# Heating water manually

Is it possible to heat a small amount of water, may be a cup of it, only by human force, using a small scale version of a device like Joule used to prove the heat mechanical equivalence, just by inner liquid molecular friction? Can a human, using his/her muscles generate enough energy for make a cup of water go from environment temperature to hot or boiling water? I had never seen anything alike, and I am very curious to know if there is such a thing.

I would also want to know what mathematical equations are involved here. Thanks in advance.

• If you try this, definitely use the best vacuum thermos you can find and pay attention to the integrity of the lid when you modify it. You'll also want gearing to match you muscle power output to the paddles. Sep 13 '20 at 17:12

Any kind of work done on the water eventually turns into thermal agitation (heat/temperature). It first produces waves (pressure oscillations) which is nearly adiabatic during a few seconds. But as long as the waves calm down in one way or another, the energy is turned into heat.

The energy required to heat 1.5 kg water from 20 degrees to 80 degrees (C) is around 400 kJ = 100 Wh. This is what a person can produce in one hour with all the muscles is his body.

You would need an efficient stationnary bike, sports training. Most importantly, an extremely effective insulating device otherwise the energy would be lost faster than your production rate before you can reach 100 degees (C).

Since a cup in only is only 0.05 liters, you could divide by 30, which yields 2 minutes of muscle power. But here the insulation is more a problem: you can't realistically heat only the water in the cup, you need to heat a significant part of the environment as well: the insulating appartus (cup + things around the cup...). The exact conductivity of the materials around the water would be a key to estimate numerical values.

• Where did you get the 1,5 kg value from? A cup here is an espresso cup! It has a capacity of 4 to 6 cL. So doing the average, which is 5 cL = 50 mL = 50 g water = 0.05 Kg water. Sep 13 '20 at 22:12
• Ok, I see where you got your 1,5kg value, I 've edited to fix your broken link. So the same equation for 0,05kg, using the specific heat of water value of 4182 J / kg º C gave me 12546 Joules. Sep 13 '20 at 22:33
• Ok. I hadn't seen you talked about a cup, Sep 14 '20 at 16:51

If you mean can a person supply sufficient pressure to raise the temperature of the water to its boiling point, the answer is no. Water is relatively incompressible. It takes an extraordinary amount of pressure to raise its temperature just a few degrees.

The person could, of course, turn the paddle wheel instead of using the weight and achieve the same result as the Joule experiment, which is to only raise the temperature a few degrees.

The relevant equation in the Joule experiment start with the first law of thermodynamics

$$\Delta U=Q-W$$

Where $$\Delta U$$ = the change in internal energy of the water. In this case

$$\Delta U= mC\Delta T$$

where $$m$$ is the mass of the water, $$C$$ is its specific heat, and $$\Delta T$$ is the increase in temperature.

$$Q$$ = the heat transfer between the water and surroundings, and is positive if heat transfers to the water. In this experiment, $$Q=0$$ because the water is not heated.

$$W$$ = the work done, and is negative when work is done on the water. In this case, it is the paddle work done on the water, sometimes referred to as shaft work.

So we have

$$mC\Delta T=-W$$

Since the shaft work done equals the loss of potential energy of the weight. If the weight drops a height $$h$$, then

$$mC\Delta T=Mgh$$

Where $$M$$ is the mass of the weight, and $$g$$ is the acceleration due to gravity.

So finally

$$\Delta T=\frac{Mgh}{mC}$$

All the above ignores the heat capacity of the vessel, heat losses through the vessel walls, and any friction in the apparatus.

Hope this helps.