Suppose we have a material such as salt that when added to water it will increase its boiling temperature. Can we conclude that it also has to reduce the latent heat of vaporization?
I'm asking this question based on the following thought experiment. I am interested to know if this is true or not independent of whether this thought experiment proves it or not.
Suppose we have a large amount of salt water at its boiling temperature $T_2$. Then, we add a small amount (say mass $m$) of pure water just below pure waters boiling temperature $T_1< T_2$. Then we add enough heat to bring the total salt water to the boiling temperature again and boil off the small amount of water added. To do so, we have to spend $m C_w (T_2-T_1)$ for heating plus the latent heat for vaporization. But, we could have just boiled off the pure water (without adding it to the salt water) and then heat it up to higher temperature $T_2$ and end up with the same outcome. That would cost a different latent heat plus $mC_v(T_2-T_1)$. But $C_v<C_w$, so the difference should come from the difference in the latent heats.
Is this argument correct?
Is there a clearer/different way to show this?
Update: In the analysis of this thought experiment, I have ignored the heat of mixing. I do not know if that would play a significant role or not. Is the claim true without this approximation?
Best Answer
The latent heat of vaporization as a function of temperature and salinity can be found on page 9 of the Seawater Thermophysical Properties Library. (See the top-note about entries over 100C being at the seawater vapor pressure, i.e. boiling equilibrium)
For comparison to the thought experiment in the question above, note that the heat is proportional to the water in the salt+water mix. For example, with 100g/mg salt, i.e. 900g/kg water, the latent heat at 100C is 0.9*2256.5 = 2030.8.