Up to the estimation problem statement.

How Big Is The Container?

Recall that we have a reasonable estimation that one tonne of lead is emitted daily from vehicle exhausts.

We wish to work out how much volume this will occupy, so we can compute the size of a container needed to hold all this lead.

You may remember that the density of water is 1 g / cm^3, which means that one litre of water weighs exactly one kilogram.

If lead had the same density as water, it would occupy a volume of

1 cm^3/g x 1 000 000 g = 1 000 000 cm^3

Now 1 000 000 cm^3 is equal to 1 m^3 (see the exercise on conversions), so the lead would occupy a volume of 1 m^3 in this case.

We don't know what the density of lead is at the moment (of course, we could easily look it up, but let's not cheat), but we do know that it is a heavy metal. So it is reasonable to assume that it is a lot more dense than water (i.e. a given volume of lead obviously weighs more than the same volume of water). In fact, it is not too far-fetched to assume that lead has a density that is at least ten times that of water. In that case, the volume of our emitted lead would be

1/10 cm^3/g x 1 000 000 g = 100 000 cm^3 = 0.1 m^3

So how big is 0.1 m^3? Imagine an ingot of lead that has a square base, 50 cm on each side and is 40 cm tall. That is 0.1 m^3. Many microwave ovens are larger than these dimensions, which is possible an easy object to visualise.

So we see that the second claim in the article is also false ... the amount of lead emitted each day is hardly a container load.


Some comments about this exercise.
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