Watching the video$\, \!^{*}$ I'd estimate the relevant dropping height $h$ and dropping speed $v$ (close to reaching the ground, and probably nearly constant for much of the drop) conveniently at about

$h \approx$ 1.0 m, and $v \approx$ 0.1 m/s.

Thus

$W_{pot}/m = g ~ h \approx$ 10 m^2/s^2

while

$E_{kin}/m = 1/2 v^2 \approx$ 0.005 m^2/s^2.

[Additional remark: 
Joule himself would have easily been able to calculate and account for this,  of course (and likely he did); and more precisely, too, based on measuring values of $h$ and $v$ of his actual setup.]

Therefore the ratio of kinetic energy of the weight (at the end of dropping) and its initial potential energy happens to be fairly close to the relative difference between the values of "original calculation" and  "modern measurements" you provided,

(4.186 J/cal - 4.184 J/cal) / 4.186 J/cal $\approx$ 0.0005.

However, looking at http://en.wikipedia.org/wiki/Properties_of_water (Table "Constant-pressure heat capacity", which I believe is relevant), the temperature-dependence of the quantity being measured seems at least as important. 

(*: nice one!; some Germans' public-service-broadcasting fees being put to good use!)

Mechanical Equivalent of Heat


The circuit shows an IR receiver whose output pin is interfaced with a PIC. Can you explain why a resistor and capacitor were used and why those values were picked.
