More on cocktail thermodynamics

Debby found this post from Dave Arnold that resembles what I discussed in the last post. Shorter version: I’m probably wrong about the ice being close enough to thermal equilibrium for government work, but the explanation of what is going on isn’t quite the way I recall Weitz explaining it. And my old Intro Bio prof is still wrong.

Fact 1:Ice at 0°C can chill an alcoholic drink well below 0°C. This fact is counter-intuitive to many, but is an irrefutable consequence of the laws of thermodynamics.

Arnold does an experiment that is essentially what we saw on Friday night, using vodka instead of tequila. After pre-incubating ice in water, the water is drained off and vodka is added. He gets  slightly diluted vodka at -4.5 °C. He also did a measurement of how fast ice reaches thermal equilibrium:

Fact 2: Bar ice is almost always at 0°C unless it comes straight from the freezer. People have a hard time accepting this fact. As a test, I froze a large ice cube with a super-thin hypodermic thermocouple probe in the center.  I put that ice cube, along with some run-of-the-mill ice cubes for insulation, into a blast freezer for 4 hours until everything was at -20 C.  I then put the entire batch into a plastic container and waited.  In under 20 minutes, the large ice cube was within 0.5 degrees of zero.

In the comments to an earlier post, a reader wonders if he is measuring the core of the ice or surface water that develops around the temperature probe. Although the temperature might not be quite all the way up to zero for that reason, I suspect that the known conductive properties of ice mean that it’s closer to 0 than -20.

So why does this work and what’s wrong with my earlier analysis? Because I idiotically glossed over an important part of the Clausius version of the Second Law


The other change that is connected therewith to passage of heat is the breaking of bonds in the ice and the conversion of ice to water. Duh! As the vodka/tequila goes from 20 to 0, the heat is passing from the warmer liquid to the cooler solid ice, melting the latter with 333.55 J/g (80 cal/g). Melting a g of ice can chill 4 ml of water from room temp to 0.

But when the vodka/tequila is below 0 but above the depressed freezing point, melting reactions don’t stop. There are still a fraction of the total intermolecular collisions between the liquid and the ice that have sufficient energy to break the water-water bonds on the surface of the ice and melt off some of it (Not being an ideal gas, this won’t be a Maxwell-Boltzmann distribution, but there will be a mix of energies). This cools the liquid phase, just as it does when the collision occurs above 0°C.

Whether you are cooling water or liquor, the final temperature is still set by the freezing point, which is where the probability of a water molecule joining or leaving the crystal lattice is equal. So I think I’m still right that the freezing point depression is important. The ethanol isn’t doing anything to the heat exchange per se. It’s allowing us to see cooling below 0 when we measure the temperature of the liquid phase.

Arnold argues that the chilling is due to a combination of the heat that goes into melting and the entropic gain from diluting the water released by that melting into the liquid phase. In another comment he writes

Freezing point depression isn’t enough to explain why the drink gets colder than zero as you shake it. For instance, many oils have a very low freezing point but if you put an ice cube in them they will only go down to 0 degrees because there is no mixing.

I’m not crazy about this explanation, as it seems to me that it approaches Gibbs’ paradox territory with respect to the water released from the ice being diluted into the bulk vodka/tequila. Here, I think he’s seeing an effect of surface water around the ice cube limiting the temperature change. The observation about the melting points of oils does suggest a possible way to measure the internal temperature of ice cubes if one could suspend oil droplets in clear ice.