Comments on: Tales of The Cocktail: Science of Shaking II The International Culinary Center's Tech 'N Stuff Blog Thu, 09 Jan 2014 18:17:16 +0000 hourly 1 By: davearnold Mon, 28 Feb 2011 00:37:43 +0000 The entropy argument isn’t hand waving. Entropy is the true driving force of the phenomenon. The melting point of a pure substance is based on the Gibbs free energy equation. At the melting point, the amount of heat required to melt or freeze the substance (the Delta H) is offset by the change in entropy of the system (T Delta S). The ice in a cocktail is still pure ice –not an ice composed of ethanol and water (as the phase diagram of the paper you reference shows). The Enthalpy required to add a molecule of water or remove a molecule of water from the ice doesn’t change when you add it to vodka (the heat of fusion is the same). What changes is the entropy side of the equation (T delta S). Because the entropy gain of melting is greater in a water-ethanol system than a ater system at the same temperature, the ice melts and the temperature of the system drops. The drink will continue chilling and diluting till the entropy and enthalpy balance again.

By: Mike Wed, 16 Feb 2011 20:28:03 +0000 Just to clarify a bit:

The freezing point of a pure compound is defined as the temperature at which a solid form of the compound is in equilibrium with a liquid form of the compound. Adding energy to the system will decrease the amount of solid as it melts without increasing the temperature of the system. Removing energy will increase the amount of solid without decreasing the temperature. Consequently, adding ice at 0C to water at 0C will not change the temperature of either. Any energy absorbed by ice that melts will be replaced by water freezing, and the energy balance of the system will be unchanged.

However, ice in vodka is not a pure compound. The ice has a freezing point of 0C, but the freezing point of vodka is considerably less. Consequently, energy absorbed by the ice melting will be replaced by the temperature of the liquid dropping rather than by new ice forming as it would if it was at the freezing point. The melting of the ice is driven by entropy so it will continue to melt (diluting the vodka and lowering the temperature of the entire system) until a new equilibrium is established. This equilibrium will be established at the freezing point of the ethanol-water mixture surrounding the ice, slightly higher than the freezing point of the vodka because the resulting solution, now that some ice has melted, is somewhat more dilute. This will still be higher than the freezing point of pure ethanol, and the ice formed will still be pure water (barring inclusions and other physical processes).

The hand-waving argument is that the entropy of the water molecules bound up in the ice is the same regardless of whether the ice is floating in vodka or in water, but the entropy of the water molecules in the vodka is less than it is in pure water. (Etropy is a measure of the amount of order, and there is more order in a system where water molecules are surrounded only by other water molecules than there is when some of them may have ethanol molecules nearby). For water in the vodka to associate into crystal structures, it must move around those pesky ethanol molecules until it is surrounded by all water molecules, which takes energy and means that it won’t happen until the temperature gets lower.

There is a full phase diagram in this paper:

Actually, I see this as the strongest argument for using ice that is below the freezing point of the drink you are trying to make to cool your drinks. That would give you the most consistent dilution. The amount of ice melted would be a direct measure of the original temperature of the drink and its volume – any additional ice would remain in solid form, and as long as the volume of ice was at least large enough to reach that equilibrium, no additional ice would be melted. Using 0C ice, you have to consider not only the reduction in temperature of the drink to its freezing point, but also the reduction in temperature of the ice to the freezing point of the drink. Any additional ice added would require additional melting and therefore dilution to bring the total mass of ice into thermal equilibrium with the solution.

By: Dave A Wed, 18 Nov 2009 16:51:33 +0000 Ice cubes have three separate problems:
1. Water contains impurities that cause disruptions in the crystal structure and cause cloudiness. Distillation will help with this.
2. Water contains air which provide places for the crystal structure to be disrupted. Boiling and vacuuming helps this, but simply pouring water from one container to another can cause problems.
3. Your freezer lets the top of the ice freeze first. As the center of the ice cube freezes it expands breaking through the crystals above it and ruining everything. This is why the pros (like clinebell) freeze from the bottom and constantly circulate the water on top. This 1. stops the top from freezing and 2. keeps the impurities from settling on the bottom and providing a place for cloudiness to happen.
I once put a circulator in a chest freezer and froze a 30 gallon block of ice. It looked great except the ice froze from the sides. Most of it was super clear except for a concentrated dirt tornado in the very center. Antigriddles don’t have enough power to freeze a really big cube. Iv’e heard antigriddle plus chest freezer plus circulator works.
At home try using boiled distilled water in super clean large containers. Agitiate the top every once and a while to prevent freezing or put an aquarium pump at the top. Be prepared to lose the top portion of the ice.
Hope this helps.

By: Rob Wed, 18 Nov 2009 15:43:31 +0000 Dave… do you know how to make the ice cubes that people use at home clear instead of cloudy? I heard that boiling purified water, letting it cool, boiling it once more and letting it cool again does the job. This technique supposedly removes impurities. However, after trying this technique, I can say that it doesn’t make the ice cube fully translucent. It only helps slightly. Your thoughts, methods? Are clear ice cubes made from distilled water perhaps?

By: Dave A Mon, 02 Nov 2009 15:33:17 +0000 Interesting.

By: galin Sat, 31 Oct 2009 03:36:05 +0000 Hi Dave,
the experiment you guys made is absolutely agreeable with what i have experienced for the last 11 years as a bartender (although big and solid ice cubes tend to make the drink nice and frothy if it contains pulp or juices or other proteins). the thing that really puzzles me though is your ” using different shaker configurations” graph and the conclusion that metal shaker achieved lower final temperature than glass. in my experience at work and here i must say i never approached this scientifically, when shaking in the glass-steel boston shaker frost always forms on the outside of the thin, but whenever i shake that same cocktail between two thins i can never get them to form frost on the outside.
however your post is as always brilliant and educational. Thank you

By: blair frodelius Mon, 14 Sep 2009 17:19:45 +0000 One thought I just had. You may want to shake using gloves to insulate the shaker from body heat. Of course, if you are going for typical results, then you’d want to forego the gloves.


By: Dave A Mon, 14 Sep 2009 17:00:09 +0000 I think the deviation in ice melted comes from the shaking loss we experienced. The ice melted number is a calculated number equal to the weight of the inputs minus the weight of the drinks. That number includes the shake loss. The abv follows the expected trend because the loss in liquid was probably early in the shake rather than later. I’d like to run the test again in a completely sealed system instead of actual cocktail shakers but haven’t had the time. ABV, temperature, and drink weight are not calculated, they are measured.

By: slkinsey Mon, 31 Aug 2009 14:05:00 +0000 Something has been puzzling me about your combined chart showing ice melted, %ABV and temperature. . . Looking at the data points, we see that amount of standard ice melted increases sharply between 8 and 9.5 seconds. In addition, see that the %ABV for the standard ice sample also increases sharply between 8 and 9.5 seconds. How is this possible? More water melted in equals lower %ABV, not higher. Indeed, I don’t quite understand any of the reverse turns in dilution and ice melted. Is it possible that the sample size was too small for a good average and these values are being unduly influenced by outliers? With a sufficiently large sample size, I would expect to see more or less smooth curves showing a steep incline or decline at the beginning and then tapering out over time to some asymptote.

By: Eben Tue, 25 Aug 2009 20:47:48 +0000 And as for your post beginning “The ice dropping in temperature”…no. Unless I am totally misreading what you are trying to say, the ice does not cool down the drink and then get recooled by the drink. That would be perpetual motion.