Here comes some math, beware !!!

H = delta T * mCp

(heat) = (the change in temperature) * (the amount of energy required to change the temperature, per degree C, per unit of mass) * (mass)

10 Gallons = ~ 40 L

40 L = 40 KG (mass)

specific heatcapacity of water (Cp)= 4.186 kJÂ·kg−1Â·K−1

H = T * 40 KG * 4.18 KJ/(KG*k) = H = ~ 16.8 KJ/ T

T = (16.8KJ) / H

T is really delta T, I dropped the notation because of typing ^_^

that is to say, your 10 gallon reservoir requires 16.8 KJ to increase its temperature by one degree.

let us say you load 150 W from cpu + 48 W from pump,

heat dump (power)= 198W = 198 J/s

198 J / 16800J = 0.01178 S

that is to say, the increase in temperature will be 0.01178 K per second, if we assume no loss of heat to the outside world.

to continue, over one hour, your water will increase by 39.6 Degrees C after the addition of 712.8 KJ of heat.

This is all innaccurate as you will be losing heat to your surroundings, I have no way of estimating it though...If you use a steel resevoir and place some heat-pipes through it, to the top. I expect it could handle it

Edited by wcdolphin - 4/10/10 at 5:50pm