7
$\begingroup$

I want to create a steam bath for a reaction the proceeds well at 105℃. I cannot use pure (demineralized, anyway) water for this, since it would boil at 100℃ at STP.

As such I figured that maybe I could get the required temperature by mixing water with another liquid (with which water is miscible) which has a higher boiling point (e.g. vinegar at BP around 118.5℃ or vegetable glycerin with BP at 190℃).

Does the boiling point of the mixture of liquids proceed in a linear manner between the two temperatures? E.G. would a volume of 50% water & 50% VG boil at 145℃?

As it is being done right on the boiling point of the combined liquids, I expect to lose some of the liquid to steam and need to add more. Will the lower temperature liquid boil out first?

$\endgroup$
2
$\begingroup$

Except if molecular composition of your two liquids are similar, the boiling point of the mixture will not proceeds linearly but like this for instance :

enter image description here

With the lower blue curve the boiling curve. In between the two curve you'll then have a mixture of vapor and liquid and only vapor on the upper part the binary diagram.

So you should find out an already drawn binary diagram for your mixture or draw it out by yourself first.

$\endgroup$
  • $\begingroup$ I was looking at a few of the diagrams and they all seemed to follow the same basic shape as seen above. I'm thinking that if I can draw a grid over one of them and scale my temperatures (along with 105℃) onto the diagram, it should give me a rough idea of the mix required. Does that sound feasible, or was I seeing a limited subset of them that followed 'nice logical curves'? $\endgroup$ – Andrew Thompson Feb 16 '15 at 11:52
  • $\begingroup$ The shape will often be similar yes. But I doubt the values will. $\endgroup$ – Babounet Feb 16 '15 at 18:35
  • $\begingroup$ This can't be responsive to the question: If the liquids are merely mixed then when the boiling point of the first is hit it is will predominately vaporize. Yes, some of the second will be evaporating in the process, but essentially at the temperature above lower liquid's boiling point it is leaving the mix en mass. This is the basis of distillation! So replenishment would not be proportional to the initial mix, and I don't see how it makes sense to characterize the mix as having a unified boiling point. $\endgroup$ – feetwet Jul 31 '15 at 23:07

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.