Quote:
Originally Posted by SGFBeerBuzz
Ouch. Math hurts me tiny brain.

It's actually more complicated than that.
I worked out some numbers to cool 5 gallons from 212 to an equilibrium temperature using various amounts of water and ice. Assumptions are:
heat capacity of wort is almost the same as water (very nearly true)
water at boiling weighs 8.0 lbs per gallon
water at room temp. weighs 8.33 lbs per gallon
ice has a latent heat of fusion of 143.9 BTU per lb
ice has a heat capacity just about 1/2 that of water
Assume we start with W gallons of water at 65º F and I lbs of ice at 0º F
BTU to cool 5 gallons of water from 212º F to equilibrium temp. X =
5(8.0)(212X) = 40(212X)
BTU to heat I lbs of ice from 0º F to 32º F = (1/2)(I)(32) = 16I
BTU to melt I lbs of ice = 143.9 I
BTU to heat the water from I lbs of melted ice to equil. temp. X =
(I)(X32)
BTU to heat W gallons of water from 65º F to equilibrium temp. X =
(W)(8.33)(X65)
therefore:
40(212X) = 16I + 143.9I + I(X32) + 8.33W(X65)
or
X = [(40)(212)+(8.33)(65)W  128I]/(8.33W+40+I)
W (gallons)......I (lbs)......Equilibrium temp. X (º F)

......5...............10.................108.1
....10...............10...................94.6
....15...............10...................87.6
....20...............10...................83.2
......5...............20...................84.9
....10...............20...................79.1
....15...............20...................75.9
....20...............20...................73.9
......5...............30...................65.8
....10...............30...................65.6
....15...............30...................65.5
....20...............30...................65.4
So you can see how hard it is to cool 5 gallons of 212º wort to pitching temp.