3.6 Buoyant force  (Page 5/9)

Calculating density: is the coin authentic?

The mass of an ancient Greek coin is determined in air to be 8.630 g. When the coin is submerged in water as shown in [link] , its apparent mass is 7.800 g. Calculate its density, given that water has a density of $1\text{.}\text{000}{\text{g/cm}}^3$ and that effects caused by the wire suspending the coin are negligible.

Strategy

To calculate the coin’s density, we need its mass (which is given) and its volume. The volume of the coin equals the volume of water displaced. The volume of water displaced $V_{\text{w}}$ can be found by solving the equation for density $\rho =\frac{m}{V}$ for $V$ .

Solution

The volume of water is $V_{\text{w}}=\frac{m_{\text{w}}}{{\rho }_{\text{w}}}$ where $m_{\text{w}}$ is the mass of water displaced. As noted, the mass of the water displaced equals the apparent mass loss, which is $m_{\text{w}}=8\text{.}\text{630 g}-7\text{.}\text{800 g}=0\text{.}\text{830 g}$ . Thus the volume of water is $V_{\text{w}}=\frac{0\text{.}\text{830 g}}{1\text{.}\text{000 g}{\text{/cm}}^3}=0\text{.}\text{830}{\text{cm}}^3$ . This is also the volume of the coin, since it is completely submerged. We can now find the density of the coin using the definition of density:

Discussion

You can see from [link] that this density is very close to that of pure silver, appropriate for this type of ancient coin. Most modern counterfeits are not pure silver.