Calibrating temperature probes

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?

Oh yeah, to make this an aviation related topic consider calibrating a
temperature probe or thermometer to use as a reference to check the OAT
gauge in your plane once in a while. You all do that now and then, don't
you?

"Casey Wilson" wrote in message
...
Question: Why doesn't the same pressure effect occur for the freezing
point?

I believe it does. If I recall correctly, the pressure from the blade of an
ice skate is what melts the ice and allows the skate to glide easily across
the ice. However, it requires a much greater change in pressure to make the
same change in freezing point, so it's not relevant for your purpose.

All off the top of my head...look it up if you really want to know.

Pete

"Peter Duniho" wrote in message
...

I believe it does. If I recall correctly, the pressure from the blade of
an
ice skate is what melts the ice and allows the skate to glide easily
across
the ice.

I don't doubt your recall, Peter, as I'm sure I was taught the same.
However, check out:

http://www.physlink.com/Education/AskExperts/ae357.cfm

Julian Scarfe

"Julian Scarfe" wrote in message
...
I don't doubt your recall, Peter, as I'm sure I was taught the same.
However, check out:

http://www.physlink.com/Education/AskExperts/ae357.cfm

Interesting. Oh well...

I'm at least relieved to find that it is indeed the case that the melting
point changes with pressure, and that it changes very little. It just
changes a lot less than I even suspected.

The explanation on that page *is* a little confusing though. I think they
got their math right, but the wording is odd. They start out talking about
the reduction of freezing point, but then later talk about an increase in
the temperature of the ice. The two are not really the same, even if they
result in the same effect.

Pete

"Julian Scarfe" wrote in message
...
"Peter Duniho" wrote in message
...

I believe it does. If I recall correctly, the pressure from the blade
of
an
ice skate is what melts the ice and allows the skate to glide easily
across
the ice.

I don't doubt your recall, Peter, as I'm sure I was taught the same.
However, check out:

http://www.physlink.com/Education/AskExperts/ae357.cfm

Julian Scarfe


I don't buy the physicist's argument. Blades are curved, so there is likely
ten times less surface area on the ice.
--
Jim in NC

http://www.physlink.com/Education/AskExperts/ae357.cfm


"Morgans" wrote in message
...

I don't buy the physicist's argument. Blades are curved, so there is
likely
ten times less surface area on the ice.

....which would make the depression of freezing point 0.2 degC. Still not
enough.

Julian

He is all wet.... Good thing for his employer he is retired.. I hope I
don't use any products he was part of designing...

Skate blades have a rocker bottom, like a boat hull, limiting the area of
steel in contact with the ice, which radically raises the pounds per square
inch of pressure, applied (PSI) at the contact patch... Further, the skating
edge is ground concave, not flat, so that two knife edges raise the PSI to
huge levels at the minute points of contact between the edge(s) and the ice,
liquifying the ice instantly, and the wedging action of the inside profile
of the concavity then squirts the liquid towards the center of the concavity
raising the blade up onto a hydrostatic wedge of water... Same hydrostatic
phenomena that keeps your crankshaft from welding to the rod bearings...
Same logarithmic rise in pressure at the wedge point phenomena that allows a
sharp knife to cut more easily than a dull knife...

As far as unsatisfied hyrdrogen bonds at the interface between ice and air,
I suspect that is true but that is not what makes a skater glide...

Get your kid on skates and take a magnifying glass with you and look at the
fresh skate track and you will see how the ice liquified and then refroze
instantly leaving a different ice surface in the track than on the ice
adjacent to the track... Try the same test with a skate blade ground flat,
or even a little convex, on the bottom (make sure your kid is well
padded)...

Experts!!! ya gotta love em...

Denny
"
Morgans" wrote in message I don't buy the
physicist's argument. Blades are curved, so there is likely
ten times less surface area on the ice.
--
Jim in NC

"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

Julian, that's the complex answer. Remember PV=nRT?
This is what I use...
Change it to PV/T=nR
From this you can write:

(P1V1)/T1=(P2V2)/T2

Julian Scarfe wrote:

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

"john smith" wrote in message
...
Julian, that's the complex answer. Remember PV=nRT?

PV=nRT does not apply to solids or liquids. Hell, it barely applies to
gasses, since they rarely are "ideal".

Pete