Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10 5/3/83; site cubsvax.UUCP Path: utzoo!linus!decvax!harpo!floyd!cmcl2!rocky2!cubsvax!peters From: peters@cubsvax.UUCP Newsgroups: net.physics,net.followup Subject: More on Cold Bottles of Coke Message-ID: <134@cubsvax.UUCP> Date: Tue, 3-Jan-84 14:59:48 EST Article-I.D.: cubsvax.134 Posted: Tue Jan 3 14:59:48 1984 Date-Received: Thu, 5-Jan-84 01:26:03 EST Organization: Columbia Univ Biology, New York City Lines: 77 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)