### Video Transcript

Diagram a) is a Sankey diagram of a 1.8-kilowatt kettle. Diagram b) is a Sankey diagram of an 80-watt light bulb. Which device is more efficient?

Taking a look at both diagram a) and diagram b), we see this grid with a total input power shown in each one. Then, depending on the device, whether it’s the kettle or the light bulb, some fraction of that total input power is used to either heat the water or light the light. And likewise, some fraction is lost due to inefficiencies. In the case of the kettle, it’s lost as sound and heat through evaporation. And in the case of the light bulb, energy is wasted as heat.

The first question in this exercise asks, which device, the kettle or the light bulb, is more efficient? To figure this out, let’s recall what efficiency is from a mathematical perspective. Simply stated, efficiency is a ratio of output power to input power. We could think about it this way. If we had a device that was perfectly efficient, then all of the input power, 100 percent of it, would manifest as output power. None would be lost through heat, or sound, or other energy loss processes. In that case, when our output power is equal to our input power, our efficiency is one. That’s a perfect machine.

But in reality, machines aren’t perfect. They do lose some energy to unintended causes. We see in our diagrams, for example, that both these devices lose energy as heat. So they’re not perfectly efficient. Neither one has an efficiency of one. But our question simply asks, which device is more efficient? To figure that out, we’ll solve for the ratio of output power to input power for each of these two devices. In order to do that, we’ll use these Sankey diagrams. And in particular, we’ll use the grid marks on each diagram.

Sankey diagrams are designed to give us graphical information. They always exist on a grid. And if we count the number of grid boxes that go into the input side of the diagram, we can say that that represents the input power, 100 percent of it. Then, as we see, some fraction of that power is dissipated through sound and heat in the case of the kettle. And another fraction goes into its intended purpose of heating the water in the kettle. It’s the power used for this purpose, the intended purpose of the kettle heating the water, that we consider the output power in our efficiency equation.

So the way to figure out this ratio, output power to input power, is to count the number of blocks that appear vertically in the output power side and divide that by the number of blocks that appear in the input power side. All this means we’ll have to concentrate very carefully on the grid in this diagram.

Let’s start out by calculating the efficiency of the kettle, we’ll call it 𝐸 sub 𝑘. And we’ll do this by studying diagram a). If we first look at the input-power side of this diagram, we’ll want to count the number of vertical blocks in this side. So we can start right down here at the bottom. There is one, two, three, four, five, six, then on to seven, eight, nine, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20. This means that the input power for this process is represented by 20 blocks, or 20 units. In our equation for kettle efficiency, we’ll write that this way. We’ll say that this efficiency is equal to sum output power, we haven’t solved for that yet, divided by the input power of 20 units, or 20 blocks.

And now that we figured out input power for the kettle, we’ll solve for output power using the same process. And recall that output power is considered useful, in this case it goes towards heating the water. So let’s count the number of vertical blocks on that side of the diagram. Starting at the bottom of our diagram, we have one, two, three, four, five, six, seven, eight, nine, 10, 11, then 12, 13, 14, 15, 16, 17, 18. 18 total blocks for output power for the kettle. So we’ll write that into our equation. We’ve now solved for the efficiency of the kettle. It’s 18 divided by 20. Or, if we write this as a decimal, 0.90.

Next, let’s calculate the efficiency of the light bulb, we’ll call that 𝐸 sub 𝑙. Just like the kettle, the light bulb has 20 units of input power. That is, if we count the number of blocks on the left side of this diagram, we get 20. But then, looking at the useful output part of this diagram, we see that much less of this power goes into making light. If we count the vertical blocks, we have one, two, three, four. So the efficiency of the light bulb is only four divided by 20. As a decimal, that’s equal to 0.20. Comparing these two efficiencies for the kettle and the light bulb, we see which one is more efficient. By a wide margin, it’s the kettle. Let’s move on now to the next two questions in this exercise.

The next two questions ask, what is the efficiency of the kettle? And what is the efficiency of the light bulb? Now, in answering the first question, to a large extent, we’ve answered these as well. We calculated the kettle’s efficiency to be 0.90 and the light bulb’s efficiency to be 0.20. All we need to do now, to report these as our final answers, is to convert them into percentages. That is, what percent does 0.90 represent? And what percent does 0.20 represent?

Well, considering the efficiency of the kettle, nine-tenths of a value is 90 percent of that value. This tells us the kettle efficiency is 90 percent. And similarly for the light bulb, two-tenths of a value is equal to 20 percent of that value. And we write that down as the efficiency of the light bulb, 20 percent. This concludes our look into these Sankey diagrams.