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?

  • $\begingroup$ How would we be able to conclude that? $\endgroup$ – JMac Jan 15 '18 at 18:23
  • $\begingroup$ Not getting how you boil off then heat to a higher temperature. $\endgroup$ – paparazzo Jan 15 '18 at 18:58
  • $\begingroup$ Is the heat of solution approximately equal to zero? $\endgroup$ – Chet Miller Jan 15 '18 at 19:08
  • $\begingroup$ Assuming it is a good approximation, what are the initial and final thermodynamic equilibrium states for which you want the change in enthalpy? $\endgroup$ – Chet Miller Jan 15 '18 at 19:17
  • $\begingroup$ 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. $\endgroup$ – paparazzo Jan 15 '18 at 19:34

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)

enter image description here

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.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.