Evaporative cooling can occur in an air-liquid water system initially at the same temperature. If the air is not saturated with water vapor, the liquid water will evaporate until the partial pressure reaches its vapor pressure. In this process, the temperature of the liquid water decreases. This is because there is a distribution of speeds among the water molecules, and only the water molecules moving fast enough with enough kinetic energy to break their hydrogen bonds will escape into the gas phase. This selection based on kinetic energy leaves the slower molecules in the liquid phase, thereby lowering the temperature of the liquid.

My question concerns the parallel process of an ice cube melting. If the ice cube is in thermal equilibrium with a super-cooled salty brine, say -20 degrees C, there will still be a distribution of speeds among the water molecules in the ice lattice. Since only the fastest moving water molecules in the ice lattice will have enough energy to break the hydrogen bonds, would the melting of the ice cube cause a similar cut-off of the speed distribution curve and lower the ice cube temperature? If the volume of the brine solution was large enough so that the brine temperature doesn't rise much, could there be a situation where the ice cube both melts and also get cooler?

  • $\begingroup$ I'd call your first paragraph a simplistic explanation (don't worry - it is used a lot). The real, thermodynamic answer is that the entropy gain of the system by making some liquid in to vapor compensates for the enthalpy required to do so. Just how the system does that is a kinetics question. A comparison of the liquid/gas situation to the solid/liquid phase balance is left as an exercise for the reader. $\endgroup$ – Jon Custer Oct 13 '16 at 19:31
  • $\begingroup$ Thanks, I'm a teacher and trying to help students grasp some thermodynamic phenomena without invoking energy and state functions. $\endgroup$ – lamplamp Oct 16 '16 at 15:41

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