CD:
Now, I'm going to go into my old chemistry/thermodynamics here (and it's been a bloody while). If I'm off on some/any/all of these equations, please, correct me, don't flame me.

Anyways, first find out the total loop volume in mL. Then, add a heat source. ANYthing will work. Measure the temp of the water as it enters, and then as it leaves.

Now then, with that, you can calculate the heat loss of the water. The specific heat of water is 4.18 joules per grams degree celcius (j/gC). (note: 1g of water=1mL of water).
When you figure the joules of heat changed by the system you can convert joules to watts. Joules=Watts*seconds. Use a 1 second time, and there ya go.

Now then, please excuse me while I go find my kevlar flame suit (I probably got something wrong)...
-Ghent