Anyone care to give me a comment on this?  I'm having a discussion with 
someone about moisture produced within the hive during winter and he 
laid this on me.  Been away from calculating this stuff for forty years. 
Doesn't seem right.

Dave Tharle
Ardmore, AB
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Here's some math on 100lbs of honey with 20% moisture... nice round 
numbers to keep the math simple:
20% moisture on 100lbs (45.4kg) of honey is 9.08kg of water which at 1L 
per kg is 9.08kg of water.
Now for the rest of the honey:
This quantity of honey is 80% (80lbs) of fructose. Molar mass of 
fructose (C6H12O6) = 180.72g/mol

6 Carbon -> 12.01g/mol x 6 = 72.60g/mol
12 Hydrogen -> 1.01g/mol x 12 = 12.12g/mol
6 Oxygen -> 16.00g/mol x 6 = 96.00g/mol
Total = 72.60g/mol + 12.12g/mol + 96.00g/mol

80lbs of fructose = 36287g
36287g of fructose / 180.72g/mol = 201 mol of fructose.
For future ease, lets round this to 200 mol of fructose.

Fructose is consumed by the bees and burnt with the oxygen they consume 
to release carbon dioxide and water. Here's the balanced formula:

C6H12O6 + 6O2 -> 6CO2 +6H2O

Since 80lbs of fructose is roughly 200mol of fructose we need 1200 mol 
of oxygen to produce 1200mol of carbon dioxide and 1200mol of water.

200[C6H12O6] + 1200[O2] -> 1200[CO2] +1200[H2O]

Water as a molar mass of 18.02g/mol.

So 1200mol of water x 18.02g/mol = 21.6kg of water.
At 1L per kg we get 21.6kg of water released in the consumption of 80lbs 
of fructose.

So the total water in 100lbs of honey at 20% moisture is 9.08L + 21.6L = 
30.68 liters of water.

If getting over 30L of water off of 31.5L (110lbs) of honey still sounds 
crazy, realize that the bees will have to consume 38.4kg of oxygen to 
metabolize the honey. So 45.4kg of honey and 38.4kg of oxygen combine - 
through the wonders of cellular respiration - to release 30.7 liters of 
water inside the hive.
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