# Does adding salt to water decrease the latent heat of vaporization?

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?

• How would we be able to conclude that?
– JMac
Jan 15, 2018 at 18:23
• Not getting how you boil off then heat to a higher temperature. Jan 15, 2018 at 18:58
• Is the heat of solution approximately equal to zero? Jan 15, 2018 at 19:08
• Assuming it is a good approximation, what are the initial and final thermodynamic equilibrium states for which you want the change in enthalpy? Jan 15, 2018 at 19:17
• Fail to understand how your scenario proves what you assert. You do know that heating water or vapor to T2 should not be included the heat of vaporization. If you can to a certain state with different energy then that is a perpetual motion machine and dismissed by thermodynamics. Jan 15, 2018 at 19:34