[Dix Dogfight]

Well, a small part of the answer anyway.
There has been alot of discussion about how much heat the pump adds to the water and does it matter if i run it inline or submerged.

The pump consumes a certain amount of energy per second, around 20-30 Watts (W).
Some of that energy is transformed into waterflow and some into heatlosses in the pump itself.
The energy transformed into flow is eventually transformed into heat because of the friction in the system. If this were not the case then the velocity of the flow would increase untill it reaches lightspeed. And it doesn't so therfore all flowenrgy is transformed into heat in the water.
(And for all you nonbelievers pleace study THE FIRST LAW OF THERMODYNAMICS before you contradict this statement)

If i place my pump inline only the "flowheat" is added to the water and if i place it submerged all consumed energy is added to the water.

The question everybody wants to know is:
How much of the total amount of consumed energy is transformed into flow and how much into heatloss in pump.
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Here comes a calculation=answer of that question:
(I've used data from my Eheim 1250 as input)

E_tot=28W (total consumed power)
F_max=1200l/h=0.33l/s (maximum waterflow)
h_max=2m (maximum height)
E_flow=? power transformed to flow
E_loss=? powerlosses in pump
g=9.80665N/kg (gravityconstant)
rho=0.998kg/l (density of water in 20C)

(note that F_max and h_max don't occur simultaneusly but using them together gives the "best case senario")

E_flow=F_max*rho*g*h_max=
=0.33(l/s)*1(kg/l)*9.81(N/kg)*2(m)=6.5(Nm/s)=
=6.5W

E_loss=E_tot-E_flow=21.5W

Conclusion:
If the pump was able to pump 1200l/h to a height of 2m the powerinput into the water would be 6.5W and the powerloss in the pump would be 20.5W.
So by placing the pump under water you add 20.5W of unnecessery heat to the water.
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I hope someone can use this info and if you are of a different oppinion pleace
respond with an INTELLIGENT answer.
(i hope my english isn't to unbearable)

CU