danb said:
I wonder how much ice (wet) it would take to make a significant temp change in say a 15000 gal ag pool.
Well you significant heat absorber there would be the transformation energy it takes to actually change the ice from a solid to a liquid.
So, an example...
for a 15000 gal pool, thats 15000gal X 3.7854118 kg/gal = 56781 kg (approx) or 56781000g (approx)
Lets say the starting temperature of the pool is 92deg F
Target temperature is 85 deg f
We will ignore the heat loss/gain through the sides and bottom of the pool for this example, as well as to-from the air.
Change in temperature desired is 7 deg F
Thats 33.33 deg C to 29.44 deg C a change of 3.89 deg C
lets also assume that the ice is at -10 deg C starting.
The target temperature for the ice is also 85 deg F (29.44 deg C)
The specific heat of water is 4.186 joule/gram °C
The heat of transformation for water ice is 333 kJ/kg or 333 J/G
The specific heat of water type ice is 2.1 j/g °C at 1 atmosphere (we are going to assume sea level here as well)
so we are going to wind up with some unknown mass amount at 29.44 °C . that mass will be Z
We need an energy budget, so taking water through a 3.89 °C change will require 56781000g (mass of pool water) X 4.186 J/g deg C X 3.89 = 924595685 joules to be absorbed by the ice on it way to 85 deg F
so we have 3 components to getting the ice to 85 deg F. Heating the ice to 0 deg C, melting the ice, then taking the now cold water to 85 deg F.
heating the ice 2.1 j/g deg C X 10 deg C X Z g = 21Z
melting the ice 333 j/g X Zg = 333Z
heating the cold water 4.186 j/g deg C X 29.44 X Zg = 1232.3584Z
Add all those up =1536.3584Z which has to equal 924595685, so solving for Z is 601810g (rounding)
so thats 601 Kg of ice.
or 1326.7617354 lb of ice.
if you buy it in 20 lb bags, thats 66 bags of ice. :blah: