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So I've got a pump, block (incoming Supreme LT) and may use a free res i received with the pump. I also got a 120mm Tt rad but it's aluminium, so i don't want to use that with my shiny LT
Therefore i need some other method of cooling. The original plan was to use a horizontal volenti cooler or at worst my already built vertical bong, however the incredible heat capacity of a mass of water has been brought to my attention courtesy of MCPetrolhead and Year 11 Chemistry.
So time for some maths. Some of you may know this formula:
q = m C (delta)T
Where q = quantity of heat (joules), m = mass (of water, grams), C = specific heat capacity (J/k/g), deltaT = Change in temperature (celcius)
Q = 150W (approximate CPU output) x 60 seconds x 60 minutes x 10 hours = 5,400,000
C = 4.2 J/k/g (just a fact for water. You can research how this is calculated if you like)
m = 10 Litre res, may change to 5L or 20L or something.
t = ?
Therefore, transposing,
T = q / C m
= 5,400,000J / 4.2 J/k/g * 10000g
= 128.5°
TL;DR (or to save brain asplosions): 128°
Over 1 hour
T = q / C*m
= (150J*60s*60m) / 4.2*10000g
= 540,000J / 42000
= 12.8° over 1 hour of 150W load.
This is far more manageable. I think close to 12 degrees can be dissipated naturally
Obviously 128 degrees isn't too good. However there are many variables to consider:
1: the water/res will dissipate heat. If i have an opening (maybe a grill too?) in the top of the res some evaporation cooling will also take place
2: the load will not be a constant 150W. Much of the time will be at idle.
3: i can cycle reusable gel "hotpaks" - you know those things you can stick in the freezer or microwave?
Predictions:
The water and CPU temperature will rise exponentially as greater heat is added.
Comments, opinions, contradictions of my bucket chemistry?

Therefore i need some other method of cooling. The original plan was to use a horizontal volenti cooler or at worst my already built vertical bong, however the incredible heat capacity of a mass of water has been brought to my attention courtesy of MCPetrolhead and Year 11 Chemistry.

So time for some maths. Some of you may know this formula:
q = m C (delta)T
Where q = quantity of heat (joules), m = mass (of water, grams), C = specific heat capacity (J/k/g), deltaT = Change in temperature (celcius)
Q = 150W (approximate CPU output) x 60 seconds x 60 minutes x 10 hours = 5,400,000
C = 4.2 J/k/g (just a fact for water. You can research how this is calculated if you like)
m = 10 Litre res, may change to 5L or 20L or something.
t = ?
Therefore, transposing,
T = q / C m
= 5,400,000J / 4.2 J/k/g * 10000g
= 128.5°
TL;DR (or to save brain asplosions): 128°
Over 1 hour
T = q / C*m
= (150J*60s*60m) / 4.2*10000g
= 540,000J / 42000
= 12.8° over 1 hour of 150W load.
This is far more manageable. I think close to 12 degrees can be dissipated naturally
Obviously 128 degrees isn't too good. However there are many variables to consider:
1: the water/res will dissipate heat. If i have an opening (maybe a grill too?) in the top of the res some evaporation cooling will also take place
2: the load will not be a constant 150W. Much of the time will be at idle.
3: i can cycle reusable gel "hotpaks" - you know those things you can stick in the freezer or microwave?
Predictions:
The water and CPU temperature will rise exponentially as greater heat is added.
Comments, opinions, contradictions of my bucket chemistry?