Cocktail thermodynamics

I enjoy the occasional cocktail. My colleague Chris enjoys, makes and criticises cocktails. Door 34 in Bath makes really excellent cocktails. I have been known to indulge in a little thermodynamics; it was inevitable that one day these positions would collide in “the Boston Shaker Problem”. Chris swore on his reputation as a mixologist that:

  1. When cocktail ingredients are shaken with ice in a Boston shaker, the temperature drops to well below 0°C
  2. Using more ice results in less dilution
  3. Ice used in cocktail bars has been draining on the side, and is not below 0°C

This flies in the face of what we think we know about ice and water at atmospheric pressure, namely that:

  • If you attempt to raise the temperature of ice above 0°C, it will melt.
  • If you attempt to lower the temperature of water to below 0°C, it will freeze.
  • If you have a system of ice in contact with water, it will tend to equilibrate at 0°C, and resist changes to the temperature by melting and freezing as necessary.

If the ice starts at zero, and the other ingredients start above zero, where is the mechanism to lower the overall temperature? Seemingly the only available way is to take heat energy into the solid ice phase and melt. This isn’t going to work with non-melting ice cubes. It also doesn’t explain why we don’t hit our usual equilibrium temperature of 0°C.

The second part is just as confusing: there is a fixed amount of energy required to convert water between solid and liquid phases at a given temperature and pressure. This is known as the latent heat of fusion, \Delta H. This is an amount of energy per mass of water. As the total energy needed to cool your drink is fixed, the amount of ice that needs to be melted should also be fixed. (It’s not quite proportional, as you will need a little bit of extra melting to cool the extra water – and so on.) How does a greater bulk of ice help?
At this point we had a few ideas,  but turned to the internet for help – surely this is a well-known system? We did find an article at Cooking Matters, who not only hit on a good theory, but actually tested the effect with a thermocouple on their cocktail shaker. (It’s a pretty entertaining blog, if you’re interested in kitchen centrifuges and lobster-killing technique.) The key point is this: the system in a Boston shaker is not an ice-water equilibrium. The alcohol shifts the equilibrium to a lower temperature. Why? Entropy.

Door 34 asked for a mathematical description of the effect, and we gave them this:

\displaystyle 0 \geq \sum\limits_\textrm{mixer}^\textrm{ice} \left[\Delta U + \int mC_p \textrm{d}T \right] - \int S_\textrm{mixing} \textrm{d}T

It is worth giving a brief breakdown. The right-hand side of the equation is equal to a change in the Gibb’s free energy of the system. For a spontaneous process, this must be negative; at equilibrium it will equal zero (hence the ≥  sign.)  I’ve neglected the pressure-volume work as it is not expected to significantly contribute; strictly speaking there will be a little work associated with volume change as ice has higher specific volume than water, and water/ethanol mixtures have a lower specific volume than the sum of components.  I’ve also been a bit cheeky here in separating the vibrational contribution from the internal energy (strictly U should be U_{\textrm{pot}} or something.) This is to draw attention to Kirchoff-equation related goodness; the difference in heat capacities determines how much ice must melt. As the ice melts, \Delta U increases as water is a higher energy state than ice. Assuming no heat input, the enthalpy (in square brackets) must be conserved, and so the heat capacity C_p is integrated over a negative temperature difference to compensate; the temperature drops. This is driven by the entropy term on the far-right; as long as more entropy can be gained by melting ice to mix water and alcohol, the ice will continue to melt, dropping the temperature further, and well below 0°C. (Note that T in the equation is relative to absolute zero, and will never be negative.)

A problem remains– why does more ice mean less dilution? According to this, the equilibrium amount of melting is pretty much fixed. If anything, adding more ice should drive entropy in the direction of more melting. My best guess is that this is to do with the speed of the operation; a Boston shaker isn’t quite an isolated system. The longer it takes to reach equilibrium, the more heat will have entered the system and the “wetter” that equilibrium will be. This seems to agree with the account of mixologists, that shaking with more ice is a quicker experience that “feels” different. Still, I’m left feeling that this system isn’t quite solved yet…

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