Still a decision must be made. If you aim to use the 80W your system won't be able to run 24/7. If you set your temp 10C below ambient you can use the PC until you hit maybe 5C above ambient, after that You'd be better off with aircooling alone.
So you've got a temp delta of 15C to play with.
If your CPU radiates 100W it will take 360L of water 1 hour to heat up 1C.
That is if you would shut the TEC off when you start your PC. Since the TEC cools the water at them same time as the CPU warms it the time will be much longer before temperature rises 1C.
How effective the TEC cooling will be is tricky to caculate, since we don't know what kind of cooling it has. Only thing you can bet on is that you won't get 80W of cooling from it.
If you get 50W cooling from it, then you've doubled the time to heat up the water (if your CPU generates 100W heat).
So, back to our example. 360L is alot of water and it has to be well insulated!
Let's say you plan to use your PC max 8 hours a day.
Since you've got 50W cooling(from an aircooled 80W TEC) at hand we can half the water.
360 / 2 = 180L
We can take an increase of temp of 15C....
180L / 15 C = 12 liter (now we are getting somewhere!!)
Oh no... we need the PC 8 hours a day though....
8 x 12L = 96 L !!!
Okay.. so 96 L (whats that? 25 US gallons) if your PC runs at 100% at 100W 8 hours a day with a 15 C temp increase... (we haven't even considered pumpheat and insulation leakage)
(Hmm... sounds like alot of water.. I hope someone can correct if my math or formula is failing !!!!! )
(I used the formula 1 Joule can heat up 1 gram of water 1 celsius. And 1kW/hour = 3 600 000 joules)
So... it's doable but bulky (unless my calc is dead wrong somewhere.... arrgh.. "learn first, then speak of it"... someone will probably make me eat my own post...)