While I will appreciate the equation for this calculation,
I will really appreciate a reference for looking this up
myself.
No, I am not a student looking for homework. This is for
alcohol recipe which calls for 4 cups of sugar in 5 cups
of water and I want to convert it to metric.
Dick
Believe it or not, the answer is 8 liters.
Try it. Try it on a small scale if you think
I'm kidding. Fill a glass to the brim, and
see how much sugar you can dissolve in it.
The question is how much the final volume will be.
You can easily dissolve 4 l of sugar in 5 l of water,
for a final volume of 5 l.
I know that sounds crazy if you've never quantitatively
dissolved sugar in water, but it's true. Try it.
> Believe it or not, the answer is 8 liters.
By experience with large amounts of sugar and water, I must
respectfully disagree.
I have been making Limoncello for 2.5 years now and there is
always excess sugar water which gets tossed because it is of
no other use to me.
Tomorrow I will add 16.6 lbs to 11 liters boiling water and
measure the output in a graded fermenter pail. My money says
there will be more than 14 liters. But I'd like to have an
equation with which to work.
One day I will con my wive into my need for a 25 lb metric
scale until then it's U.S. weights.
I got an "A" in Chemistry 101, but the only chemistry I recall
has to do with the young women in the Chem Lab.
Dick
The volume of the water will be somewhat greater, but not equal to the
sum of the water and the volume of the sugar (ie, (wt of sugar)/
(density of solid sucrose)).
There may be tables do to this,eg, for winemaking. The term 'partial
molal volume' may help in the search.
DB
I did and the volume was somewhat greater. It wasn't the worlds most
accurate quantitative test, but it was more than the water added.
I suspect you might be right for dissolving anhydrous sucrose in water.
Regards,
Martin Brown
I did and the volume was somewhat greater. It wasn't the worlds most
It is predictable if you know the state of hydration of the sugar you
are using. This varies from soft brown unrefined sugar that is already
visibly wet (which I used) to pure hard dry crystal refined sucrose. I
don't offhand know the water of crystallisation of pure sucrose.
1L water + 1L ethanol is a good mixing liquids demo. And might in part
be relevant to the OPs question in the manufacture of lemoncello. The
sum of these two components mixed is less than you might naively expect.
Regards,
Martin Brown
Anybody have a table of sucrose gradients and densities?
Mine aren't handy. Ah. Google is my friend.
books.google.com, Beet Sugar Handbook, Table A.1, page 779.
Your 1 kg of sucrose in 8 L (8 kg) water is an 11.1
weight-% solution. Table A.1 says that 11% sucrose
has a density of 1044.13 kg/m^3 at 20 C. Let's call that
1.044 g / mL.
1 mL 11% sucrose solution = 1.044 g
0.89X g water + 0.11X g sucrose = 1.044g
1.0X = 1.044
X = 1.044
0.89(1.044) g water + 0.11(1.044) g sucrose = 1 mL = 1.044 g
0.929 g water + 0.115 g sucrose = 1.044 g = 1 mL
0.929 mL water + 0.115 g sucrose = 1.044 g = 1 mL
The scaling factor from 0.929 mL water to 8000 mL water
is ~8611.
The scaling factor from 0.115 g sucrose to 1000 g sucrose
is ~8695. (The numbers SHOULD be the same but are probably
off due to my rounding errors.)
Taking the average scaling factor as ~8500,
0.929 mL water + 0.115 g sucrose -----> 1 mL
8000 mL water + 1000 g sucrose -----> ~8500 mL
There is no chemical theory to that. It's algebra
plus the known (empirical) density of the weight-%
sucrose solutions at 20 C.
Let us know if that's close to what you get when you
brew your next batch.
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The anonymizer might post things out of sequence, so if
you don't see my PREVIOUS post based on 1 kg sucrose + 8 L
water, please wait for it to show up. Same table of
sucrose densities; same calculations except different
amounts of sugar and water:
16.6 lb = 7.53 kg sucrose
11 L water = 11 kg water
7.53 / (7.53 + 11) = 7.53 / 18.53 ~= 0.40 = 40% solution.
Density of a 40 weight-% sucrose solution at 20 C = 1.17853 g/mL
0.6X g water + 0.4X g sucrose = 1.1785 g = 1 mL
1.0X = 1.1785
X = 1.1785
0.6X = 0.707 g water
0.4X = 0.471 g sucrose
0.707 mL water + 0.471 g sucrose = 1.1785 g = 1 mL
11 L = 11000 mL
11000 / 0.707 = 15,559 (scale factor)
7.53 kg = 7530 g
7530 / 0.471 = 15,987 (scale factor)
Avg Scale Factor ~= 15773
0.707 mL water + 0.471 g sucrose = 1.1785 g = 1 mL
scale by 15773:
11000 mL water + 7530 g sucrose = 18530 g = 15773 mL
You should get approx 15.77 L.
Let us know your actual result (at 20 C).
There is a recipe for Limoncello. It calls for
1.50 liters of Vodka
1.25 liters of water
800 grams of table sugar
Plus the zests of lemons
I make this recipe 2-3 times a year with 14 liter
of Vodka (eight 1.75 liter bottles). It all goes
into a 29 liter HDPE pail. My constant problem
has been excess sugar water. I have solved that
problem by getting a 45 liter HDPE pail.
BUT I still want to know the expected volume of
1250 ml of water plus 800 grams of sugar. Much
more important than that, I'd like to know where
to look it up for myself.
EXPERIMENT:
-----------
Using a 2 liter container premarked every 200 ml
a metric measuring cup, I put 1250 ml of room
temperature water into the container. Using an
11 lb scale marked every 200 grams. I added 800
grams to the container. After shaking several
times over a three hour period, my eyeball
estimate is 1750 ml or a 40% increase in volume.
NOTE:
-----
U.S. weights and measures are my native tongue.
They must have value somewhere, but it is not in
cooking or brewing.
Dick
The earlier post (Nomen Nescio; Sept 10) gave the details
of the calculation. The density of a 40 w-% sucrose
solution is 1.17853 g/mL. From there, it's just algebra.
Once you compute the "1 mL scaling factor" for a 40 w-%
solution using your values:
1250 / 0.707 ~= 1768
800 / 0.471 ~= 1698
Avg ~= 1733.
0.707 x 1773 ~= 1250 mL H2O
0.471 x 1773 ~= 800 g sucrose
-----------------------------------------
1.178 g = 1 mL solution x 1773 ~= 1773 mL
1773 mL (at 20 C) is very close to your value
estimated to be 1750 mL (temp?).
>From the known density of 40 w-% sucrose, you can now
calculate the amounts to use to make any desired volume and
minimize waste.