Question

Grains of fine California beach sand are approximately spheres with an average radius of$50\mu m$and are made of silicon dioxide, which has a density of$\mathbf{2600}\mathbf{kg}/{\mathbf{m}}^{\mathbf{3}}$.What mass of sand grains would have a total surface area (the total area of all the individual spheres) equal to the surface area of a cube 1.00 m on an edge?

Short answer

The mass of sand grains is 0.260 kg.

Step-by-step solution

Given data

The density of silicon dioxide,${\text{ρ=2600kg/m}}^3$

The average radius of the sphere, $\text{r=50μm}$

The edge length of cube, 1.00 m

Understanding the density of material

The density of a material is given by mass per unit volume. In order to calculate the mass of sand grains on the total surface of the cube, the total number of sand grains on the total surface of the cube should be known.

The expression for density is given as:

$\rho =\frac{m}{V}$ … (i)

Here, $\rho$ is the density, $m$ is the mass and$V$ is the volume.

Determination of the volume of a sand grain

First, convert radius from into using the conversion factor.

${\text{1μm=1×10}}^{-6}m$

Then,

The volume of the sand grain is calculated as:

$$\begin{gathered} V=\frac{4\pi r^3}{3} \\ =\frac{4\pi {\left(50\times {10}^{-6}\text{m}\right)}^3}{3} \\ =5.24\times {10}^{-13}{\text{m}}^3 \end{gathered}$$

Thus, the volume of the sand grain is$5.24\times {10}^{-13}{\text{m}}^3$

Determination ofthe mass of sand grains

Using equation (i), the mass of sand grains is calculated as:

$$\begin{gathered} m=\rho \times V \\ =2600{\text{kg/m}}^3\times 5.24\times {10}^{-13}{\text{m}}^{\text{3}} \\ =1.36\times {10}^{-9}\text{kg} \end{gathered}$$

Thus, the mass of the sand grain is$1.36\times {10}^{-9}\text{kg}$

Determination of surface area of sand grain

The surface area of sand grain is calculated as:

Thus, the surface area of sand grain is$3.14\times {10}^{-8}{\text{m}}^{\text{2}}$

Step 6: Determination of the total number of sands grains

The surface area of cube of side 1 m is,

$A_c=6{\text{m}}^2$

Take the ratio of the area of a cube to the area of a sand grain to find the total number of sand grains.

$N=\frac{A_c}{A_s}$

Substitute the values in the above expression

$$\begin{gathered} N=\frac{6{\text{m}}^{\text{2}}}{3.14\times {10}^{-8}{\text{m}}^{\text{2}}} \\ =1.91\times {10}^8 \end{gathered}$$

Thus, the total number of sand grains are$1.91\times {10}^8$

Determination of the total mass

The total mass equals to the product of number of sand grains and the mass of each sand grain.

$$\begin{gathered} M=N\times m \\ =1.36\times {10}^{-9}\text{kg}\times 1.91\times {10}^8 \\ =0.260\text{kg} \end{gathered}$$

Thus, sand grains have mass of$0.260\text{kg}$