|MadSci Network: Physics|
Greetings, Ron: Your co-worker is partly correct, but not completely correct. Before considering water in a vacuum, it is appropriate to consider some of it under ordinary conditions. Think about a container full of water molecules; at any ordinary temperature, the molecules are moving about randomly. Some are moving faster than average; some are moving slower than average. As they collide and interact with each other, the faster ones mostly slow down and the slower ones mostly speed up. Most of the average ones will stay average, but some will speed up and some will slow down. All of that will happen constantly. Now suppose that this container of water is an ordinary drinking glass, surrounded by ordinary air pressure. As you know, if you let the glass of water rest on a table long enough, the water will gradually evaporate away. What is going on, at the molecular level, during evaporation? Well, randomly moving water molecules are by definition moving in all possible directions. One of those directions is upwards, towards the surface of the water, where it meets the air. Air molecules are also moving randomly, some of them downwards against the surface of the water. For the most part, the air molecules and the water molecules bounce off each other, and this is what normally keeps most of the water in the glass. However, because some of the water molecules are moving at higher speed than average, those particular molecules can be said to have a higher temperature than the average. The fastest-moving ones are quite truly and literally "boiling hot". THEY can escape the glass of water -- and they do. Now, since those escaped water molecules are no longer in the glass, the energy they took with them is 'lost' from the viewpoint of the glass. Remember that the temperature of the water in the glass WAS an average of ALL the molecules -- including the ones that have now escaped. Well, since those boiling-hot molecules HAVE left the glass, the temperature of the water in the glass must now be recomputed without them -- and this will be a lower temperature than before. As you know, evaporation is associated with cooling, and the mechanism is as simple as this: If the hotter molecules escape, then the colder ones are left behind, and the average temperature of what was left behind must go down! With respect to the whole glass of water evaporating, there is a little more to the story. Those boiling-hot water molecules that escaped the glass start losing energy to the air just as soon as they enter the air. This makes the average temperature of the air go up a bit, of course. Well, if we assumed that the air temperature was originally the same as the water temperature, then we now have a difference in that the air has warmed a bit, and the water has cooled a bit. Obviously, with random motion still going on, that difference is not going to persist. Higher- than-average-temperature air molecules will collide with lower- than-average-temperature water molecules, and the original overall average will be restored. Which means this description of evaporation has come full circle, because now the glass of water once more has the energy it needs, such that more boiling-hot water molecules can escape. It should now be obvious that after enough time passes, all the water in the glass can evaporate. (Let me note here that considerations of atmospheric humidity have been ignored in this description. Humid air will slow the rate of evaporation, or even inhibit it altogether. But I am ignoring it because it is not very relevant to your Question.) ------------- Now, what happens when the air pressure is reduced? Remember that the air acts as a kind of barrier to the water in the glass, such that water molecules bounce off the air and mostly stay in the glass. Only the hottest water molecules can escape. Well, if there is less air pressure, then there is less AIR above the water in the glass. This makes it easier for water molecules to escape, and another way to say that is: Water molecules that are LESS than the very hottest can now escape the glass. The lessened air pressure will be directly related to an increase in the rate of evaporation of the water in the glass. As before, however, the colder molecules will remain in the glass. Now, there is a pretty good correlation between the amount that the air pressure is reduced, and the boiling point of water; it is well known that the boiling point is lower at high altitudes. That trend will persist all the way to practically zero air pressure (a vacuum). In a vacuum, it is so easy for average-temperature liquid water molecules to escape a glass, that only the very coldest ones do NOT escape. Instead, because they ARE truly cold molecules of water, they clump together to form ice! Therefore your co-worker is correct in saying that the water will freeze, even though it started out at a warm temperature. He is incorrect in implying that the frozen water is warm. The two key things to keep in mind are: (1) Only part of the original liquid water will freeze in a vacuum; (2) The frozen water was ORIGINALLY that cold all along! Only because it was mixed with hot-water molecules (before exposure to vacuum), and there was an average temperature, did we not notice the frozen molecules in the glass of liquid water.
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