Question

In 1983, the United States began coining the one cent piece out of copper clad zinc rather than pure copper. The mass of the old copper penny is $\mathbf{3}\mathbf{.}\mathbf{083}\mathbf{}\mathit{g}$ and that of the new cent is $\mathbf{2}\mathbf{.}\mathbf{517}\mathbf{}\mathit{g}$. The density of copper is $\mathbf{8}\mathbf{.}\mathbf{920}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$ and that of zinc is $\mathbf{7}\mathbf{.}\mathbf{133}\mathbf{}\mathit{g}\mathbf{/}\mathit{c}{\mathit{m}}^{\mathbf{3}}$. The new and old coins have the same volume. Calculate the percent of zinc (by volume) in the new cent.

Short answer

The percent of zinc (by volume) in the new cent is $f=91.64\%$.

Step-by-step solution

A concept:

The mass is given by:

$m=\rho V$

….. (1)

Where, $V$ is the volume of the body, $\rho$ is the Density of body, and $m$ is the Mass of the body.

When an object is partially or completely submerged in a fluid, the fluid exerts an upward force on the object called buoyancy. According to Archimedes' principle, the magnitude of the buoyant force is equal to the weight of the fluid displaced by the object.

$B={\rho }_{fluid}Vg$

(2) Here, is the volume and is gravity.

Calculate the percent of zinc (by volume) in the new cent:

Let $f$represent the fraction of the volume $V$ occupied by zinc in the new coin.

You have, the mass for both coins:

$$\begin{gathered} m=\rho V \\ 3.083g=(8.920g/c{m}^3) \end{gathered}$$

By substitution,

$$\begin{gathered} 2.517g=(7.133g/c{m}^3)\left(fV\right)+(8.920g/c{m}^3)(1-f)V \\ 2.517g=(7.133g/c{m}^3)\left(fV\right)+3.083-8.920g/c{m}^3\times fV \\ fV=\frac{3.083g-2.517g}{8.920g/c{m}^3-7.133g/c{m}^3} \end{gathered}$$

And again, substituting to eliminate the volume,

$$\begin{gathered} 2.517g=(7.133g/c{m}^3)\left(fV\right)+3.083g-8.920g/c{m}^{3}fV \\ f=\frac{0.566g}{1.787g/c{m}^3}\left(\frac{8.920g/c{m}^3}{3.083g}\right) \\ f=0.9164 \\ f=91.64\% \end{gathered}$$