23 minutes ago, paulraphael said:
I'm not 100% sure I understand your point here. It's true that once the cocktail reaches its equilibrium point (around -5°C for the drinks of the strength used in DA's experiments there will be no more melting. And there will likewise be no more chilling.
Are you talking about what happens when you put extra cold ice into a cocktail? The experiments address that. You get a very small additional amount of cooling from the colder ice. 1/2 calorie per degree for each gram of ice, vs. the 80 calories per gram you get from the ice melting. This ends up making a minute difference.
You are stuck in the cocktail shaker. Let’s do the Duveltini experiment 2.0 (extreme case, just to demonstrate):
Dewar vessel, temperature equilibrium, 1000 g of ice, -20 oC. We add 10 g of a suitable alcoholic drink with 20 oC.
What will happen ?
1) Will the ice melt:
Nope. Heat introduced into the system by the drink is not sufficient to raise the temperature above its melting point.
2) Will there be any dilution ?
(Apart from some surface phenomena) Nope. Ice doesn’t melt significantly. Truth is, there will be some, but it is not significant.
3) How does the content of the Dewar vessel cool the drink ?
Thermal mass transfer. Nothing magical, no phase transition, just plain heat transfer - just when you put a pot of water on a hot stove.
