If you want to know a lot more about Fleischmann's calorimetry, for example, see:
http://lenr-canr.org/acrobat/MilesMisoperibol.pdf
Miles is better at explaining these things then Fleischmann was. I believe your focus in recent discussions here was on the effects of vaporization. That is an important issue, and as McKubre wrote here, it is measured and included in the calorimetric equations. See especially Appendix I, and look for the term L. See, for example, p. 26:
"The term involving the enthalpy of vaporization (L) contributes -4.9103 mW and -52.9857 mW, respectively, at these two cell temperatures, (Ref. 2)."
[The two temperatures are 40 deg C and 80 deg C. The cell is usually at 40 deg C, and ~5 mW is much smaller than the excess heat in the weeks leading up to the boil-off event. Obviously, vaporization dominates during the boil-off.]
Yes, obviously, vaporization dominates during the boil-off. And boiling was exactly the condition that Fleischmann's "major paper" was aimed to investigate, as stated at the beginning of the abstract (1): "We present here one aspect of our recent research on the calorimetry of the Pd/D2O system which has been concerned with high rates of specific excess enthalpy generation (> 1kWcm-3) at temperatures close to (or at) the boiling point of the electrolyte solution."
And the specific enthalpy generation greater than 1kWcm-3 (namely 3700 Wcm-3) was calculated on page 16 by means of a formula that doesn't appear in either the F&P paper of 1992, or in the Miles paper of 2008. In all the equations presented in their documents, the enthalpy of vaporization L is multiplied by P/(P*-P), whose denominator goes to zero at boiling, so all those sophisticated formulas are unuseful to calculate the enthalpy lost in this condition. In fact on page 16 of (1), they used a much simpler formula, such as:
[1] ΔEboil = ΔmW,ev L
Where:
ΔEboil is the enthalpy loss due to boiling
ΔmW,ev is the mass of water effectively evaporated
L is the latent heat of vaporization of water
The big problem is that they implicitly assumed that all the missing water was evaporated. On the contrary, they should have added the following mass balance as well:
[2] ΔmW,ev = ΔmW,cell – ΔmW,drop
Where:
ΔmW,cell is the mass of water lost by the cell not due to electrolysis
ΔmW,drop is the mass of water exiting the cell as water droplets!
If they really believed that the mass of water exiting the cell as droplets was zero, they should have declared it and prove it adequately.
Quoteit could only cause an error in the milliwatt range (or probably micro-watts), whereas the excess was 133 W, so this could not possibly be a significant factor, and it is not relevant.
This only happens when the error calculated at 40°C is arbitrarily compared to the excess heat allegedly generated at 100°C!