Question

A standard gold bar stored at Fort Knox, Kentucky, is 7.00 inches long, 3.63 inches wide, and 1.75 inches tall. Gold has a density of 19,300 kg/m3. What is the mass of such a gold bar?

Answer 1

Concepts and reason

The concept required to solve this problem is density.

Initially, calculate the volume of the gold bar. Then, calculate the mass by using the equation of density.

Fundamentals

The volume of cuboid is,

$V=lwt$

Here, $l$ is length, $w$ is width, and $t$ is height.

The density of a material is,

$\rho =\genfrac{}{}{0ex}{}{m}{V}$

Here, $m$ is mass and $V$ is volume.

Substitute $7.00\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}$ for $l$ , $3.63\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}$ for $w$ , and $1.75\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}$ for $t$ in the equation $V=lwt$ .

$V=\left(7.00\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\right)\left(3.63\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\right)\left(1.75\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\right)$

Convert inches to meter by multiplying with $\left(\genfrac{}{}{0ex}{}{1\mathrm{m}}{39.37\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}}\right)$ .

$\begin{array}{c}V=\left(\begin{array}{l}\left(7.00\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\left(\genfrac{}{}{0ex}{}{1\mathrm{m}}{39.37\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}}\right)\right)\left(3.63\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\left(\genfrac{}{}{0ex}{}{1\mathrm{m}}{39.37\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}}\right)\right)\\ \left(1.75\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}\left(\genfrac{}{}{0ex}{}{1\mathrm{m}}{39.37\mathrm{i}\mathrm{n}\mathrm{c}\mathrm{h}\mathrm{e}\mathrm{s}}\right)\right)\end{array}\right)\\ =7.29\times 10^{-4}{\mathrm{m}}^3\end{array}$

Rearrange the density equation to solve for mass.

$\begin{array}{c}\rho =\genfrac{}{}{0ex}{}{m}{V}\\ m=\rho V\end{array}$

Substitute $19300\mathrm{k}\mathrm{g}/{\mathrm{m}}^3$ for $\rho$ and $7.29\times 10^{-4}{\mathrm{m}}^3$ for $V$ in the equation $m=\rho V$ .

$\begin{array}{c}m=\left(19300\mathrm{k}\mathrm{g}/{\mathrm{m}}^3\right)\left(7.29\times {10}^{-4}{\mathrm{m}}^3\right)\\ =14.07\mathrm{k}\mathrm{g}\end{array}$

Ans:

The mass of the gold bar is $14.07\mathrm{k}\mathrm{g}$ .