Calibrating temperature probes

[Casey Wilson]

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?

[Peter Duniho]

"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

[Julian Scarfe]

"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 Scarfe]

"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

[Peter Duniho]

"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

[L Smith]

Casey Wilson wrote:

> 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. It's just that the melting point curve is a lot steeper than
the boiling point curve,
so it takes a lot more pressure for the same effect.

An interesting experiment can demonstrate the effect. Suspend an ice
cub between
two supports. Loop a fine wire over the ice cube, and use it to suspend
some significant
weight (say 0.5-1 pounds). After some time the wire will be in the
middle of the ice
cube, with ice above and below it. After even more time, the wire will
"cut" through
the cube and fall out the bottem, leaving behind a solid block of ice.
(The pressure
of the wire on the ice raises the melting point, so that the ice in
immediate contact
with the wire becomes liquid. After the wire passes that spot, the
melting point
returns to normal and the water re-freezes.

Rich Lemert

>
>

[L Smith]

Nomen Nescio wrote:

>
>> Question: Why doesn't the same pressure effect occur for the freezing
>>point?
>>
>>
>
>The vapor pressure of ice (and ice -water mix) is effectively 0.
>
>
Irrelevent! The melting point involves an equilibrium between the
solid ice and the liquid
water. The vapor is totally irrelevent (except at the triple point,
where all three phases
are in equilibrium).

Rich Lemert

[john smith]

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

[Morgans]

"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

[Peter Duniho]

"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

[Julian Scarfe]

> > 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

[Dennis O'Connor]

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
>
>

[Julian Scarfe]

"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