peters@cubsvax.UUCP (01/03/84)
My reply to message from fortune!norskog (Note: I'm sending it to Lance by mail, but also posting it to net.physics & net.followup because I think people following the discussion will find it interesting.) >>fortune!norskog Jan 1 13:03:00 1984 >> >>I am confused by cubsvax!peters response. I had assumed that the >>phenomenon described occurred because under pressure, the freezing >>temperature dropped below ambient temperature, which was slightly below >>0 C. Is this in fact what the happened? (Is this what supercooling means?) >> >>Please reply by mail, with polite language, >> Lance Norskog >> Fortune Systems >> {hpda,harpo,sri-unix,amd70,ihnp4,allegra}!fortune!norskog The Clapeyron equation gives the relation between pressure and melting point, given the heat of fusion and the volume change on fusion. (Note that fusion = melting.) This equation is: (dp / d ln Tf) = (delta Hf) / (delta Vf) Since the enthalpy (heat) of fusion is always positive, the sign of the derivative on the left depends on the sign of delta Vf, which is negative for water (a model for Coca Cola!), though positive for most things. Negative delta Vf means the substance contracts on melting (ie, the volume change is from a larger to a smaller volume on melting). This means that raising the pressure will decrease the melting point, and the equation tells how much. Here is a table of computed values based on integration of the above: Pressure (atm) Tf=Melting Point (deg C) 1 0 50 -0.4 100 -0.8 500 -3.9 1000 -7.7 1500 -11.5 Now, I believe the pressure of a Coke bottle is closest to 50 atm of the numbers listed above (that's 750 psi; I really think it's closer to 150 psi). According to this, the bottle should freeze at -.4 deg C, or about 31 deg F; if it does not, it is supercooled. This is a phenomenon whereby the thermodynamic state of minimum energy can not be achieved because of kinetic (rate) factors, such as the rate of rearrangement of the molecules of a syrupy (viscous) liquid to the positions they occupy in the crystal. Under proper conditions, pure water can be supercooled to -80 deg C. Glass is a supercooled liquid. Honey on a cold day is a supersaturated solution -- similar, except the phenomenon is phase separation rather than freezing. Actually, the dissolved salts and sugars in Coke further lower the freezing point, but not by a terribly large amount, so the above numbers, for pure water, do not apply exactly. The argument still holds, however. Given appropriate disturbance, such as the agitation of the liquid or the introduction of an active surface -- such as a bubble or (especially) a seed crystal -- the supercooled state will "collapse" into the thermo- dyanamically favored crystalline state. I contend this is what happens when the top is opened and bubbles of CO2 are released. ********************************************** Even more incidentally, it is sometimes asserted that the pressure of a skater on the ice melts some of the ice beneath the skate, causing a relatively frictionless surface to be continually formed. The above equation and table shows that this notion is false. A 300 lb skater on a single blade 1/16 by 10 in exerts a pressure of 480 psi, or 33 atm. If the ice started out colder than about 30 def F, the pressure would not be enough to melt it. The frictional heating of the surface by the moving skate may be the real answer. {philabs,cmcl2!rocky2}!cubsvax!peters (Peter S Shenkin; Dept of Biol Sci; Columbia Univ; NY, NY 10027; 212-280-5517)
halle1@houxz.UUCP (J.HALLE) (01/04/84)
The comment about the skater is wrong on two counts. First, the pressure exerted is much greater than given. Skates are hollow ground, making the edge much less than 1/16". That explains what is meant by "inside edge", eg. Why else does sharpening the skates work so well? Friction cannot be the process, since that would melt the ice you're leaving, not the stuff you're on. Second, the ice is very close to the freezing point. The water underneath is above the freezing point, so the ice cannot be much colder. The heat loss comes from the water, not the ice. (ultimately, of course, not directly) Even if the ice is not over water, in most instances it will still be very close to the freezing point.
usenet@abnjh.UUCP (usenet) (01/04/84)
In fact, the area of contact of a skate blade with the ice it is
skating on is very small indeed. 10 inches long by 1/16th inch wide is not
even a good estimate. The actual area of contact is one or the other
of the sharp edges of the blade, not the flat surface. Schematicly,
it looks like this:
\ \ | |
\ \ not like this: | |
\ / | |
\ / | |
-----------\/------------- --------+-----+-----
[[Boy, that's the last time I try to use VI for drawing pictures.]]
That's why a 'dull' skate does not move as smoothly or as fast.
I don't know how to even guestimate the true area of contact, but
you can see it will be very much smaller than 0.625 inches.
Rick Thomas
ihnp4!abnji!rbt
ihnp4!abnjh!usenetbarryg@sdcrdcf.UUCP (Barry Gold) (01/05/84)
There is more involved in ice skating than just the effect of pressure on
the freezing point of water. Normal ice found on ponds, etc. is ice I
("ice one"). A relatively small amount of pressure will convert it to ice
II. (By relatively small, I mean an automobile tire - while the weight is
much larger, the surface area is enormously larger than a hollow-ground ice
skate). Ice II has a melting point several degrees lower than ice I and
consequently melts. Heat loss to the surrounding ice I and air will
rapidly freeze it again, but meanwhile the skater gets hiser low friction
surface. (As does the unfortunate fellow without chains on his car.)
barrypeters@cubsvax.UUCP (01/05/84)
Whoops! I just replied to my mail copy of a followup to my comments
about skaters and the Clapeyron equation; now I see it's also been
posted, so I'll try to reconstruct my mailed reply, which I didn't save.
Basically, on a cold day, the temperature of the ice surface is the same as
that of the air, though it may be colder deep below, at the ice/water
interface. Ice is a poor conductor of heat. So, since we all know that
it's possible to skate at 0 deg F, we'd have to achieve enough pressure
to melt the ice at that temp, if the pressure-melting hypothesis is correct.
But even if we could only skate down to 11 deg F, we'd need to produce
1500 atm (about 22,000 psi) of pressure.
Someone pointed out to me (by mail) that skates are not only hollow-ground,
but also convex, so let's consider a contact *length* of 3 in; this would
require each blade to make contact over a width of about .0023 in
for our 300 lb skater. I don't believe it. Blades cut into the ice,
as one can see since they leave tracks, and this increases the surface
area of contact, reducing the pressure.
Furthermore, it's easy to slide on ice in one's shoes -- and you just *can't*
say that you're melting the ice by pressure there. In reconsideration, I
don't think frictional heating is the answer either. It's just probably
that ice is slippery! Sharp skates work well because they enable one to
direct ones momentum parallel to the blade, instead of skidding all over
the place.
Note that I never said that skating is impossible -- only that pressure-
melting of the ice is unlikely in the process. I believe that even
more strongly now, because of what I mentioned in the last paragraph.
Incidentally, someone from a Bell Labs site sent me mail about the
pressure in a Coke bottle; my mail reply was returned because of a
routing glitch, so let me say here that I agree (as a careful reading
of my first posting on this subject will reveal) that the pressure in
a coke bottle is "more like 150 than 750 psi." But that confirms my notion
-- also proposed by the individual who first brought up the Coke bottle
on a cold day phenomenon -- that, in fact, supercooling, not pressure-
induced freezing-point depression, was responsible.
Also to this individual: the problems of sealing a pressure vessel are very
different from those encountered when sealing a vacuum vessel. With high
pressures -- say above 1000 atm -- one can use "unsupported area seals,"
which use the pressurized fluid to make the seal better, so that the
higher the pressure, the better the seal. In such equipment, leakage
is encountered at the lower pressures -- a few hundred bars -- only,
unless a seal is damaged. See Bridgman's book on high pressure (Dover).
Coke bottles may be closer to the way vacuum vessels are sealed.
{philabs,cmcl2!rocky2}!cubsvax!peters
(Peter S Shenkin; Dept of Biol Sci; Columbia Univ; NY, NY 10027; 212-280-5517)jdd@allegra.UUCP (John DeTreville) (01/07/84)
Net.physics is the appropriate newsgroup for people who don't know much
about physics. Let's move this discussion out of net.followup so I don't
have to unsubscribe to it too.
Cheers,
John ("I'm in a \Bad/ Mood Today") DeTreville
Bell Labs, Murray Hill