OK, look at it this way. The total power applied to the circuit must be dissipated (or radiated). You add up all contributions, and they must equal the applied power. If you are applying 5 watts, then you will be dissipating+radiating 5 watts. In such a small coil, you can ignore the radiation component.
The dissipations will primarily be in the wire of the coil and in the specimen (if the specimen is large enough). There may also be significant dissipation in the capacitors you use to get a match to your 50 ohm generator.
The coil, by itself, can be modelled pretty accurately as a pure inductive reactance in series with a pure resistance. Taking Win's example of 0.42uH, and assuming a coil Q (unloaded Q) of 40 (about right, probably, for a 1cm ID coil), that model becomes Z=0.066+j2.639. Note: 2.639/0.066 = Q = 40. If you put 5 watts of 1MHz RF into that coil (using a lossless matching network composed of a couple of good capacitors, from your generator), then the coil must be dissipating 5 watts, and the current must be such that i^2*R=5W. In other words, the current must be sqrt(5/0.066) = 8.7 amps RMS. The dissipation is all in the resistive part of the model. The inductive part only stores energy and does not dissipate it.
If you put your sample inside the coil, it will lower the Q by some amount, because it adds to the dissipation. Let's say that the Q is lowered to 30. Now the model for the inductor + sample is Z =
0.088+j2.639. For 5 watts input as before, the current is sqrt(5/0.088) = 7.53A RMS. But you can also reasonably assume that the contribution of the coil itself is still 0.066 ohms, and the sample is effectively 0.022 ohms. As Rich pointed out, the sample acts like the secondary of a transformer, but the effect of that reflected back to the primary, your coil, is 0.022 ohms. [Note: the sample may also affect the inductance slightly--I'm ignoring that in this calc.] So the power dissipated in the sample (including the liquid! Don't discount that...) is i^2*0.022 = 1.25 watts.
Once you understand that, you can see that if the Q drops by y percent from its coil-only value when you introduce the sample, then y percent of the power will be dissipated in the sample.
Does that help you out of your circular reasoning? Hope so!
Oh, and I really think you should re-read Win's original posting in this thread--I think you will actually find a low inductance coil just as easy or easier to match than one with lots and lots of turns, and it will be easier to make and probably easier to keep cool. For example, if you went with Z=50+j2000 (about 320uH), at 5 watts you'd have .316A and 630 V RMS across the coil. I'd rather work with a bit lower voltage... If you went with Z=0.031+j1.25, so that if you resonate it with a good capacitor it looks like 50 ohms, you get about 16 VRMS, but
12.7 amps RMS, and you'll have trouble finding a capacitor that can take that much current. The important thing to note here is that you will get essentially THE SAME field strength inside the coil at a given power level, independent of the coil's inductance. Pick a coil that's practical in terms of the currents and voltages and matching network complexity.
Cheers, Tom