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Hair of the Dog Dave Clone


read 11204 times • 89 replies • posted 6/2/2009 8:50:03 AM

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tgncc 2751:77
Originally posted by puzzl
Originally posted by goldtwins
Originally posted by puzzl
Originally posted by joet
Originally posted by puzzl
Iím going to go ahead and ice some crappy beer and drink the frozen solution and let you guys know what happens.


I think youíre in for a treat!


Considering eisbier is a style using freezing to greater concentrate alcohol then one might assume that under certain or most conditions icing will produce disproportionate icing of water and alcohol.


Iíve decided on an 06 Black Chocolate Stout. Iím going to pour the beer into a bowl and stick it in the freezer along with a small strainer, and will skim the ice every 10-15 minutes until Iíve got about 50% off. That sound like it make sense to people who know more about this stuff than me?

Matt TBS and myself did this exact thing with this exact beer. We were supposed to give a lecture on eisbeers and provide one. Well the day before the meeting we still had not brewed anything. So we stuck a bowl of a BCS 6 pack in the freezer and kept straining. the removed ice was an odd color and tasted gross so we didnít try drinking the whole thing for alcohol. The beer was stronger and concentrated yes but no ABV numbers will be provided as I donít remember any.


Iíve frozen out 2 ounces so far, though my freezer blows so the last hour was totally wasted as it warmed up to about 36. Iíve moved it to a colder section, hopefully I can have it half out by 9 or something so I can have ample time to experiment. Whatís even worse is that I canít have a beer while I wait as killing my sobriety will make the tipsy test far more difficult.


I tried this once with Lagunitas Hairy Eyeball. I racked off ice very hour for about 5 hours. Came out about 1/2 the volume, and I definitely caught a buzz.
6/2/2009 10:19:35 PM

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TheBeerSommelier
Originally posted by JoeMcPhee
It has to do with the fact that mixtures donít behave like equal proportions of their constituent parts. Water and ethanol form an azeotrope which will not purify beyond a certain point. For heat distillation, this occurs at a mixture of about 98% ethanol and 2% water... for freeze distillation, it is closer to 30% with a home setup. The situation you are describing though is one that is even more complicated because of the presence of sugar. Nobody (to my knowledge) has done a systematic analysis of this though.

When you freeze beer, you donít remove all of the ethanol from the frozen section... a 5% beer will freeze solid. You do certainly concentrate it, but you also concentrate the sugars. By the same token the slurry left behind still has some ethanol and sugar in it. Like I said, you donít concentrate these at the same rate, because sugar and ethanol have very different properties. If you want to know the ABV of your beer, you need to have it tested, either by GC or LC.

I donít really care if you donít believe it... keep doing math to convince yourself that itís right... Iíll believe the lab report.


There’s no question that nothing’s as accurate as a lab testing.
6/2/2009 10:26:58 PM

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OldSock
Originally posted by JoeMcPhee
It has to do with the fact that mixtures donít behave like equal proportions of their constituent parts. Water and ethanol form an azeotrope which will not purify beyond a certain point. For heat distillation, this occurs at a mixture of about 98% ethanol and 2% water... for freeze distillation, it is closer to 30% with a home setup. The situation you are describing though is one that is even more complicated because of the presence of sugar. Nobody (to my knowledge) has done a systematic analysis of this though.

When you freeze beer, you donít remove all of the ethanol from the frozen section... a 5% beer will freeze solid. You do certainly concentrate it, but you also concentrate the sugars. By the same token the slurry left behind still has some ethanol and sugar in it. Like I said, you donít concentrate these at the same rate, because sugar and ethanol have very different properties. If you want to know the ABV of your beer, you need to have it tested, either by GC or LC.

I donít really care if you donít believe it... keep doing math to convince yourself that itís right... Iíll believe the lab report.


Thanks for posting an explanation (thatís all you had to do 3 pages back). I don’t have enough (or care enough) to get it tested so I guess I’ll just call it a beer of unestimateable strength...
6/3/2009 8:42:09 AM

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puzzl 3249:138
Originally posted by OldSock
Originally posted by JoeMcPhee
It has to do with the fact that mixtures donít behave like equal proportions of their constituent parts. Water and ethanol form an azeotrope which will not purify beyond a certain point. For heat distillation, this occurs at a mixture of about 98% ethanol and 2% water... for freeze distillation, it is closer to 30% with a home setup. The situation you are describing though is one that is even more complicated because of the presence of sugar. Nobody (to my knowledge) has done a systematic analysis of this though.

When you freeze beer, you donít remove all of the ethanol from the frozen section... a 5% beer will freeze solid. You do certainly concentrate it, but you also concentrate the sugars. By the same token the slurry left behind still has some ethanol and sugar in it. Like I said, you donít concentrate these at the same rate, because sugar and ethanol have very different properties. If you want to know the ABV of your beer, you need to have it tested, either by GC or LC.

I donít really care if you donít believe it... keep doing math to convince yourself that itís right... Iíll believe the lab report.


Thanks for posting an explanation (thatís all you had to do 3 pages back). I donít have enough (or care enough) to get it tested so I guess Iíll just call it a beer of unestimateable strength...



Don’t you only need a few ounces to get it tested?


I think it would actually be really helpful to the community to have even this first data point to work with. This issue comes up over and over again, and to be able to say even for once instance "hey, this guy froze off 50% of a 10% beer and ended up with a 16% beer" would be extremely helpful.
6/3/2009 11:35:50 AM

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puzzl 3249:138
By the way, there is some great info on freeze distillation on wikipedia:



Such enrichment by freezing of a solution in water is sometimes oversimplified by saying that, for instance, because of the difference in freezing points of water (0 įC/32 įF), and ethyl alcohol (-114 įC/-173 įF), ďthe water freezes into ice...while the ethyl alcohol remains liquid.Ē This is false, and although some of the implications of that description are true and useful, other conclusions drawn from it would be false.

The detailed situation is the subject of thermodynamics, a subdivision of physics of importance to chemistry. Without resorting to mathematics, the following can be said:

* Freezing in this scenario would begin at a temperature significantly below 0 įC.
* The first material to freeze would not be water, but a dilute solution of alcohol in water.
* The liquid left behind would be richer in alcohol, and as a consequence, further freezing would take place at progressively lower temperatures, and the frozen material, while always poorer in alcohol than the (increasingly rich) liquid, would become progressively richer in alcohol.
* Further stages of removing frozen material and waiting for more freezing will come to nought once the liquid uniformly cools to the temperature of whatever is cooling it.
* If progressively colder temperatures are available,
o the frozen material will contain progressively larger concentrations of alcohol, and
o the fraction of the original alcohol removed with the solid material will increase.
* In practice, unless the removal of solid material carries away liquid, the degree of concentration will depend on the final temperature rather than on the number of cycles of removing solid material and chilling.

* Thermodynamics gives fair assurance, even without more information about alcohol and water than that they freely dissolve in each other, that
o even if temperatures somewhat below the freezing point of ethyl alcohol are achieved, there will still be alcohol and water mixed as a liquid, and
o at some still lower temperature, the remaining alcohol-and-water solution will freeze without an alcohol-poor solid being separable.
6/3/2009 11:36:19 AM

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ryan 3186:13
Originally posted by puzzl
By the way, there is some great info on freeze distillation on wikipedia:



Such enrichment by freezing of a solution in water is sometimes oversimplified by saying that, for instance, because of the difference in freezing points of water (0 įC/32 įF), and ethyl alcohol (-114 įC/-173 įF), ďthe water freezes into ice...while the ethyl alcohol remains liquid.Ē This is false, and although some of the implications of that description are true and useful, other conclusions drawn from it would be false.

The detailed situation is the subject of thermodynamics, a subdivision of physics of importance to chemistry. Without resorting to mathematics, the following can be said:

* Freezing in this scenario would begin at a temperature significantly below 0 įC.
* The first material to freeze would not be water, but a dilute solution of alcohol in water.
* The liquid left behind would be richer in alcohol, and as a consequence, further freezing would take place at progressively lower temperatures, and the frozen material, while always poorer in alcohol than the (increasingly rich) liquid, would become progressively richer in alcohol.
* Further stages of removing frozen material and waiting for more freezing will come to nought once the liquid uniformly cools to the temperature of whatever is cooling it.
* If progressively colder temperatures are available,
o the frozen material will contain progressively larger concentrations of alcohol, and
o the fraction of the original alcohol removed with the solid material will increase.
* In practice, unless the removal of solid material carries away liquid, the degree of concentration will depend on the final temperature rather than on the number of cycles of removing solid material and chilling.

* Thermodynamics gives fair assurance, even without more information about alcohol and water than that they freely dissolve in each other, that
o even if temperatures somewhat below the freezing point of ethyl alcohol are achieved, there will still be alcohol and water mixed as a liquid, and
o at some still lower temperature, the remaining alcohol-and-water solution will freeze without an alcohol-poor solid being separable.



This is a very good idea. Simply measure the freezing point of the beer (the temp at which ice and liquid are both present) and then use that to calculate the amount of alcohol present, since the sugars have a minimal effect on the freezing point. You could however take this into account with an estimation from your gravity readings.
6/3/2009 12:11:12 PM

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OldSock
Originally posted by puzzl
Donít you only need a few ounces to get it tested?


I think it would actually be really helpful to the community to have even this first data point to work with. This issue comes up over and over again, and to be able to say even for once instance "hey, this guy froze off 50% of a 10% beer and ended up with a 16% beer" would be extremely helpful.


Does anyone know where I could get this sort of testing done? Cost?
6/3/2009 1:20:11 PM

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JoeMcPhee 8314:502
Originally posted by ryan
Originally posted by puzzl
By the way, there is some great info on freeze distillation on wikipedia:



Such enrichment by freezing of a solution in water is sometimes oversimplified by saying that, for instance, because of the difference in freezing points of water (0 įC/32 įF), and ethyl alcohol (-114 įC/-173 įF), ďthe water freezes into ice...while the ethyl alcohol remains liquid.Ē This is false, and although some of the implications of that description are true and useful, other conclusions drawn from it would be false.

The detailed situation is the subject of thermodynamics, a subdivision of physics of importance to chemistry. Without resorting to mathematics, the following can be said:

* Freezing in this scenario would begin at a temperature significantly below 0 įC.
* The first material to freeze would not be water, but a dilute solution of alcohol in water.
* The liquid left behind would be richer in alcohol, and as a consequence, further freezing would take place at progressively lower temperatures, and the frozen material, while always poorer in alcohol than the (increasingly rich) liquid, would become progressively richer in alcohol.
* Further stages of removing frozen material and waiting for more freezing will come to nought once the liquid uniformly cools to the temperature of whatever is cooling it.
* If progressively colder temperatures are available,
o the frozen material will contain progressively larger concentrations of alcohol, and
o the fraction of the original alcohol removed with the solid material will increase.
* In practice, unless the removal of solid material carries away liquid, the degree of concentration will depend on the final temperature rather than on the number of cycles of removing solid material and chilling.

* Thermodynamics gives fair assurance, even without more information about alcohol and water than that they freely dissolve in each other, that
o even if temperatures somewhat below the freezing point of ethyl alcohol are achieved, there will still be alcohol and water mixed as a liquid, and
o at some still lower temperature, the remaining alcohol-and-water solution will freeze without an alcohol-poor solid being separable.



This is a very good idea. Simply measure the freezing point of the beer (the temp at which ice and liquid are both present) and then use that to calculate the amount of alcohol present, since the sugars have a minimal effect on the freezing point. You could however take this into account with an estimation from your gravity readings.

Sugars have a very large effect on the freezing point though, as does any other dissolved solid. That’s why pure water-ice mixtures will never be colder than 0C, while a salt water solution can be gotten down as low as -20 or so.
6/3/2009 1:22:42 PM

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ryan 3186:13
Originally posted by JoeMcPhee
Originally posted by ryan
Originally posted by puzzl
By the way, there is some great info on freeze distillation on wikipedia:



Such enrichment by freezing of a solution in water is sometimes oversimplified by saying that, for instance, because of the difference in freezing points of water (0 įC/32 įF), and ethyl alcohol (-114 įC/-173 įF), ďthe water freezes into ice...while the ethyl alcohol remains liquid.Ē This is false, and although some of the implications of that description are true and useful, other conclusions drawn from it would be false.

The detailed situation is the subject of thermodynamics, a subdivision of physics of importance to chemistry. Without resorting to mathematics, the following can be said:

* Freezing in this scenario would begin at a temperature significantly below 0 įC.
* The first material to freeze would not be water, but a dilute solution of alcohol in water.
* The liquid left behind would be richer in alcohol, and as a consequence, further freezing would take place at progressively lower temperatures, and the frozen material, while always poorer in alcohol than the (increasingly rich) liquid, would become progressively richer in alcohol.
* Further stages of removing frozen material and waiting for more freezing will come to nought once the liquid uniformly cools to the temperature of whatever is cooling it.
* If progressively colder temperatures are available,
o the frozen material will contain progressively larger concentrations of alcohol, and
o the fraction of the original alcohol removed with the solid material will increase.
* In practice, unless the removal of solid material carries away liquid, the degree of concentration will depend on the final temperature rather than on the number of cycles of removing solid material and chilling.

* Thermodynamics gives fair assurance, even without more information about alcohol and water than that they freely dissolve in each other, that
o even if temperatures somewhat below the freezing point of ethyl alcohol are achieved, there will still be alcohol and water mixed as a liquid, and
o at some still lower temperature, the remaining alcohol-and-water solution will freeze without an alcohol-poor solid being separable.



This is a very good idea. Simply measure the freezing point of the beer (the temp at which ice and liquid are both present) and then use that to calculate the amount of alcohol present, since the sugars have a minimal effect on the freezing point. You could however take this into account with an estimation from your gravity readings.

Sugars have a very large effect on the freezing point though, as does any other dissolved solid. Thatís why pure water-ice mixtures will never be colder than 0C, while a salt water solution can be gotten down as low as -20 or so.


Hummingbird food freezes at about 25F, water freezes at 32F and alcohol freezes -173F. I really hope you can realize why the amount of sugar doesn’t matter.

Again though, you can further refine your calculations by using the measured gravity as a measurement of the dissolved sugar and get a modified value for the freezing point of your water/sugar solution.
6/3/2009 1:28:51 PM

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JoeMcPhee 8314:502
Originally posted by ryan
Originally posted by JoeMcPhee

Sugars have a very large effect on the freezing point though, as does any other dissolved solid. Thatís why pure water-ice mixtures will never be colder than 0C, while a salt water solution can be gotten down as low as -20 or so.


Hummingbird food freezes at about 25F, water freezes at 32F and alcohol freezes -173F. I really hope you can realize why the amount of sugar doesnít matter.

Again though, you can further refine your calculations by using the measured gravity as a measurement of the dissolved sugar and get a modified value for the freezing point of your water/sugar solution.


You are always assuming that there is no ethanol in the liquid left behind though, which is clearly false. The freezing point of ethanol isn’t the question... what’s the freezing point of a 50% solution of water and ethanol... by your calculations, you’ll tell me it’s the average of -173 and 32, which it isn’t.
Freezing points of ethanol-water mixtures (by volume) in degrees C

0% 0
10% -4
20% -9
30% -15
40% -23
50% -32
60% -37
70% -48
80% -59
90% -73
100% -115

You don’t just derive physical properties of mixtures from an average of the physical properties of the pure compounds. With sugar it’s even more complicated because the freezing point of a sugar/water solution doesn’t even vary in a straight line.

From the Merck Index, the freezing points of glycerol (a sugar) and water mixtures.
0% 0
10% -1,6
30% -9,5
50% -23
67% -46,5
80% -20,3
90% -1,6
100% 17,8

When you add ethanol to this mixture, the system becomes even more complicated and I can’t find any information on it (although it has probably been done at some point).
6/3/2009 1:56:14 PM

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