# Joule's “Work-Heat Equivalent” Experiment

I have been reading about the Joule's experiment that is supposed to have shown that work is the equivalent of heat. I can't really find the original paper and I am finding trouble understanding how the measurements really happened.

So the forces that are acted upon the masses that we let fall are their weights and the tensions that is acted by the rope. Therefore the work that is produced by these forces to the masses is:

$$W_{total}=W_W - |W_T|= U_{initial}-U_{final} - |W_T|$$

By the work-energy theorem we know that:

$$W_{total} = K_{final}- K_{initial}$$

Therefore if we define $U_{final} = 0$

$$|W_T|= U_{initial} - K_{final}$$

The work that is produced to the system (water-thermometer-paddles etc) is $|W_T|$

In this similar question they answer that $δθ$ is proportional to $δh$ however according to the above expressions this is possible only when the Kinetic energy in the final position is 0 as its is also pointed out in this question. But in the answers that it got says that the Kinetic energy although small it was probably measured by Joule. Therefore $δh$ and $δθ$ were not proportional.

So what am i missing? If it is better try to explain with math.

• If you care about the kinetic energy, you can measure the velocity of the falling mass and then you just apply ${1 \over 2}mv^2$ as a correction to the work done by the falling mass. – CuriousOne Jun 14 '16 at 0:03
• So what the experiment showed was that the amount of temperature rise is propirtional to the work that is performed and not to $Δh$ – George Smyridis Jun 14 '16 at 0:07
• Yes, that's hopefully what it showed, but you are correct that one should look at the original publication. – CuriousOne Jun 14 '16 at 2:39
• – Farcher Jun 14 '16 at 6:59