also freeze much more quickly than a larger droplet.

What would happen if room temperature water were used instead of boiling water? In all likelihood, a big cold bunch of water would hit the ground. Why? It turns out that both the viscosity and the surface tension of water decrease with increasing temperature. This means that a pot of hot water will tend to break into smaller droplets when tossed than the cold water would. Smaller droplets means less mass to freeze per droplet and a larger surface area (adding up all the surface area of all the droplets) exposed. Hence, faster freezing!

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