Kevin just reminded me that I better get back to the library and study for the boards but this is much more fun, so go ahead and get some more coffee Andy (and whatever else Kev) as this is interesting stuff (in response to Guy's question) ==>
Conclusion #2: In order to cool Andy's tank from 88F to 78F by putting 2liter bottles full of ice in the sump rather than using 1 gallon of evaporation you would need 10.66 two liter bottles of ice!!! Glad you brought that up Guy, I hadn't thought about it that way...man, I really appreciate my 2.3 gallons per day of evaporation as I would need another refrigerator to hold the 24.52 bottles of ice each day!!!
Calculations #2:
-Assume that the ice is at 32F and will be melted and then heated to 78F
-heat of fusion of ice tells us that it would take 333 KJ to turn one KG of ice at 32F into one KG of liquid water at 32F. Then to heat the water we would use the specific heat of water to determine that it would take 106.8 kJ to then heat that water to 78F. So per liter it would take 439840.8 J to turn the ice into water at 78F.
-Recall, we needed 8060000J to cool Andy's tank 10F, so we would need 19.55KG of ice to do this.
-Since ice is less dense than water we also have to take into account that we could only put 1.834 liters of water in each bottle to end up with 2L bottles completely filled with ice (otherwise we would have a bunch of broken bottles).
-So as I said above, with 1.834 KG of ice in each bottle at 32F and with time to allow this to completely melt and then warm up to 78F you would end up needing 10.66 bottles to replace one gallon of evaporation!
Man, I'm sure that this stuff isn't on the boards!!!!
Talk to you guys later.
Todd