# Why do we heat salt water to desalinate it rather than freeze it?

Most sea water desalination plants rely on the basic principle of heating water to evaporate it, after which the salt-free vapor is condensed. According to Wikipedia, over half the cost of desalination is made up of energy cost, since heating water is very energy intensive. As a back-of-the-envelope estimate, the energy required to bring $$1~kg$$ of room temperature ($$20$$°) to a boil under atmospheric pressure is about $$620~kcal$$. $$80~kcal$$ to heat it to $$100$$°C and another $$540~kcal$$ to bring it to a boil.

Conversely, cooling it down to $$0$$° takes $$20~kcal$$ and $$80~kcal$$ to freeze it, in total $$100~kcal$$, less than $$1/6$$ of the energy requirement of the heating method.

This kind of desalination procedure happens in nature - icebergs are largely frozen fresh water. Nevertheless, it seems this is not common in commercial desalination.

Is there a physical reason why freezing water is less suited for desalination compared to heating it?

• Icebergs are made from precipitation, coming from evaporated water. They do not freeze out of the salt water oceans. Freezing a concentrated brine will, ultimately, result in a portion of the solid with less salt concentration. You would then need to separate that piece out, and it cannot be refined further (you are at the solubility limit). Evaporation, since the salt does not go into the gas phase, is a process that can be run continually to keep making distilled water. Commented May 16, 2019 at 13:43
• However, both methods can be used to separate alcohol... @JonCuster you should make this an answer.
– user207455
Commented May 16, 2019 at 14:16
• The energy computation for cooling in the OP is not correct. The $100 kcal$ is the amount of heat that must be removed from the water (that number is only an approximation, by the way, since the water is presumed not to be pure). But removing heat from water requires energy, otherwise we wouldn't need electricity to run our freezers. Commented May 16, 2019 at 14:41
• @RickGoldstein I am aware that this is energy that is leaving the system. If we used a heat pump with $\eta = 1$, we would need to expend that amount of energy. Since this is in reality more like $\eta = 0.4$, it would rather be $250~kcal$, which is still only about half of the energy for boiling the same amount of water. Commented May 16, 2019 at 15:46