Freeze Point calculation?

[Silver_Is_Money]

Is there an easy mathematical means whereby to compute the nominal freeze point for a beer if given its OG and FG (and thereby, its calculated ABV and ABW)?

This might come in handy for lagering.

 

[ajdelange]

Sure, approximately. In water the depression is 1.853 K·kg/mol. Multiply this by the moles of solutes in the beer per kg solvent. To do this you will have to work out the weights of alcohol and extract and water from the the Balling formula. Converting weight of alcohol to moles is trivial but for extract less so. What to use for the molecular weight? At least the molecular weight of triose. If we do that it gives a molecular weight almost 10 times that of ethanol. And we know the average molecular weight of extract is going to be more than that because of 4 unit and up sugars in it. An RDF of 50 - 60% is typical so that a beer with OE of 12 might exhibit an apparent extract of extract of 2 - 3 °P and have TE of 4.8 to 5 °P meaning a liter of it contains 48 to 50 grams of extract. Using 3*180 as the molecular weight of the extract we'd have 0.09 to 0.092 moles of extract. Such a beer would be about 6% ABV or about 4.7 ABW and thus contain about 47 grams of alcohol per liter which is 47/46 which is a tad over a mole. Thus it looks as if the alcohol will be the dominant factor. Our liter of beer is 6% alcohol and thus contains 940 cc of water with 47 grams of extract in it. We could work out the details and get the exact amount of water but for illustration lets wag it at 930 grams water. The depression is then

dT = 1.853*0.092/0.940 + 1.853*1/0.940 = 0.18 + 1.97 °C

 

[Silver_Is_Money]

AJ, you make it all sound easy, but I'll have to admit that I'm confused. Are there any online calculators that put all of this together for the intelligence impaired?

 

[ajdelange]

Probably. I forgot to point you at the Wikipedia article on freeze point depression.

 

[Silver_Is_Money]

I don't know how accurate it is, but I found this chart:

http://2.bp.blogspot.com/-uS2e_hWtUjM/UsMGdxJdd9I/AAAAAAAAA30/KYlbsd730CU/s400/table+prism.jpg

 

[ajdelange]

It agrees pretty well with the ROM calculation we did in #2 with the message being that for beers of nominal strength the depression will be around 2 °C and that most of that is due to the alcohol though the residual extract does have a small contributory effect.

 

[Silver_Is_Money]

Via linear regression analysis I have come up with this tentative relationship:

Freeze_Point_F = (-0.804255319148936 * ABW) + 32.1244680851064

It only seems to be potentially valid through the typical range of beer and wine percent alcohol concentrations.

R^2 = 0.999485025071148 (assuming that the data I fed it is valid to begin with)

No guarantees. Use at your own risk.

 

[ajdelange]

Looks good. You could use -0.80425*0.791 for the slope and then enter ABV rather than ABW if you find that convenient. You could also subtract off 1.8*1.853*10*TE/(3*180)/(1 - ABV/100) if you wanted to account for the extract but as we've noted that will be small compared to the depression from the ethanol.