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Old 08-22-2010, 03:31 PM   #1
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Default Table of Freezing Points of Alcohol Solutions

There was a question in another thread about freezing points
of beer, so I thought I'd post this info for reference. I wrote
a program to spit out the numbers.

%Alc Temp (F)
by
vol

3.0 30.2
3.1 30.2
3.2 30.1
3.3 30.0
3.4 30.0
3.5 29.9
3.6 29.8
3.7 29.8
3.8 29.7
3.9 29.7
4.0 29.6
4.1 29.5
4.2 29.5
4.3 29.4
4.4 29.3
4.5 29.3
4.6 29.2
4.7 29.2
4.8 29.1
4.9 29.0
5.0 29.0
5.1 28.9
5.2 28.8
5.3 28.8
5.4 28.7
5.5 28.7
5.6 28.6
5.7 28.5
5.8 28.5
5.9 28.4
6.0 28.3
6.1 28.3
6.2 28.2
6.3 28.1
6.4 28.1
6.5 28.0
6.6 27.9
6.7 27.9
6.8 27.8
6.9 27.8
7.0 27.7
7.1 27.6
7.2 27.6
7.3 27.5
7.4 27.4
7.5 27.4
7.6 27.3
7.7 27.2
7.8 27.2
7.9 27.1
8.0 27.0
8.1 27.0
8.2 26.9
8.3 26.8
8.4 26.8
8.5 26.7
8.6 26.6
8.7 26.6
8.8 26.5
8.9 26.4
9.0 26.4
9.1 26.3
9.2 26.2
9.3 26.2
9.4 26.1
9.5 26.0
9.6 26.0
9.7 25.9
9.8 25.8
9.9 25.7
10.0 25.7
10.1 25.6
10.2 25.5
10.3 25.5
10.4 25.4
10.5 25.3
10.6 25.3
10.7 25.2
10.8 25.1
10.9 25.1
11.0 25.0
11.1 24.9
11.2 24.8
11.3 24.8
11.4 24.7
11.5 24.6
11.6 24.6
11.7 24.5
11.8 24.4
11.9 24.4
12.0 24.3


Ray

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Old 08-22-2010, 03:33 PM   #2
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Cool, you got anything simple like that for boiling points/solution for making N/a beer? There's debate about boiling being a viable means of doing it, and it looks like something I'll be needing to play with for healt reasons.

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Old 08-22-2010, 04:27 PM   #3
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Quote:
Originally Posted by Revvy View Post
Cool, you got anything simple like that for boiling points/solution for making N/a beer? There's debate about boiling being a viable means of doing it, and it looks like something I'll be needing to play with for healt reasons.

Do you mean what the boiling point would be at reduced pressure?
If you boil beer at atmospheric pressure not only would ruin the
taste of the beer (that's why flash pasteurization is used when it
is used) but the boiling point of any typical beer isn't going to
be much different from 100C. What would happen is you would start
to collect vapor at say, 95C, then the temperature in the pot would
slowly rise until the ethanol was all gone, at which point the temp
would read 100C, or maybe a little higher because the dissolved solids
in the beer cause a boiling point elevation, unlike liquids where the
dissolved solids cause a freezing point depression. Go to Google
and type in "ethanol water phase diagram" in the exact phrase
box to see the behaviour of ethanol/water solutions.

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Old 08-22-2010, 05:02 PM   #4
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I get somewhat different answers. An 8% ABW solution contains 80 g alcohol per kG for a molality of 80/46.07 = 1.7365 mol/kg so its freezing point depression should be 1.858*1.7365 = 3.23 °C ~ 5.81 °F and the freezing point should thus be 26.19 °F. Pure ethanol has a density of 0.789 g/cc (at 20 C) and so these 80 grams correspond to a volume of 80/0.789 = 101.363 cc. An 8% ABW solution has density 0.98473 (20°C) and thus 1 kg of it a volume 1015.5 cc and the ABV is thus 9.98 percent - very close to 10. In your table for 10% ABV the freezing point is given as 25.7 - about half a degree lower.

Now I note that if the true extract of a beer is, say 5°P and I model it as sucrose, with a molecular weight of 342.1 and I assume that the low pH of the beer inverts each sucrose molecule into a fructose and a glucose that will increase the molality of the solution by 2*50/342.1 = 0.292 for an extra depression of 1.858*0.292 = 0.5 C ~ 0.98 °F and the freezing point would be 25.21°F , about half a degree lower than the figure you post. Lots of assumptions go into that last calculation because 1)many beers don't have TE as high as 5 °P 2) the TE is made up of, presumably, higher molecular weight dextrines and oligosaccharides (which leads to smaller depression than fructose/glucose would) 3)sucrose might not be fully inverted 4) a kg of 5 °P solution would contains less water than a kg of pure water.

So the questions really are: 1)Are your numbers based on a pure water ethanol solution and if so where do we differ? 2)If they account for the other stuff dissolved in beer, what was the model (how much of what)?

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Old 08-23-2010, 05:25 PM   #5
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Quote:
Originally Posted by ajdelange View Post
I get somewhat different answers. An 8% ABW solution contains 80 g alcohol per kG for a molality of 80/46.07 = 1.7365 mol/kg so its freezing point depression should be 1.858*1.7365 = 3.23 °C ~ 5.81 °F and the freezing point should thus be 26.19 °F. Pure ethanol has a density of 0.789 g/cc (at 20 C) and so these 80 grams correspond to a volume of 80/0.789 = 101.363 cc. An 8% ABW solution has density 0.98473 (20°C) and thus 1 kg of it a volume 1015.5 cc and the ABV is thus 9.98 percent - very close to 10. In your table for 10% ABV the freezing point is given as 25.7 - about half a degree lower.
First, there is an error in calculating the molality. Molality is moles
per kg of SOLVENT, not total volume of solution. 80 gm of ethanol
leaves 920 gm of water, so the correct molality is 1.7365/.92 = 1.8875.
I used the rounded value of 1.86 for the molal freezing point depression,
so that comes to 3.51C or 6.32F, which is more than the half degree
you report but it gives the same value in the table for 10% by volume.
I didn't check all the rest of your calculation, but I just
converted volume % to weight % by dividing volume% by 1.25.


The values in the table are for pure ethanol solutions, so they
are a maximum. There is no way for a homebrewer to know the
weight% of dissolved solids in any of his beers, since that requires
laboratory analysis. Even a guess would depend upon accurate
values for initial and final gravities. There is no way
to know exactly what the average molecular weight of those
solids is either (I suspect it's pretty high, higher than sucrose)
There is no way for a homebrewer to even know what the exact percent
alcohol he has. The use of the table is so you have some general
idea how low you can go, but there is no guarantee you won't
still freeze your beer, but in general the actual freezing point
should be lower (but not by much) than that in table, so you
should be able to go at least that low.

Ray
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Old 08-23-2010, 05:58 PM   #6
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Quote:
Originally Posted by rayg View Post
First, there is an error in calculating the molality. Molality is moles
per kg of SOLVENT,
Absolutely right! If I had a nickle for every time I've made that mistake I'd be a rich man.
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Old 08-23-2010, 07:01 PM   #7
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This thread got me thinking

[quote=rayg;2232962]There is no way for a homebrewer to know the
weight% of dissolved solids in any of his beers, since that requires
laboratory analysis.

The analysis is so simple that any motivated (probably very few) home brewer should be able to carry it out. Simply evaporate the beer down to about half or a bit less of its original volume then make back up to full volume with DI water and measure the SG. Convert this to Plato. Then the TE (True Extract) is TE = Plato*SG/SG_beer (SG_beer is the SG of the beer before evaporation). The only equipment required is a volumetric flask, a beaker and a hydrometer.

Then the other half:

Quote:
Originally Posted by rayg View Post
There is no way
to know exactly what the average molecular weight of those
solids is either
How did the earliest chemists measure molecular weight? Freezing point depression and/or boiling point elevation and that should work here too. If p= TE/100 the weight of the solids in V mL of beer is Ws= V*dens*p
and the weight of the water is Ww= V*dens*(1-p). dens is the density of the dealcoholized beer but it doesn't need to be calculated as it cancels out. The molality, call it M, is the number of moles of solids divided by the weight of the solvent (got that right this time): M = mA*p/(1-p) with mA being the average molecular weight. With the depression (or raise) dT = k*mA*p/(1-p) it's clear that mA = dT*(1-p)/(k*p) (with k being the appropriate constant depending on whether it's freezing point or boiling point being considered).

All this is not of very much practical use, of course, but I'm intrigued enough about the surviving sugar (and protein) that I may check it out on the next beer I analyze (in the fermenter now). So watch this space (if you're interested).
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Old 08-23-2010, 08:01 PM   #8
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[quote=ajdelange;2233214]This thread got me thinking

Quote:
Originally Posted by rayg View Post
There is no way for a homebrewer to know the
weight% of dissolved solids in any of his beers, since that requires
laboratory analysis.

The analysis is so simple that any motivated (probably very few) home brewer should be able to carry it out. Simply evaporate the beer down to about half or a bit less of its original volume then make back up to full volume with DI water and measure the SG. Convert this to Plato. Then the TE (True Extract) is TE = Plato*SG/SG_beer (SG_beer is the SG of the beer before evaporation). The only equipment required is a volumetric flask, a beaker and a hydrometer.
I don't see how this would be any more accurate for
homebrewers than simply estimating the true extract by
calculation. Most homebrewers don't have precision equipment.


Quote:
Originally Posted by ajdelange View Post
How did the earliest chemists measure molecular weight? Freezing point depression and/or boiling point elevation and that should work here too. If p= TE/100 the weight of the solids in V mL of beer is Ws= V*dens*p
and the weight of the water is Ww= V*dens*(1-p). dens is the density of the dealcoholized beer but it doesn't need to be calculated as it cancels out. The molality, call it M, is the number of moles of solids divided by the weight of the solvent (got that right this time): M = mA*p/(1-p) with mA being the average molecular weight. With the depression (or raise) dT = k*mA*p/(1-p) it's clear that mA = dT*(1-p)/(k*p) (with k being the appropriate constant depending on whether it's freezing point or boiling point being considered).

All this is not of very much practical use, of course, but I'm intrigued enough about the surviving sugar (and protein) that I may check it out on the next beer I analyze (in the fermenter now). So watch this space (if you're interested).
How are you going to measure the freezing point accurately?

I know that I have seen analyses of beer with a "percent
residual solids" reported in one of my brewing books.
I would then just assume that the average molecular weight
is that of maltose, and calculate a depression for
that, then do the calculation again assuming that
the molecular weight is higher, say that of maltoheptose,
and use those numbers as a range.

All I need to know is how low I can go before it freezes,
for lagering purposes. There is really no other use
for the freezing point table, I put it there mostly
just to satisfy anyone's curiosity. If I wanted to make
ice beer, for example, I'd have to watch it freeze because you
can't let it all freeze, you have to collect the
unfrozen portion after only some of it freezes.

Ray
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Old 08-23-2010, 08:39 PM   #9
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[quote=rayg;2233377]
I don't see how this would be any more accurate for
homebrewers than simply estimating the true extract by
calculation. Most homebrewers don't have precision equipment.[?QUOTE]

Well, as I said it's more of academic interest than practical significance. But it doesn't take precision equipment (unless you consider the hydrometers sold in sets of three by B^3 "precision"). Obviously from the practical consideration standpoint the best way to see what temperature your beer freezes at is to put it in the freezer and measure the temp. when slush starts to form.


Quote:
Originally Posted by rayg View Post
How are you going to measure the freezing point accurately?
With a thermometer or RTD. Actually, now that I think of it ebulliometer's thermometers are narrow range so that boiling point elevation might be the way to go.


Quote:
Originally Posted by rayg View Post
I know that I have seen analyses of beer with a "percent
residual solids" reported in one of my brewing books.
That's true extract. Your books might have used the term "residual" because TE is usually measured on the residue left behind in the distillation flask when alcohol is measured. Brewers (not homebrewers) use TE and alcohol content to calculate effective original gravity (using the same equation that homebrewers use to calculate alcohol from OE and apparent extract.


Quote:
Originally Posted by rayg View Post
I would then just assume that the average molecular weight
is that of maltose, and calculate a depression for
that, then do the calculation again assuming that
the molecular weight is higher, say that of maltoheptose,
and use those numbers as a range.
I could do that but that wouldn't be much fun. Given I've got the distillation apparatus set up and the dealcoholized, reconstituted beer it would be a snap to take the portion of that left over from residue gravity determination, put it back in the distilling flask and measure the boiling point. And no, I don't expect that this has much relevance to the average homebrewer.

Quote:
Originally Posted by rayg View Post
All I need to know is how low I can go before it freezes, for lagering purposes.
Understood. But this is a Brewing Science forum so I have no qualms about discussing brewing science (the TE method comes right out of the ASBC's MOAs).
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Old 08-24-2010, 07:42 PM   #10
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I found an analysis of a stout in a book I have,
the alcohol content of which was 6.37% by w, which
is 7.96% by v.

The amount of other material dissolved is pretty small,
the most being protein 5.8g/L and glycerol 2 g/L. Other
things include:

K .468g/L
P .227 g/L
So4 .130 g/L
Cl .250 g/L
Pyruvate .128 g/L
citrate .187 g/L
Acetate .219 g/L
Polyphenols .199 g/L

There are lots of other things present in the microgram/L range
but I'm not including them here.

I used the exact molecular weights for everything
except protein (I just used 115, which is the weight
of proline, as an average) and polyphenols (I used
250, which is about the weight of a biphenyl with
4 hydroxy groups). The total number of moles comes
out to be tiny (0.107 moles/L) compared to the ethanol
(1.38 moles/L). So the biggest change is just in the
reduced mass of the solvent which goes from
0.936 kg to 0.927 kg. The number of moles increases from
1.384 to 1.491 moles, and the molality increases
from 1.384/.936 = 1.479 to 1.491/.927 = 1.608.
That in turn means the freezing point decreases
from 27.0F to 26.6F.

Ray

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