109 workshop 1 density question

In 109 workshop 1 a lot of you struggled with the density conversion question . So here is a step-by-step approach to this question that might help you...

The question: What is the density of mercury (13.6 g cm-3) in units of kg m-3 ?

First of all, think about what 13.6g cm-3 actually means. This means that in one cmthere are 13.6 grams. So this is like saying a cube of mercury with dimensions of 1cm x 1cm x 1cm  (around the size of a sugar cube) weighs 13.6 grams

So that is what 13.6g cm-3 means.

Now the question wants you to convert the units g cm-3 to kg m-3 . Or, in other words, you need to find out how much a cube of dimensions 1m x 1m x 1m of mercury would weigh in kilograms.

So, to start with think about a cube with dimensions 1cm x 1cm x 1cm - this is about the size of a sugar cube. Now think of a cube with dimensions 1m x 1m x 1m - which could also be written as 100cm x 100cm x 100cm ...




So, you can see from the above diagram that there are 100 x 100 x 100 = 1000000 1 cm3 cubes in a m3 cube. Or in other words, we could fit 1000000 of our little 1cm3 cubes into a m3 cube.

Right, so we know that in one cmthere are 13.6 grams, and we need to find out how much mercury there is in one m3 . To find this out we just have to do 13.6 x (100 x 100 x 100) = 13600000 - so this now means that we have 13600000 grams of mercury in one m3 , or 13600000 gm-3 .

So we now know that a m3  box of mercury weighs 13600000g, also written as 13600000 g m-3 , but the question wanted the units to be kg m-3 .

This means that we have to convert our answer to kg. There are 1000 g in 1 kg, so we will have to divide our answer by 1000 to convert it to kg:

13600000 / 1000 = 13600 kg m-3

This is not the only way you could go about answering this question, but I think it is a nice visual approach - I hope it helps.

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