Answer
\[\underline{49 percent}\]
Work Step by Step
Since \[1\text{ kg}=1000\text{ g}\], convert the mass into grams as follows:
\[\left( 4.36\text{ kg} \right)\left( \frac{1000\text{ g}}{1\text{ kg}} \right)=4360\text{ g}\]
Now, assume that the volume of copper is \[x\text{ c}{{\text{m}}^{3}}\]. Thus, the volume of lead is calculated as follows:
\[\begin{align}
& {{V}_{\text{T}}}={{V}_{\text{Pb}}}+{{V}_{\text{Cu}}} \\
& 427\text{ c}{{\text{m}}^{3}}={{V}_{\text{Pb}}}+x\text{ c}{{\text{m}}^{3}} \\
& {{V}_{\text{Pb}}}=\left( 427-x \right)\text{ c}{{\text{m}}^{3}}
\end{align}\]
Now, the mass of copper is calculated as follows:
\[{{m}_{\text{Cu}}}=\left( {{d}_{\text{Cu}}} \right)\left( {{V}_{\text{Cu}}} \right)\]
Here, \[{{d}_{\text{Cu}}}\] is the density of copper and \[{{V}_{\text{Cu}}}\] is the volume of copper.
Substitute \[8.96\text{ g/c}{{\text{m}}^{3}}\] for \[{{d}_{\text{Cu}}}\] and \[x\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\]:
\[\begin{align}
& {{m}_{\text{Cu}}}=\left( 8.96\text{ g/c}{{\text{m}}^{3}} \right)\left( x\text{ c}{{\text{m}}^{3}} \right) \\
& =8.96x\text{ g}
\end{align}\]
Now, the density \[\left( {{d}_{\text{Pb}}} \right)\] of lead is \[\text{11}\text{.4 g/c}{{\text{m}}^{3}}\]. Calculate the mass of the lead as follows:
\[{{m}_{\text{Pb}}}=\left( {{d}_{\text{Pb}}} \right)\left( {{V}_{\text{Pb}}} \right)\]
Substitute \[\text{11}\text{.4 g/c}{{\text{m}}^{\text{3}}}\] for \[{{d}_{\text{Cu}}}\] and \[\left( 427-x \right)\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\]:
\[\begin{align}
& {{m}_{\text{Cu}}}=\left( \text{11}\text{.4 g/c}{{\text{m}}^{3}} \right)\left( 427-x \right)\text{c}{{\text{m}}^{3}} \\
& =11.4\left( 427-x \right)\text{ g}
\end{align}\]
The total mass is written as follows:
\[\begin{align}
& {{m}_{\text{T}}}={{m}_{\text{Pb}}}+{{m}_{\text{Cu}}} \\
& 4360\text{ g}=8.96x\text{ g}+11.4\left( 427-x \right)\text{ g} \\
& x=\frac{4867.8-4360}{2.44} \\
& =208.11
\end{align}\]
Thus, the volume of copper is \[208.11\text{ c}{{\text{m}}^{3}}\]. The percentage of the total volume of the sphere that is copper is as follows:
\[\text{percentage}=\frac{{{V}_{\text{Cu}}}}{{{V}_{\text{T}}}}\times 100\]
Substitute \[208.11\text{ c}{{\text{m}}^{3}}\] for \[{{V}_{\text{Cu}}}\] and \[\text{427 c}{{\text{m}}^{3}}\] for \[{{V}_{\text{T}}}\]:
\[\begin{align}
& \text{percentage}=\frac{208.11\text{ c}{{\text{m}}^{3}}}{427\text{ c}{{\text{m}}^{3}}\text{ }}\times 100 \\
& =49\,\text{percent}
\end{align}\]
The percentage of the total volume of the sphere is \[49\,percent\].
