Calibrating temperature probes

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

"Nomen Nescio" ] wrote in message
...

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

The key word here is SAME.

As temp. approaches 0 deg.C, vapor pressure approaches 0 mmHg.

No it doesn't. The vapor pressure of water (and ice) at 0 degC is about 6
mbar (about 4 mmHg). There's nothing special about 0 degC as far as the
vapor pressure is concerned.

Since the boiling
point is the temp where vapor pressure equals atmospheric pressure. The
"SAME
PRESSURE EFFECT" is effectively nonexistent.
Over an atmospheric pressure ranging from say 5 - 14.7 psi, the change in
the melting point
of ice is virtually nonexistent and can be ignored for all practical
purposes in answering
the original question.
You're trying to complicate a question that can be answered quite simply.

But the whole question is *why* is "the change in the melting point of ice
virtually nonexistent". As far as I can see it has nothing to do with the
vapor pressure.

While Rich's comment was not the friendliest I've seen on Usenet, he does
have a point.

Julian Scarfe