
12102012, 02:06 AM

#1

Mar 2012
Crown Point, Indiana
Posts: 29
Liked 1 Times on 1 Posts

Ok, maybe I'm making this more complicated than I have to but could I get a little help figuring this out? This year we've made "Apple Pie Liquor".. we bottle it in mason jars and give it as gifts for Christmas. Tastes sooo good! I want to make labels this year with the ABV on it..
the recipe consists of..
1.5 gal Apple Juice @ 0% ABV
1 gal Apple Cider @ 0% ABV
6 cups Everclear @ 95% ABV
1 cup Buttershots @ 15% ABV
1 cup Gold Schlager @ 43.5% ABV
So what would my total ABV be? My calculations put me at about 13% but that sounds too low.
If anyone's interested in the rest of the recipe (yields about 12 qts)..
Bring Apple Juice + Cider + Spices to boil, simmer 20 minutes
(Spices: 2 tsp apple pie spice, 10 cinnamon sticks, 2 cup brown sugar)
Cool to room temp, add alcohol + 2 tbs vanilla, mix thoroughly, transfer to jars.



12102012, 02:10 AM

#2

Jan 2009
Kingston, PA
Posts: 1,761
Liked 41 Times on 35 Posts

Quote:
Originally Posted by nullvalue
Ok, maybe I'm making this more complicated than I have to but could I get a little help figuring this out? This year we've made "Apple Pie Liquor".. we bottle it in mason jars and give it as gifts for Christmas. Tastes sooo good! I want to make labels this year with the ABV on it..
the recipe consists of..
1.5 gal Apple Juice @ 0% ABV
1 gal Apple Cider @ 0% ABV
6 cups Everclear @ 95% ABV
1 cup Buttershots @ 15% ABV
1 cup Gold Schlager @ 43.5% ABV
So what would my total ABV be? My calculations put me at about 13% but that sounds too low.
If anyone's interested in the rest of the recipe (yields about 12 qts)..
Bring Apple Juice + Cider + Spices to boil, simmer 20 minutes
(Spices: 2 tsp apple pie spice, 10 cinnamon sticks, 2 cup brown sugar)
Cool to room temp, add alcohol + 2 tbs vanilla, mix thoroughly, transfer to jars.

That's what I come up with, too.
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12102012, 02:21 AM

#3

Stay Rude, Stay Rebel, Stay SHARP
Apr 2011
Arlington (DC), VA
Posts: 13,506
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Same here.
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12102012, 02:40 AM

#4

Mar 2012
Crown Point, Indiana
Posts: 29
Liked 1 Times on 1 Posts

Awesome, I guess I can do math! If you tasted this stuff you'd swear it was higher! thanks guys



12102012, 04:13 AM

#5

Nov 2004
Posts: 2,621
Liked 205 Times on 171 Posts

Quote:
Originally Posted by nullvalue
Ok, maybe I'm making this more complicated than I have to but could I get a little help figuring this out? This year we've made "Apple Pie Liquor".. we bottle it in mason jars and give it as gifts for Christmas. Tastes sooo good! I want to make labels this year with the ABV on it..
the recipe consists of..
1.5 gal Apple Juice @ 0% ABV
1 gal Apple Cider @ 0% ABV
6 cups Everclear @ 95% ABV
1 cup Buttershots @ 15% ABV
1 cup Gold Schlager @ 43.5% ABV
So what would my total ABV be? My calculations put me at about 13% but that sounds too low.
If anyone's interested in the rest of the recipe (yields about 12 qts)..
Bring Apple Juice + Cider + Spices to boil, simmer 20 minutes
(Spices: 2 tsp apple pie spice, 10 cinnamon sticks, 2 cup brown sugar)
Cool to room temp, add alcohol + 2 tbs vanilla, mix thoroughly, transfer to jars.

Total volume = 384 oz
Total alcohol = 25.14 oz
25.14/384 = .0654687
.0654687 X 2 = 13.0974 Proof



12102012, 04:20 AM

#6

Dec 2011
Durham, NC
Posts: 1,553
Liked 92 Times on 80 Posts

Quote:
Originally Posted by BigEd
Total volume = 384 oz
Total alcohol = 25.14 oz
25.14/384 = .0654687
.0654687 X 2 = 13.0974 Proof

Nah, it's ~50oz of ethanol. 13% ABV, 26 proof.
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12102012, 01:58 PM

#7

Aug 2010
McLean/Ogden, Virginia/Quebec
Posts: 9,408
Liked 1550 Times on 1181 Posts

You need to take into account the fact that if you add 20 mL of ethanol (15.785 grams @20 °C) to 80 mL of water (79.852 grams @ 20 °C) you get an 20% ABV solution with volume 98.238 cc i.e. not 100 cc. The alcohol molecules slip in between the water molecules to some extent so the volumes don't add.
6 cups of Everclear gives 6*236.6*.95*0.789239 = 1064.38 grams EtOH
1 cup of Buttershots gives 1*236.6*0.15*0.789239 = 28.0101 grams
1 cup of Gold Schlager gives 1*236.6*0.435*0.789239 = 81.2293 grams
for a total alcohol content of 1173.62 grams EtOH.
Now you need the total weight of the mixture. For the sake of getting some numbers we'll make some guesses. You should either weigh all the components (before or after mixing) or measure the specific gravity and volume and get the weight from SG*0.998203*Volume. You'll need specific gravity in any event. In the calculations below all the extra decimal places are there because I am pasting from a calculation utility program. They are not justified:
1.5 gal cider at 6 °P weighs 1.5*3.785*1.02369*0.998203 = 5.80156 kg
1 gal juice at 6 °P weighs 1*3.785*1.02369*0.998203 = 3.8677 kg
6 cups Everclear weighs 6*0.2366*0.81138 = 1.15184 kg
1 cup Buttershots weighs 1*0.2366*(1?)= 0.2366 kg
1 cup Gold Schlager weighs 1*0.2366(0.96?) = 0.2272 kg
for a total of 11.285 kg. Thus the ABW is 100*1.17362/11.2848 = 10.4%. To find ABW we need the specific gravity of the mixture. 10.4% ABW means 100 grams of solution has 10.4 grams of ethanol in it. The volume of 10.4 grams of ethanol is 10.4/(SG_EtOH*0.998203) and the volume of 100 grams of solution is 100/(SG_solution*0.998203) thus the ABV is ABW*(SG_solution/SG_EtOH). The specific gravity (20/20) of EtOH is 0.79066 so
ABV = ABW*(SG_solution/0.79066)
Assuming the SG of the mix to be essentially that of 6°P apple juice (and that's a WAG, of course) which is 1.02369 you would have
ABV = ABW*(1.02369/0.79066) = 10.4*(1.02369/0.79066) = 13.4652
or 27 proof. Now the first thing you should note after all this is that the numbers don't differ appreciably from what you calculated by ignoring the volume effect so that even if your measured SG and volume differ from the estimates I made the result won't be that much different. Obviously if one were to do this frequently he'd want to put the calculations in a spreadsheet.
One other note: all the EtOH density numbers I used were taken from the OIML polynomial for 20 °C. The TTB uses AOAC density data which is at 60 °F.
Edit: Just noted that the apple cider/juice are simmered. This will result in water loss and thus higher final ABV. Be sure to measure volume and SG after this simmering.



12112012, 05:32 PM

#8

Mar 2012
Crown Point, Indiana
Posts: 29
Liked 1 Times on 1 Posts

Quote:
Originally Posted by ajdelange
You need to take into account the fact that if you add 20 mL of ethanol (15.785 grams @20 °C) to 80 mL of water (79.852 grams @ 20 °C) you get an 20% ABV solution with volume 98.238 cc i.e. not 100 cc. The alcohol molecules slip in between the water molecules to some extent so the volumes don't add.
6 cups of Everclear gives 6*236.6*.95*0.789239 = 1064.38 grams EtOH
1 cup of Buttershots gives 1*236.6*0.15*0.789239 = 28.0101 grams
1 cup of Gold Schlager gives 1*236.6*0.435*0.789239 = 81.2293 grams
for a total alcohol content of 1173.62 grams EtOH.
Now you need the total weight of the mixture. For the sake of getting some numbers we'll make some guesses. You should either weigh all the components (before or after mixing) or measure the specific gravity and volume and get the weight from SG*0.998203*Volume. You'll need specific gravity in any event. In the calculations below all the extra decimal places are there because I am pasting from a calculation utility program. They are not justified:
1.5 gal cider at 6 °P weighs 1.5*3.785*1.02369*0.998203 = 5.80156 kg
1 gal juice at 6 °P weighs 1*3.785*1.02369*0.998203 = 3.8677 kg
6 cups Everclear weighs 6*0.2366*0.81138 = 1.15184 kg
1 cup Buttershots weighs 1*0.2366*(1?)= 0.2366 kg
1 cup Gold Schlager weighs 1*0.2366(0.96?) = 0.2272 kg
for a total of 11.285 kg. Thus the ABW is 100*1.17362/11.2848 = 10.4%. To find ABW we need the specific gravity of the mixture. 10.4% ABW means 100 grams of solution has 10.4 grams of ethanol in it. The volume of 10.4 grams of ethanol is 10.4/(SG_EtOH*0.998203) and the volume of 100 grams of solution is 100/(SG_solution*0.998203) thus the ABV is ABW*(SG_solution/SG_EtOH). The specific gravity (20/20) of EtOH is 0.79066 so
ABV = ABW*(SG_solution/0.79066)
Assuming the SG of the mix to be essentially that of 6°P apple juice (and that's a WAG, of course) which is 1.02369 you would have
ABV = ABW*(1.02369/0.79066) = 10.4*(1.02369/0.79066) = 13.4652
or 27 proof. Now the first thing you should note after all this is that the numbers don't differ appreciably from what you calculated by ignoring the volume effect so that even if your measured SG and volume differ from the estimates I made the result won't be that much different. Obviously if one were to do this frequently he'd want to put the calculations in a spreadsheet.
One other note: all the EtOH density numbers I used were taken from the OIML polynomial for 20 °C. The TTB uses AOAC density data which is at 60 °F.
Edit: Just noted that the apple cider/juice are simmered. This will result in water loss and thus higher final ABV. Be sure to measure volume and SG after this simmering.

Wow! I knew someone would come up with an answer like this! Next year I'll calculate it like this. Thank you



12112012, 05:43 PM

#9

I'm no atheist scientist, but...
Oct 2009
Thiensville, Wisconsin
Posts: 8,370
Liked 502 Times on 390 Posts

i would also ensure your everclear is indeed the 190 proof stuff  here in Wisconsin we can only get the 151 proof
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