### Re: Will H2O freeze at a warmer temperature if it is in a vacuum (low press.)?

Date: Sun Mar 11 15:27:38 2001
Posted By: Vernon Nemitz, , NONE, NONE
Area of science: Physics
ID: 984085738.Ph
Message:
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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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