It seems I keep finding things to plug into this Watts Up Pro – a device that can record the power that household devices use. This particular model is nice in that you can use it with Vernier’s data acquisition stuff.

If it seems like a good idea to measure the power for a refrigerator, why not a coffee pot?

## Power and Energy

Before talking about the physics, let me just show the fun stuff. Here is a plot of the power, and two temperatures. The first reading shows the temperature of the water in the coffee pot (I didn’t waste valuable coffee). The second temperature reading is just some water in a different coffee pot at room temperature (for reference).

This is a plot of the power.

This shows the coffee pot doing two things. First, it takes water and makes coffee. Boom. Just like that. On the graph, this is the mostly constant power consumption portion. After the coffee is finished brewing, the coffee heater turns on and off in order to keep the coffee at a constant temperature. Oh, 670 Watts. A little bit higher than I expected. I will need to remember this the next time the power goes out and I want to brew some coffee with my portable generator.

From this, can I determine the total energy the coffee pot uses to make coffee. Remember the definition of power:

I could use this and the graph above to find the total energy. In fact, it would mostly work since the power is approximately constant. However, I want better than that. And here is the trick. If I break this data up into a whole bunch of short time intervals, I can use the above equation to find the energy. Then I just need to add up the energy for all these intervals. This is what a numerical integration actually does. Logger Pro has the feature built in. So, the energy to make the coffee can be seen here.

Logger Pro lists this as 4.102 x 10^{5} Watts*seconds which happens to be the same units as Joules.

## Energy and Temperature

Ok, but what does all that energy do? I am not absolutely sure, but I think the coffee pot has to bring all that water to a boil. How does the water get in the holding tank and up to fall through the coffee grounds? There isn’t a pump in the pot. The only thing I could think of is that the pot boils some of the water at a time and that causes it to rise up some tube or something.

Here is a plot of the temperature of the stuff in the coffee pot as well as a plot of room temperature.

From this, it looks like the water starts at about 21.5°C and ends at about 81°C. I suspect that in between it did get to 100°C but cooled off a bit as the water fell through the coffee filter.

## The Calculation

Here is what I want to look at. How much energy would it take to bring 1 liter of water (that is what I started with) from 21.5°C to boiling at 100°C? How does this compare to the total amount of energy the coffee pot used?

To boil water, you need to do two things. First, increase the temperature to 100°C. Next, you need to keep adding energy to cause a phase change from water to a gas. The amount of energy to do this depends on the mass of water (*m*), the specific heat of water (*C*) and the latent heat of vaporization (*L*).

The energy required to boil the water would then be:

For water, the specific heat capacity is 4200 Joules/(kg*°C) and the latent heat of vaporization for water is 2.27 x 10^{6} J/kg. Yep. It takes a ton of energy to turn water into a gas.

One last thing. How much water did I use? It was 1 liter (0.001 m^{3}) in volume. The density of water is 1000 kg/m^{3}. This means the mass of the water is about 1 kg.

With these values, I get a required energy of 2.6 x 10^{6} Joules. Notice that most of this energy would just go into the phase change of the water.

But wait. If the coffee pot uses 4.102 x 10^{5} Joules and it takes 2.6 x 10^{6}Joules to boil the water, something isn’t right. Either not all of the water is boiled, or there is some hidden power supply in the coffee pot. Maybe this hidden power supply could be used for the good of all. Maybe we could have coffee pot-powered cars in the future? It might be quicker to implement than nuclear fusion.

Ok. So maybe I was wrong about the boiling. Maybe just enough of the water in the holding tank boils to kind of “push” the other water up a tube so that it can go through the coffee grounds. If I assume 100% efficient coffee pot (which clearly isn’t true), how much of the water would be boiled?

Let me rewrite my energy equation like this:

Here I am using *m*_{b} to represent the mass of water that boils. Solving for this, I get:

So, less than 1% of the water is boiled, but the other 99% of the water gets a free ride up to the top of the coffee pot.

Authors: