17.7 Solid gold?

Before you read this, I suggest you read posts 16.44 and 17.6.

Suppose you are asked to find out whether a gold ring is made of solid gold or a cheaper metal with a gold coating. You need to do this without damaging the ring. How are you going to do it?

The simplest way is to measure its density. Cheaper metals have a lower density than gold (post 16.44). So, a cheap ring with a gold coating will have a lower average density than a ring of solid gold.

picture-17-7-solid-gold-cropped

One way of measuring the density is to use Archimedes principle (post 17.6). You need to weigh the ring surrounded by air and then surrounded by water, as shown in the pictures above. When you weigh something, the device that you use will measure the gravitational force but will tell you the mass of the object (post 16.17). To weigh the ring in water you will need to suspend it with a thread (post 17.6); so you will need to weigh the thread to make sure that its mass is negligible compared to the mass of the ring.

Suppose the mass of the ring is 5.3 g (remember that g is the abbreviation for a gramme, post 16.13) and appears to be 5.0 g when surrounded by water. We expect the apparent mass to be less when the ring is surrounded by water because its weight is less – its weight is less because the water gives some support to the water (see post 17.6). This support is sometimes called buoyancy.

According to Archimedes principle, if the ring has a volume V and an average density ρ its weight in air will be

Wair = Vρg

and its weight in water will be

Wwater = Vg(ρρo).

This is explained in post 17.6. Here ρo is the density of water and g is the modulus of the gravitational field. It’s important to distinguish g, the standard abbreviation for a gramme, and g (note the italics) which is the symbol I’m using to represent the gravitational field (see post 16.13 to find out more).

If we divide the second equation by the first, we get the result that

Wwater/Wair = (ρρo)/ρ.

Some simple algebra (see appendix) gives the result that

ρ = [Wair/(WairWwater)]ρo.

Since weight is proportional to mass (post 16.17) and the density of water is 1.0 ×103 kg.m-3 (post 16.44) the average density of the ring is

[5.3/(5.3 – 5.0)] × 1.0 × 103 = 18 × 103 kg.m-3.

This is less than the density of pure gold (see post 16.44).

However, solid gold jewellery is not pure gold; the gold is mixed with other metals to make it less soft (http://www.hsamuel.co.uk/webstore/jewellery/metalGuide/carats.cdo). The density of gold in jewellery varies from 18 × 103 kg.m-3 (22 carat gold which is 91.6% pure gold) to 11 × 103 kg.m-3 (9 carat gold which is 38% pure) (https://24carat.co.uk/densityofgoldandothermetalsframe.html).

So, the ring appears to be made of solid 22 carat gold.

In practice, you would need to investigate the precision of this result by repeating the measurements several times and calculating a standard deviation (post 16.24). The standard deviation could also be used to calculate the confidence interval in which you are 95% or 99% sure that the true value lies (post 16.26).

Finally, there are other ways of measuring the average density of the ring. Can you think of one?

 

Related posts

17.6 Floating
16.44 Density
16.7 Weight

 

Appendix

To derive ρ = [Wair/(WairWwater)]ρo from Wwater/Wair = (ρρo)/ρ.

Multiply both sides of the second equation by ρWair to get the result

ρWwater = (ρρo)Wair.

Now subtract ρWair from both sides of this equation to get

ρWwaterρWair = – ρoWair.

Now multiply both sides of this equation by – 1 to give

– ρWwater + ρWair = ρoWair.

This result can be written as

ρ(WwaterWair) = ρoWair.

Dividing both sides of the equation by (WwaterWair) gives the required result.

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s