Hey guys, want to check my math before I run off and waste things. Review/error finding is much appreciated!
I am attempting to make CaCl2 and gypsum solutions for a water presentation this Friday. My goal is to have people add 0.5mL-1mL of premixed solution to 2oz of beer in their glass to represent an adjustment of 0.5g/gallon of either CaCl2 or gypsum) of per 1mL of solution added to the 2 oz sample.
I think this means that I need to use 3.906g in 500mL of solution to get to this goal.
In my head, adding 0.5mL of this solution to 2oz of beer would be roughly similar to adding 0.25g/gallon of CaCl2 or gypsum and adding 1mL of this would be like 0.5g/gallon of either mineral.
Hopefully this will give an idea of what adding approximate amounts of minerals would do to certain beers.
Math looks right to me. Great experiment, and one that I should run myself sometime as it might be very useful.
I’ve got to say… initially, I was a little concerned with the dilution of 1 mL to 2 oz beer, but when I realized that this is a volume increase of just 3.5%, I believe that will be barely detectable. In other words, you’ll be watering down the beer very very slightly, but probably not enough that anyone will care… Sort of like how most people probably won’t even be able to taste the extra salts in the beer! But who knows until you try.
You might want to try this blind with 3 or 4 different samples for each person, and tell them maybe that you messed with the salt additions with some but not all of the samples, and see if they can then almost blindly figure out which is which. My guess is most people probably cannot. Should be especially interesting if you were to do something like that.
Other the fact that you are not accounting for the 1ml dilution from the adjusting solution, your math checks out.
I conduct all of my lab work using the metric system because I find that it is better suited to small measurements than our modified English measurement-based system.
U.S. Gallon (a.k.a. Queen Anne’s Wine Gallon) = ~3,785ml
U.S. Fluid Ounce = ~29.6ml
A 0.5g per U.S. Gallon solution in milligrams per milliliter
milligrams_per_milliliter = 500 / 3,785 = 0.132 milligrams per milliliter
two_fluid_ounces_in_milliliters = ~59.2
beer_plus_adjusting_solution_total_milligrams = 59.2 (beer) + 1 (adjusting solution) * 0.132 = 7.95mg, which for the forum readers means that each milliliter of adjusting solution has to contain approximately 8 milligrams of adjusting salt(s).
adjusting_solution_500ml = 500 x 7.95mg = 3.975 grams (just round to 4 grams)
Yeah, I know. I figured it was a pretty good approximation to get some club members excited/informed about the possibilities of water adjustment. I can deal with a 3.5% in this setting.
My club has started keeping the PDFs of our presentations (some have audio recordings) here if people are interested in listening to myself and other club members talking about nerdy stuff. This presentation will be up probably by the end of the month.
That is correct. It was an oversight on my behalf.
Learning to be comfortable with the metric system is like learning a foreign language. The problem at first is that one wants to think in one’s native language and translate. What happens over time is that one learns to think in the foreign language, and that is what happens with the metric system. The metric system is superior to our modified English unit of measurement system. It’s a base-10 system of measurements that matches our base-10 number system.
I will give you an example; namely, weight by volume (w/v) and weight by weight (w/w) solutions. While a milliliter of water weighs one gram, a U.S. fluid ounce of water weighs 1.0432 dry ounces. A 1L (1000ml) solution that contains 100 grams of extract is a 100 / 1000 * 100 = 10% w/v solution. A 1 U.S. quart 10% w/v solution contains 3.2 * 1.0432 = 3.34 dry ounces of extract. Which solution is easier to calculate in one’s head?
By the way, a 10% w/v solution has a specific gravity of 1.040, which why one sees it along with 5% w/v (1.020) solutions used so frequently when preparing laboratory media. A 10% w/v solution is a 10 Degree Plato solution when the solute is water because 1ml of water weighs 1 gram, and Degrees Plato is a weight by weight system. A beautiful thing about working with weight by weight when the solute is water is that one can determine the S.G. of a solution at 16C/60F by knowing the weight of the solution and the weight of the extract mixed into solution. No hydrometer is required.
