# Mechanical Equivalent of Heat

Recently I have been looking up James Joule's experiment regarding the mechanical equivalent of heat. After viewing some drawings of the apparatus, I assumed that the lines holding the weights would have to be attached to the drum. They would then be allowed to fall completely through a certain distance until the lines were at full length and the mechanical motion of the fluid would have to be allowed to continue to flow, causing the weights to be pulled up a bit and drop again, oscillating up and down until the motion stopped and all of the mechanical energy from the weights would be transferred into the fluid.

However, upon watching some videos, (this video is about 14 minutes long; but, if you advance to about 5 min 30 sec, it will show the experiment in question.) I noticed that the weights were allowed to be stopped by a level surface (an external force) instead of doing what I have described above. To me, this means the weights were still in motion and, therefore, still had KE that was not transferred into the fluid.

Of course, the weights are moving very slowly, much more slowly than if they had been allowed to free-fall through the distance; so, their kinetic energy is very small compared to the potential energy they lost, meaning the vast majority of the potential energy was transferred into the fluid and into other, small energy losses.

Should I assume the speeds of the weights were small enough to ignore the KE in the weights?

The assumption seems to be that the potential energy lost by the falling weights was all transferred into the fluid. However, in the experiments, the lines have to be wound back up and the weights dropped several times in order to make a measurable difference in temperature. Each time the weights fall, they are stopped by an external force, meaning they have kinetic energy that is not transferred into the fluid after falling through a measured distance. It seems to me that some of the potential energy from the weights is transferred into the fluid; but, some of it is transferred into the kinetic energy of the weights falling.

• youtube.com/watch?v=f9eVagCUPAE This video is about 14 minutes long; but, if you advance to about 5 min 30 sec, it will show the experiment in question. Of course, the weights are moving very slowly, much more slowly than if they had been allowed to free-fall through the distance; so, their kinetic energy is very small compared to the potential energy they lost, meaning the vast majority of the potential energy was transferred into the fluid and into other, small energy losses. – Michael E. Mall May 13 '13 at 21:28
• The original calculations of Joule's constant were 4.186 J/cal. Modern measurements put it at 4.184 J/cal. If all of the potential energy lost by the falling weights is assumed to go into the fluid; then, that would result in an over-estimation of this energy, which is suggested by these results. I just don't have a feel for how to "sense" the magnitude of this error and how much of it is due to ignoring the kinetic energy versus other losses due to friction, the heating of the paddles instead of the fluid, etc. – Michael E. Mall May 13 '13 at 21:31
• en.wikipedia.org/wiki/Mechanical_equivalent_of_heat This links shows a more complete of an earlier version of the apparatus, using a single weight. – Michael E. Mall May 13 '13 at 21:33
• I've edited your link and remarks into the body of the question which is the preferred way to make clarifications (that way future readers don't have to parse a bunch of comments to figure out what's going on). Notice that comments can be deleted very easily: they are second class citizen on Stack Exchange sites and could go away. – dmckee May 13 '13 at 21:34

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!)

Only the change in kinetic energy or work done in slowing down the falling mass should be considered. The work done on the falling masses when they hit the floor or table has to be omitted because the energy involved is not transferred to the water.

## protected by Qmechanic♦Mar 8 '14 at 21:42

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