Consider the following situation: We have a pot of strong brine (for concreteness let's say 25g of NaCl per 100ml of water, at standard conditions) and heat it to a boil. How will the salt content affect the resulting steam, in comparison with pure water? I.e. what temperature and which approximate ratio of droplets to water vapor will the steam above the pot have? (Perhaps we should also assume a lid on the pot to make convection less relevant. )

I understand how and why the boiling point is raised when having dissolved any nonvolatile substance, but I'm confused because I have several conflicting lines of thought in regards to the steam [feel free to ignore those and just explain the situation itself]:

  • The boiling point is raised and the steam must have the same temperature as the liquid; so steam will be hotter.
  • The evaporation needs more energy (since it adds an entropically unfavorable distillation to the evaporation) so the steam is still at 100°C. The temperature drop at the border is reminiscent of a thermocouple.
  • The steam bubbles are initially at 100° C, but superheat a bit whilst rising through the liquid, thus resulting in dry steam.

(Note: Whilst I stated it as a concrete example for clarity, general(-izable) explanations would be preferred; I'm only interested in the result to build intuition)

  • $\begingroup$ Is the question totally theoretical or are you concerned about an application? $\endgroup$ Commented Nov 6, 2019 at 16:10
  • $\begingroup$ Entirely theoretical/didactic; though it came to mind whilst steaming vegetables and might be relevant for vapor-compression distillation machines. $\endgroup$
    – caconyrn
    Commented Nov 6, 2019 at 18:38

1 Answer 1


Since boiling-point elevation is understood I would start with model likes this:

a vessel with constant pressure (i.e. volume will increase) containing the liquid and some gas (air+vapor), both are in thermal equilibrium. Then you heat up so that liquid so slowly that it remains in remain in equilibrium (or very near) with the gas. Finally, the liquid will reach the boiling point: per definition that is the point where the vapor pressure of a liquid equals the pressure surrounding the liquid. Being in equilibrium both phases still have the same temperature. Let's go through the three bullet points:

1) As stated the vapor above the solution will be hotter than above the pure liquid

2) There is no temperature drop since we are in equilibrium (or near that)

3) Vapor bubbles have exactly the temperature of the boiling point

The question mentions droplets of water: There will be no droplets in equilibrium. And you can assume that you have equilibrium immediately above the surface of the liquid. The droplets you see in the kitchen are condensed water from the gas cooling down due to contact with cold air.

  • $\begingroup$ This does not really address the question I was trying to ask, sorry. I'm interested in how/why the presence of solute (salt) changes things compared to pure solvent (fresh water). ".. same temperature in liquid an gas." This may well be true, but I'd like an explanation of why this must be the case based on thermodynamic principles, as opposed to the other things listed in the question. Thanks for your time anyways. In case the misunderstanding is cleared up now, please suggest an edit to make the question clearer. $\endgroup$
    – caconyrn
    Commented Nov 6, 2019 at 15:47

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