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)