Units and Equations

Units of measurements that are created by combining other units often demonstrate how to perform the appropriate calculations. That may sound a bit confusing but knowing how this works can be very useful when answering exam questions.

Let’s look at what this actually means…

To calculate speed we need to measure a change in distance and the corresponding change in time. The units of distance and time are metres (m) and seconds (s) respectively. To perform the calculation we divide the distance by the time to get the speed. Similarly, we divide the unit of distance by the unit of time to get the unit of speed (m/s). The fact that the unit of speed is written as m/s tells us that a division took place to calculate the final answer – and that division process applies to both the numbers used and the units of those numbers.

The same thing also works in reverse so let’s look at how to use this trick to answer an exam question if you can’t remember the appropriate equation…

If you turn to Q7 (p230) in the CGP Complete Revision and Practice book, you will see that it is about the energy needed to boil water. Part (b) states: “Use Figure 10 to determine the specific latent heat of vaporisation for water” and provides a space for your answer, followed by the unit MJ/kg. This tells you that the answer can be obtained by dividing a value in MJ by a value in kg. All you need to do is find the appropriate values to use. The accompanying graph has MJ (energy transferred) on the y-axis and kg (mass of water boiled off) on the x-axis. There is also a straight trend line that shows the relationship between these two quantities.

The trick is to choose any point on the line and read the corresponding values for MJ and kg, then divide them to get the answer. A common mistake in this sort of situation is to do the division the wrong way around. You can avoid that pitfall by looking at the unit: you must put the MJ value over the kg value.

Let’s choose 0.3 MJ on the y-axis. If we read down from the trend line at that point we see that it corresponds with slightly over 0.13 kg. I would estimate a better value to be around 0.1325 kg (one quarter of the way from 0.13 kg to 0.14 kg). The calculation is therefore 0.3 / 0.1325, giving a final answer of 2.26 MJ/kg.

From a purist’s point of view this is a horrible approach: we should really be talking about finding the gradient of the line. But if you want a quick way to use the information in the question to help you find the answer that will get you some valuable exam marks, then this is a really useful trick to know.