Calculating time to reduce alcohol in wine using heating method

[ArtZ]

TL;DR Summary

Using rate kinetics to calculate the time to reach final CF of alcohol knowing starting C0 and heating T

Hi,

This is my first post on PF. I am engaged in a project seeking to find a simple way to reduce ethanol content in wine for those wishing to minimize alcohol intake for health or personal reasons. The alcohol-reduced wine will be used for cooking. Seems heating would do the trick however, the temperature must be kept as low as reasonable to minimize degradation to the wine flavor.

Keep in mind that my field is EE not CE so I am feeling this out as I go.

So as I understand it, the rate of alcohol removal in wine is dependent on several factors, including temperature, time, and alcohol concentration. One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

The general equation for the first-order reaction model is:

dC/dt = -kC

where:
- dC/dt is the rate of change of alcohol concentration with time
- k is the reaction rate constant
- C is the concentration of alcohol in the wine at any given time

Assuming that the reaction follows first-order kinetics, seems that we can use the following equation to calculate the concentration of alcohol remaining in the wine after a given time:

C/C0 = exp(-kt)

where:
- C0 is the initial concentration of alcohol in the wine
- t is the time (in minutes)
- exp is the exponential function

I've defined the reaction rate constant, k, using the following:

k = A * exp(-Ea/RT)

where:

A is the pre-exponential factor or frequency factor, A = 10^13 s^-1
Ea is the activation energy, Ea = 65 kJ/mol = 65000 J/mol
R is the gas constant, R = 8.314 J/(mol*K)
T is the absolute temperature used in heating, T = 356.45 K

Substituting the values of A, Ea, R, and T into the equation, we get:
k = (10^13 s^-1) * exp(-65000 J/mol / (8.314 J/(mol*K) * 356.45 K))

k = 2.982 x 10^3 s^-1

When I solved for t using the above:

t = -ln(C/C0)/k

I found that the time, t, value returned, was too short given the Cf desired and the C0 of the unheated wine. The first-order reaction model does not account for the energy needed to overcome the heat of evaporation making this an incomplete solution to dC/dt.

As I understand it, the heat of vaporization or heat of evaporation, is the amount of energy that must be added to a liquid substance, to transform a quantity of that substance into a gas.

The heat of vaporization of ethanol (hfg) is (0.826 kJ/g). To incorporate we'll convert the heat of vaporization from kJ/g to J/mol. The molar mass of ethanol is 46.07 g/mol, so: hfg = 0.826 kJ/g * 1000 J/kJ * 1 mol/46.07 g = 17930 J/mol

In both equations below, C0 represents the starting alcohol concentration, Cf represents the desired final alcohol concentration, k is the rate constant for ethanol at the given temperature (in units of s^-1), hfg is the heat of vaporization for ethanol (in J/mol), and R is the gas constant (8.314 J/mol*K).

With some algebraic manipulation I arrived at these canonical forms of the t, and T equations, where t represents time (in seconds) and T represents temperature (in Kelvin):

To solve for time (given T):t = -ln(Cf/C0) / (k*(1 + (hfg/R*T)*k))

To solve for temperature (given t):T = hfg/(Rk) * (1/t - 1/(kln(Cf/C0)))

In both equations below, C0 represents the starting alcohol concentration, Cf represents the desired final alcohol concentration, k is the rate constant for ethanol at the given temperature (in units of s^-1), hfg is the heat of vaporization for ethanol (in J/mol), and R is the gas constant (8.314 J/mol*K).

Hopefully this makes sense so far. Like I said, am an EE not a CE. The first example that I'd like to solve for is using the expression for t above given T:

Given:

• Heating temperature of wine = 83.3C • Initial alcohol concentration, C0= = 0.15 (15%) • Final concentration desired, Cf = 0.04 (4%) • Reaction rate constant = 2.982 x 10^3 s^-1 • Heat of evaporization for ethanol = 17930 J/mol • Boiling point of ethanol =78.5C

An answer that I arrived at that seemed reasonable = 51 minutes to reduce the alcohol concentration from 15% to 4%. Everything I've presented seems reasonable, however, I don't get agreement between Excel and Mathcad which I use for units consistency checking.

Maybe my approach is way off base or just needs some fine tuning. Is there someone in the Forum with a better background in this area who could point me in the right direction?

Any help is greatly appreciated.

Thanks,

Art

 

[Hyperfine]

The alcohol-reduced wine will be used for cooking. Seems heating would do the trick however, the temperature must be kept as low as reasonable to minimize degradation to the wine flavor.

The cooking process itself should achieve your objectives starting with normal wine having normal alcohol concentrations.

 

[ArtZ]

Thanks for your response. One might think that but it's not correct. Tasters are still able to detect the alcohol signature when starting alcohol concentrations are relatively high (13-16%) especially if the food is cooked for a short time. (stir fry) I am using Chinese Shaozing wine which comes in at about 15% ABV. What most people don't understand is that the reduction of alcohol in the cooking process is not as complete as is widely thought. The attached shows a USDA study for residual nutrients and alcohol after heating.

 

[Hyperfine]

Fair enough. But I doubt that this situation should be treated as a problem in chemical kinetics.

Nor do I think that I would appreciate my wine being heated to 83^oC for long enough reduce the ethanol concentration to the desired level.

Obvious questions: What volume of wine do you need to deal with? How often do you need do this?

 

[Tom.G]

This sounds like a question for @Chestermiller, our resident thermodynamics expert; and this mention should alert him to take a look.

p.s. Interesting question

 

[Borek]

A is the pre-exponential factor or frequency factor, A = 10^13 s^-1
Ea is the activation energy, Ea = 65 kJ/mol = 65000 J/mol

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

 

[Chestermiller]

Your theoretical development is way off. What you are dealing with here is a solution of alcohol and water, and when you heat the mixture, both alcohol and water go into the gas phase, and in different proportions than in the liquid phase. So what you are dealing with is the vapor-liquid equilibrium behavior of a binary solution of alcohol and water. The process being considered is distillation. See Treybel, Mass Transfer Operations. Google VLE Alcohol Water. Consider using vacuum distillation to lower the temperature of the operation.

 

[Hyperfine]

Consider using vacuum distillation to lower the temperature of the operation.

A simple rotary evaporator might work well.

 

[ArtZ]

Fair enough. But I doubt that this situation should be treated as a problem in chemical kinetics.

Nor do I think that I would appreciate my wine being heated to 83^oC for long enough reduce the ethanol concentration to the desired level.

Obvious questions: What volume of wine do you need to deal with? How often do you need do this?

I'll start with a single 750mL bottle. Typical recipes only call out maybe a tablespoon or two at most so a single 750mL bottle should last a while.

 

[ArtZ]

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

 

[ArtZ]

You never defined what the reaction is, you never stated where these values come from. It is about as random as possible.

But in general, such processes are very difficult to describe as they typically heavily depend on poorly defined and hard to control factors like mixing and/or diffusion. That in turn means it is easier to test the result experimentally than to calculate what will happen.

Alcohol from beer is removed by reverse osmosis, which doesn't require elevated temperatures, that would be probably a good starting point.

I looked into the RO process and found a company that does this for wine makers. They require 1 case of wine and the processing fee is $850 USD. Too rich for a small time tinkerer.

 

[ArtZ]

A simple rotary evaporator might work well.

Too complicated. A large pot should work fine.

 

[ArtZ]

Your theoretical development is way off. What you are dealing with here is a solution of alcohol and water, and when you heat the mixture, both alcohol and water go into the gas phase, and in different proportions than in the liquid phase. So what you are dealing with is the vapor-liquid equilibrium behavior of a binary solution of alcohol and water. The process being considered is distillation. See Treybel, Mass Transfer Operations. Google VLE Alcohol Water. Consider using vacuum distillation to lower the temperature of the operation.

I am not distilling anything. If anything, this process would be called reverse distillation.

 

[Borek]

One common model used to describe the kinetics of alcohol removal is the first-order reaction model, which assumes that the rate of reaction is proportional to the concentration of alcohol remaining in the wine.

Removal by what means and in what conditions?

Sorry, but what you write is completely meaningless without context. The question is "how long does it take to get from point A to point B" and you answer with "my car is capable of 100 mph and we drive on the right". There is no way to connect the dots.

I am not distilling anything. If anything, this process would be called reverse distillation.

Reverse distillation is - if anything - just a distillation in which you keep not what evaporated, but what was left. It is exactly the same process and it is described by exactly the same models and math.

 

[Hyperfine]

I'll start with a single 750mL bottle. Typical recipes only call out maybe a tablespoon or two at most so a single 750mL bottle should last a while.

Simple. Add to a small pan over low heat and reduce to taste.

Cooking, not chemistry.

 

[ArtZ]

Simple. Add to a small pan over low heat and reduce to taste.

Cooking, not chemistry.

I sort of plan to do that. My wife has hypersensitivity to residual alcohol in cooked food. The plan is, this week if I get to it, is to reduce a known quantity of the Shaoxzing wine (15%) at a known temperatue and then test with a wine makers hydrometer that I bought online. I figure that a final concentration of 4% won't be noticed in a 4 serving cooked dish.

 

[DaveE]

Why don't you just buy some non-alcoholic wine? It's readily available, and done by people with better equipment than you are likely to have.

 

[hutchphd]

Absolutely @ArtZ you are overthinking this. Put the wine in a container and heat it to a low boil. The alcohol will preferentially vaporize. Reduce the volume by 20% and there will be far less than 4% residual ethanol. Probably a 10% reduction would get you to 4% residual. The Shaozing wine flavor is pretty robust and it is being cooked anyway so not a problem. Enjoy.

 

[ArtZ]

Absolutely @ArtZ you are overthinking this. Put the wine in a container and heat it to a low boil. The alcohol will preferentially vaporize. Reduce the volume by 20% and there will be far less than 4% residual ethanol. Probably a 10% reduction would get you to 4% residual. The Shaozing wine flavor is pretty robust and it is being cooked anyway so not a problem. Enjoy.

I may overthinking this, but I wanted to have some technical fun. Since I retired, I don't do much of what I did before. Since the alcohol BP is 78.5C (173.3 F) I figure that if I keep the wine at a simmer but below the boiling point of the water constituent, I should be OK. Also will need to replenish the remaining volume with water to restore the original volume.

 

[DaveE]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

 

[Chestermiller]

I am not distilling anything. If anything, this process would be called reverse distillation.

You don’t know much about distillation,, do you? As a ChE, my advice to you as an EE is to stick to Ohm's law.

 

[ArtZ]

Only what I've learned from watching Moonshiners.

 

[ArtZ]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

View attachment 323574

I don't expect to remove 100% of the alcohol, just enough so it can't be tasted in a prepared food dish. All online cooking references agree with your point. It's of course more evident when high alcohol spirits are used. Can you explain what the chart you posted is telling us?

 

[berkeman]

I figure that if I keep the wine at a simmer but below the boiling point of the water constituent, I should be OK.

Have you tried this? What does it do to the taste, bouquet, aftertaste, etc?

https://www.nytimes.com/1982/01/20/garden/the-tasting-ritual-4-basic-elements.html

 

[Chestermiller]

If you really want to remove all of the ethanol, you will essentially have to boil the wine. Heat will preferentially remove ethanol, but only to a certain concentration, as shown in the phase diagram below.

View attachment 323574

I think you are interpreting this wrong. It seems to me you are looking at the wrong side of the phase diagram. If you start off with 15 % by volume of ethanol, the first vapor to come off at 195 F will be 60 % ethanol. If you raise the temperature from 195 F to 202 F and allow the vapor and liquid to equilibrate, you will have lowered the concentration of ethanol in the liquid to 7.5 volume %, and decreased the amount of liquid by only a small amount. This is all at a pressure of 1 atm.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

 

[berkeman]

just enough so it can't be tasted in a prepared food dish.

What prepared dishes? Doesn't the cooking process already remove most of the EtOH anyway?

 

[Hyperfine]

Change the menu!

 

[berkeman]

just enough so it can't be tasted in a prepared food dish.

BTW, EtOH is tasteless.

Spoiler

Just like folks who've had way too much EtOH...

 

[Chestermiller]

Only what I've learned from watching Moonshiners.

So moonshiners don’t have low concentrations of alcohol in their reboilers?

 

[DaveE]

I think you are interpreting this wrong. It seems to me you are looking at the wrong side of the phase diagram. If you start off with 15 % by volume of ethanol, the first vapor to come off at 195 F will be 60 % ethanol. If you raise the temperature from 195 F to 202 F and allow the vapor and liquid to equilibrate, you will have lowered the concentration of ethanol in the liquid to 7.5 volume %, and decreased the amount of liquid by only a small amount. This is all at a pressure of 1 atm.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

Hum... Now I'm confused. Isn't it true that at 1atm, 202F, EtOH concentration in the liquid can't fall below 7.5% in equilibrium?

Now, if you blew fresh air in you wouldn't be at equilibrium. Does that mean you can reduce the concentration lower, or are you just removing H2O and EtOH in proportion to maintain the same concentration?

 

[DaveE]

You don’t know much about distillation,, do you? As a ChE, my advice to you as an EE is to stick to Ohm's law.

Working with a phase diagram like this is very basic stuff to us Chemical Engineers.

One of the things I really like about PF is the ability to learn about subjects in which I lack expertise. I very much appreciate all of you that take the time to educate people outside of your area of expertise.

OTOH, you don't have to help others if you don't want to. But if you do want to teach, it's not a great approach to imply they shouldn't ask questions or shouldn't risk being wrong. It's not a crime to talk about "reverse distillation", even though that's not the description that ChEs would use.

In any case, we are aware we aren't Chemical Engineers, but thanks for the reminder.

 

[Chestermiller]

Hum... Now I'm confused. Isn't it true that at 1atm, 202F, EtOH concentration in the liquid can't fall below 7.5% in equilibrium?

Yes, that’s what I said. At 202, it is 7.5% in the liquid.

Now, if you blew fresh air in you wouldn't be at equilibrium. Does that mean you can reduce the concentration lower, or are you just removing H2O and EtOH in proportion to maintain the same concentration?

This diagram is for the binary system water and alcohol; no air.

 

[DaveE]

This diagram is for the binary system water and alcohol; no air.

Oh. Yes you're right, I didn't know that part.
At a basic level, what difference would a bunch of Nitrogen molecules make?
Can you treat vapor pressures separately (non-reactive, of course), like superposition in linear systems?

 

[hutchphd]

This diagram is for the binary system water and alcohol; no air.

Oh. Yes you're right, I didn't know that part.

But if the purpose is to remove ethanol from the liquid why would you maintain the the alcohol vapor eqilibrium??? Take the lid off the boiling pot and let the ethanol diffuse away. Let the volume in the kettle decrease until 11% of the liquid is gone and/or the temperature has reached the value stipulated on the chart Equilibrium neither required nor requested.
I don't like to mix efficient engineering with physics pondering although I love them equally.

 

[DaveE]

But if the purpose is to remove ethanol from the liquid why would you maintain the the alcohol vapor eqilibrium??? Take the lid off the boiling pot and let the ethanol diffuse away. Let the volume in the kettle decrease until 11% of the liquid is gone and/or the temperature has reached the value stipulated on the chart Equilibrium neither required nor requested.
I don't like to mix efficient engineering with physics pondering although I love them equally.

Yes, which leads to my next/previous question.
At the liquid-vapor interface, is it always in equilibrium at the molecular level, no matter how much you try to blow fresh air onto it?

 

[Borek]

Equilibrium case is one for which exact calculations are possible, so it gives a good reference point. As in every other practical case (be it in chemistry of physics) real world is more complicated, typically to the point where it is easier to measure than to calculate.

So yes, this is a spherical cow diagram. But it is the best thing we have to analyze the situation, and it will tell us what are limits, what is possible and what is not (just like energy conservation makes discussions about over unity engines a moot).

 

[Chestermiller]

Here is a simple model to begin to work with.

Let m be the total number of moles of ETOH and water in the tank at any time, let x be the mole fraction ETOH, and let $\dot{m}_I$ be the molar flow rate of inert gas (insoluble also) bubbled through the liquid. Assume the liquid is agitated enough and the depth of the liquid is sufficient for the bubbles too reach vapor-liquid equilibrium with the liquid currently in the tank. Assume also that heat is added to the system at such a rate that the liquid remains at constant temperature T throughout the process. Let the total pressure of the system by constant at $P_T$, with the pure inert gas also furnished at this pressure and temperature T.

We will temporarily assume that the liquid solution is ideal such that Raoult's law is obeyed:$$p_{A}=xP^*_A(T)$$$$p_{W}=(1-x)P^*_W(T)$$where the $P^*(T)$'s are the equilibrium vapor pressures of pure water and ETOH at temperature T,, and the p's are the partial pressures iii the vapor.

Based on these considerations, the rate of change of the total number of moles of liquid in the tank any time is. $$\frac{dm}{dt}=-\frac{(p_A+p_W)}{P_T-(p_A+p_W)}\dot{m}_I\tag{1}$$The rate of change of the number of moles of ETOH in the tank is $$\frac{d(mx)}{dt}=m\frac{dx}{dt}+x\frac{dm}{dt}=-\frac{p_A}{P_T-(p_A+p_W)}\dot{m}_I$$or $$\frac{dx}{dt}=x(1-x)\frac{(P^*_W-P^*_A)}{P_T-(p_A+p_W)}\frac{\dot{m}_I}{m}\tag{2}$$Eqns. 1 and 2 can be integrated with respect to time, subject to the imposed initial conditions.

 

[hutchphd]

It ends up a rate problem but be aware that the atoms in the gas move roughly at the speed of sound. As for diffusion rates they are typically $cm^2 /s$ for atmospheric gasses. some small amounts will return, There a various laws of diffusion, but the rate at a rolling boil will not be significant here I believe.

Edit: Of course @Chestermiller knows the answer which would have taken me an hour at least!!. Notice how rate gets very large near the boiling point T of ethanol.

 

[DaveE]

Here is a simple model to begin to work with.

Thanks! I need to think about this some.

So if I were to add a third inert substance can I just extend these equations like this:

$$\frac{dm}{dt}=-\frac{(p_A+p_W+p_C)}{P_T-(p_A+p_W+p_C)}\dot{m}_I\tag{1}$$

 

[Chestermiller]

Thanks! I need to think about this some.

So if I were to add a third inert substance can I just extend these equations like this:

$$\frac{dm}{dt}=-\frac{(p_A+p_W+p_C)}{P_T-(p_A+p_W+p_C)}\dot{m}_I\tag{1}$$

That would apply only if the 3rd component were present in the wine or soluble in the wine. Otherwise, I have already assumed that an insoluble gas, such as N2, is being bubbled through the wine.

I'm wondering what a good temperature would be for the operation. I was thinking of something like 78.5 C, such that, at 1 atm. total pressure, the initial partial pressures of A, W, and N2 in the exit gas would be about 0.15 atm, 0.4 atm., and 0.45 atm. Thoughts?

 

[DaveE]

That would apply only if the 3rd component were present in the wine or soluble in the wine. Otherwise, I have already assumed that an insoluble gas, such as N2, is being bubbled through the wine.

OK, so molecules that won't transition across the liquid-vapor barrier can be ignored, except for their effect on the total pressure. That makes sense.

 

[ArtZ]

Hey folks, seems that by chipping at this, there is possibly a closed form analytical solution in the works. I will still move forward with the experimental approach. I mentioned earlier that I purchased a hydrometer.

The downside to the hydrometer is that it requires a 250mL volume of the wine to determine the ABV. That is disruptive during the heating process and will not provide accurate results since I would like to test for ABV during heating at 10 minute? intervals. Today, I purchased a Digital Refractometer for Wine/Grape Measurements (% Brix & Potential Alcohol). The Brix scale is used to measure sugar content.

https://www.thomassci.com/Instrumen...rape-Measurements-Brix-And-Potential-Alcohol#

This device will allow me to pipette a small quantity (2-3 drops) into the instrument to get the ABV at my desired sampling intervals.

Since the rain is clearing here, and the temperatures outside are starting to be seasonal, I can thinking about setting this experiment up on my patio. (don't want to smell up the house) I have a single burner butane stove that should work fine as the heat source.

The vessel will be a 12" diameter, 4.5 quart stock pot. I'll use my fast response K thermocouples for T measurement. The downside is that the meter resolution for temperature is only 1C. Before I start this, I'll pop over to Lab Pro and purchase appropriate small sample containers for this experiment.

I am still thinking of using a process temperature of ~83C. Any guesses on the time to reduce the Shaoxing wine from a C0 = .15 to a Cf =. 04? Any bets on the time? My guess from my now discredited calculations is ~ 50 minutes.

​

 

[Chestermiller]

Hey folks, seems that by chipping at this, there is possibly a closed form analytical solution in the works. I will still move forward with the experimental approach. I mentioned earlier that I purchased a hydrometer.

The downside to the hydrometer is that it requires a 250mL volume of the wine to determine the ABV. That is disruptive during the heating process and will not provide accurate results since I would like to test for ABV during heating at 10 minute? intervals. Today, I purchased a Digital Refractometer for Wine/Grape Measurements (% Brix & Potential Alcohol). The Brix scale is used to measure sugar content.

https://www.thomassci.com/Instrumen...rape-Measurements-Brix-And-Potential-Alcohol#

This device will allow me to pipette a small quantity (2-3 drops) into the instrument to get the ABV at my desired sampling intervals.

Since the rain is clearing here, and the temperatures outside are starting to be seasonal, I can thinking about setting this experiment up on my patio. (don't want to smell up the house) I have a single burner butane stove that should work fine as the heat source.

The vessel will be a 12" diameter, 4.5 quart stock pot. I'll use my fast response K thermocouples for T measurement. The downside is that the meter resolution for temperature is only 1C. Before I start this, I'll pop over to Lab Pro and purchase appropriate small sample containers for this experiment.

I am still thinking of using a process temperature of ~83C. Any guesses on the time to reduce the Shaoxing wine from a C0 = .15 to a Cf =. 04? Any bets on the time? My guess from my now discredited calculations is ~ 50 minutes.

​

Are you counting the amount of time it takes to heat up a gallon of wine from 20 C to 83 C?

 

[Dullard]

In a former life, I designed, built, and calibrated 'Breath Interlocks' - a device added to vehicles for those convicted of drunk driving (in many U.S. states). The calibration process used a 'wet-bath' simulator - air was bubbled through a defined ethanol/water solution at a controlled temperature. The product gas was used to calibrate the devices.
From that experience, I can say with confidence that Raoult's Law is of limited utility in the case of low concentration ethanol/water solutions. The inter-molecular forces (in solution) make a mess of the 'expected' ethanol fraction in the gas at a given solution concentration and also cause non-linear (WRT ethanol concentration) behavior over a range of ethanol concentrations.

Caveat: This is true for low ethanol/water concentrations - it is entirely possible that elevated ethanol concentrations behave in a more 'ideal' manner.

 

[Chestermiller]

In a former life, I designed, built, and calibrated 'Breath Interlocks' - a device added to vehicles for those convicted of drunk driving (in many U.S. states). The calibration process used a 'wet-bath' simulator - air was bubbled through a defined ethanol/water solution at a controlled temperature. The product gas was used to calibrate the devices.
From that experience, I can say with confidence that Raoult's Law is of limited utility in the case of low concentration ethanol/water solutions. The inter-molecular forces (in solution) make a mess of the 'expected' ethanol fraction in the gas at a given solution concentration and also cause non-linear (WRT ethanol concentration) behavior over a range of ethanol concentrations.

Caveat: This is true for low ethanol/water concentrations - it is entirely possible that elevated ethanol concentrations behave in a more 'ideal' manner.

Does this more ideal behavior include the region of the azeotrope?

 

[Dullard]

I don't know. My (worthless) guess is 'yes.' My (perhaps) more useful guess is that empirical data for ethanol/water is available (CRC, Perry's...)

 

[ArtZ]

Are you counting the amount of time it takes to heat up a gallon of wine from 20 C to 83 C?

I guess that I should. Under more controlled conditions, I could rely on E = m *Cp * T2-T1. Since a Joule = W*s, I could calculate that time. Since I don't have a hotplate with a known output power (a butane burner) I'll have to include the ramp time in the total time.

 

[Chestermiller]

I don't know. My (worthless) guess is 'yes.' My (perhaps) more useful guess is that empirical data for ethanol/water is available (CRC, Perry's...)

The azeotrope is a non-ideal solution effect.

 

[ArtZ]

Aaah... my CRC Handbook. I may have given it away. Good thought. It has tables for binary and ternary mixtures. Maybe I can locate a copy online.

 

[Borek]

Engineering toolbox site has plenty of tables, I will be surprised if they don't have ethanol/water data.