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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

= 128.5Â°

TL;DR (or to save brain asplosions): 128Â°

Over 1 hour

Over 1 hour

T = q / C*m

= (150J*60s*60m) / 4.2*10000g

= 540,000J / 42000

=

**12.8Â°**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?