"Casey Wilson" wrote in message
...
Background first: Two significant points exist for calibrating a
temperature probe or thermometer without comparing it to a known probe or
thermometer -- freezing and boiling points of water. For the freezing
point, one mixes up a slush of ice and water, the probe should read zero
degrees centigrade when immersed in the liquid. For the boiling point,
stick
the probe in a pot of boiling water and it should read 100 degrees.
Except,
of course, for the barometric pressure. We all know water boils at lower
and
lower temperatures (?? 3 degrees per 1,000 feet??) as altitude is
increased.

Question: Why doesn't the same pressure effect occur for the freezing
point?

It does. For a phase change, there's an equation called the Clapeyron
equation that shows how the temperature varies with pressure.

dT/dP = T deltaV / deltaH

where deltaV is the volume change and deltaH is the enthalpy change (think
of it as energy and you're not too far off) of the phase change.

(It sort of makes sense that the pressure variation of the phase change
depends on what the volume change is. If there's no volume change it
doesn't really matter what the pressure is, as everything happens in the
same space.)

For water - steam, deltaV is very large, so the variation of T with P is
significant.
For ice-water, deltaV is tiny, so the variation of T with P is miniscule.

At about 130 atmospheres, the melting temperature of water drops by 1 degC.

Julian Scarfe