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 3.083 g and that of the new cent is 2.517 g. The density of copper is 8.920 g/cm3 and that of zinc is 7.133 g/cm3. The new and old coins have the same volume. Calculate the percent of zinc (by volume) in the new cent.

Answer 1

The percent of zinc by volume in the new cent is approximately 50.51%.

Step-by-step explanation:

The percent of zinc in the new cent can be calculated by comparing the masses and densities of copper and zinc. We can use the formula for density, which is mass divided by volume, to find the volume of each metal in the coins. Since the new and old coins have the same volume, we can set up an equation to find the percent of zinc by volume in the new cent.

Let's denote the volume of copper as Vc and the volume of zinc as Vz. We know that:

Density of copper = mass of copper / volume of copper = 8.920 g/cm³

Density of zinc = mass of zinc / volume of zinc = 7.133 g/cm³

Given the mass of the old penny (3.083 g) and the new cent (2.517 g), we can calculate the volumes of copper and zinc using their respective densities. We have:

Volume of copper = mass of copper / density of copper = 3.083 g / 8.920 g/cm³

Volume of zinc = mass of zinc / density of zinc = 2.517 g / 7.133 g/cm³

Substituting the actual values, we find that the volume of copper is approximately 0.346 cm³ and the volume of zinc is approximately 0.353 cm³.

To find the percent of zinc by volume in the new cent, we can use the formula: Percent zinc by volume = (Volume of zinc / Total volume) × 100%

The total volume is the sum of the volumes of copper and zinc, which is 0.346 cm³ + 0.353 cm³ = 0.699 cm³.

Finally, applying the formula, we find the percent of zinc by volume in the new cent is (0.353 cm³ / 0.699 cm³) × 100% approximately equal to 50.51%.

Answer 2

To solve this problem it is necessary to use the concepts related to density as a function of mass and volume. And apply this relationship to know the unknowns. In general, the density can be described as

`\rho = (m)/(V)`

Where,

m = mass

V= Volume

Our values are given as,

`m_c = 3.083g`

`\rho_c = 8.96g/cm^3`

`v_c = (m_c)/(\rho_c)\rightarrow v_c = (3.83)/(8.96)=0.4274cm^3`

The mass of the mixture between copper and zinc is

`m_(zc) = 2.617g`

Let mass of copper in the new coin be x and that of zinc be
`(m_(zc) -x)`

`\rho_z = 7.133g/cm^3`

Thus,

\frac{m_{zc} -x}{\rho_z}+\frac{x}{\rho_c}=v_c

Since volume of new and old coin is the same:

`((2.517-x)/(7.133))+((x)/(8.96))=((3.083)/(8.96))`

`2.517-x+0.796=2.454`

`x = 0.3088g`

Thus, mass of zinc is

`m_(zc) -x = 2.517-0.3088`

`m_(zc) -x = 2.2082g`

Thus, percentage of zinc by volume

`\% = (m_(zc)-x)/(\rho_(zc))(1)/(V)*100`

`\% = ((2.2082g)/(7.133))((1)/(3.083/8.96))(100)`

`\% = 89.97\%`

Therefore the percent of zinc in the new cent 89.97%