Wow - a three year old thread.
According to my calculations, if you start with 5 gallons of water at 60 ºF with 20 lbs of ice at 0 ºF added to it, and try to cool 5.5 gallons of water (since I don't know the heat capacity of wort vs. water) from 212 ºF, they will reach equilibrium at 77 ºF.
5 gallons water at 60 ºF = 5*8.33 lbs = 41.7 lbs = 18.9 Kg at 15.6 ºC
20 lbs ice at 0 ºF = 9.1 Kg at -18 ºC
5.5 gallons water at 212 ºF = 5.5*8 lbs = 44 lbs = 20 Kg at 100 ºC
specific heat of water = 4.2 KJ/Kg ºC
specific heat of ice = 2.0 KJ/Kg ºC
latent heat of fusion of water = 333 KJ/Kg
heat to raise ice temp to freezing + heat of fusion + heat to raise temp of melted ice and water to x degrees = heat removed when "wort" is cooled to x degrees where x is the equilibrium temperature
(2.0 KJ/Kg ºC)(18 ºC)(9.1 Kg) + (333 KJ/Kg)(9.1 Kg)
+ (4.2 KJ/Kg ºC)(9.1 Kg + 18.9 Kg)(x ºC) =
(4.2 KJ/Kg ºC)(20 Kg)(100 ºC - x ºC)
328 KJ + 3030 KJ + (118 x) KJ = 8400 KJ - (84 x) KJ
202 x = 5042
x = 25
25 ºC = 77 ºF