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Does there exist a formula I can use to calculate alcohol content of a solution after heating it for some time?

I have the following use-case: I am making a pecan pie. I have a thick sugary solution. I am not sure how much the makeup of the solution matters. Just assume its more viscous than water. I add 75 ml of 40%v alcohol and I heat it at a temperature high enough to cause boiling for 20 minutes. How much alcohol remains at the end of 20 mins.

Obviously there are a lot of variables here that would make it impossible to give a precise answer. I would be interested in an answer that solves an "easier" version of this problem, for example: having 1 liter of water, adding to it 100ml of 100%v alcohol (Ethanol), boiled for 100 minutes, how much alcohol is left?

How quickly does alcohol boil off? Is it dependent on is concentration within a solution? is there a formula for that?

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For a pure substance, to be able to calculate how much has boiled off after $N$ minutes you will need the enthalpy of vaporization, $L$ (in kJ/kg) and the heat input, $Q$ (in kJ) which will give you: $$\tag{1} m=\frac{\epsilon \, Q}{L}$$ where $m$ is the mass that has boiled off and $\epsilon$ is the fraction of the heat input that actually goes into the liquid. $1-\epsilon$ would be the fraction of heat lost to the surroundings.

The very crude story

An extremely crude calculation would be to take the mixture enthalpy of vaporization and assume that only ethanol vaporizes such that the fraction of ethanol can be directly related to the total mass loss. Note that this calculation will give you a lower bound for the time needed to get rid of the ethanol

We can calculate the mixture enthalpy of vaporization by averaging the enthalpies of vaporization of water and ethanol: $$\tag{2} L_{mixture}=x_{\ce{EtOH}} L_{\ce{EtOH}}+x_{\ce{H2O}}L_{\ce{H2O}} $$

Now we need to figure out the heat input. A precise calculation is impossible here, but we can use some reasonable estimates. According to an estimate on this (dutch) website when we are cooking we use about $164$ l/h of methane. Assuming that that is for cooking with 2 pots this translates to a gas flow rate of $\dot{V}=1.4$ l/min for a single pot. With the heat of combustion of methane, $\Delta H_c$ and the density $\rho$ we can then calculate the amount of heat released per minute: $$\tag{3} \dot{Q}=\rho \dot{V} \Delta H_c $$ plugging in the values for methane we find $\approx 50$kJ/min

The enthalpy of vaporization of the mixture will constantly change, because the composition of the mixture constantly changes over time during the boiling process. Therefore, instead of simply using Eq. 3 in Eq. 1 to find $\dot{m}$ and multiplying with $N$ we will need to do an integration: $$\tag{4} m_{boil-off}=\int_0^N \dot{m}\; dt $$ where $\dot{m}=\frac{\epsilon \, \dot{Q}}{L(t)}$. Now this is a bit problematic because $L(t)$ depends on $m_{boil-off}$, but it can be solved numerically (I feel like this should also be possible analytically, but can't figure it out right now, anybody?!).

Your example question (with $\epsilon=0.5$) leads to the following curve for the fraction of ethanol in your mixture:

enter image description here

As you can see the ethanol is gone after about 7 minutes. To do this calculation for your pecan pie mixture you will need to know the heats of vaporization of all your ingredients and the mass of each.

The more precise story

The more precise story is that not only the ethanol, but also part of the water will vaporize therefore slowing down the vaporization rate of ethanol. If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid. In that case you can use the theory available for distillation columns and something called the Ponchon-Savarit Method (the complete diagram for water-ethanol can be found here), which I will not recap here, because it is basically a book chapters worth of undergrad chemical engineering.

Note that the equilibrium vapor assumption is probably not a very good one due to air flows over the pot which keep pushing the system out of equilibrium. For the evaporation of ethanol this is good news by the way, because it will speed up the process a bit. Also note that with a true equilibrium you would run into the azeotrope of ethanol-water mixtures.

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  • $\begingroup$ ""If you are 'lucky' you could assume that the vapor above your pan is at equilibrium with the boiling liquid."" This is not luck, but obvious! If the liqiuid boils, the vapor comes directly from contact with the liquid, and must have the same temperature. The unknown in the question is the reflux conditiones. Look for fractional distillation. $\endgroup$
    – Georg
    May 21 '13 at 18:33
  • $\begingroup$ @Georg - The same temperature yes, heat transfer is fast, but not the equilibrium vapor-liquid composition right? (please explain if I'm wrong here). The VLE is only reached in steady-state conditions which I think are hard to achieve in an open pan with (I presume) the exhaust hood running $\endgroup$
    – Michiel
    May 22 '13 at 5:29

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