Ask A Scientist General Science Archive

name         Harold
status       other
grade        other
location     Manitoba
   
Question -   This morning in Thompson Manitoba, Canada (where I live)
the outside air temperature was minus 40C.  I threw a mug of water very
near 100C into the air.  Almost none of the water hit the
ground.  Instead, it formed an instant cloud and disappeared.
Is this evaporation, or freezing followed by sublimation, or a combination
of the two?
From reading your site (post by Jim Swenson in November 2004), I
understand that this is likely evaporation is a function of both the
vapour pressure which is a function of the water temperature, and
exchange rate of the air.  Is there a table for the exchange rate of air
at different temperatures?
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Wow.  Great phenomenon.

There is not much difference between evaporation and sublimation, in my mind.
I think of both as "evaporation"; the only difference is whether the 
condensed phase is rigid or fluid, which does not matter much to the 
evaporation.

The only question is, does the air cool the hot water aggressively enough 
to make a frozen skin on the water?
Given the near-100C temperature of your water, I would guess that none of 
the "original bulk" of the thrown water freezes.
I think that evaporating water will keep the air a millimeter around the 
ice above freezing,
because it will release heat by re-condensing and/or freezing as small 
particles in the air,
and so the water surface itself will experience relatively low thermal flux.
I suspect this stage is the impressive one for fog formation.

Even after the air immediately touching the water reaches down to 0 degrees C,
You have the situation of comparative thermal conductivities.
The thermal conductance of the outer 1mm layer of water is about 10 times 
greater
than the thermal conductance of the closest 1mm blanket of air.
And it is likely to be a stagnant blanket, for the innermost 0.2-1mm of air,
unless the water is thrown faster than I think we can do.

Evaporation from the water surface will be a much more powerful cooling agent,
until the water is fairly cool itself.  Not sure what temperature that 
implies.
But that cooling would be why some ice comes down,
and if this happens, some skin-freezing must happen in the later stages of 
the water's arc,
despite all my arguments above.
The water has a finite thermal mass, whereas the air it flies through is 
constantly replaced.

If you poured hot water into an exposed pie tin while a cold breeze was 
blowing,
perhaps you could watch the whole evolution much slower.
You could control the time it takes by the depth of water you pour.
Perhaps there would be a noticeable change in the quantity or character of 
the fog at some point.
This point might be when the surface freezes.
Or perhaps well before the surface is frozen,
at a time when the local warming and evaporation rate is diminished,
so the fog forms as re-condensation onto ice crystals rather than 
water-droplets which later freeze.

Jim Swenson
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Ahhh! Cold is relative, a state of mind. The vapor pressure of water
at -40C (which happens to be the same as -40F) is essentially zero. When
you toss the water into the air it evaporates rapidly, so rapidly that it
disappears, except for a transient fog possibly until the vapor pressure of
the remaining (now ice for the most part) water equals the vapor pressure
of -40C. This rapid evaporation absorbs heat from the surrounding air
(about 10 kcal for every 18 ml of water evaporated). If you have enough
water initially this evaporation will cool the remaining ice to the lower
temperature, hence you see some frozen water remaining. Regarding the rate
of evaporation, this is  a much more difficult quantity to measure because
it depends upon so many variables that are difficult to control -- some of
these variables are:
temperature, quantity of water, surface area of the water air interface, the
speed of the air flow, the temperature of the air flow (which in turn
depends upon the rate of evaporation), ... The list is long. Many inquiries
are received asking for "rates of evaporation" of some substance.
Unfortunately, rates of evaporation, freezing, or melting are difficult to
make in any meaningful way -- too many uncontrollable variables come into
play.

Vince Calder
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