Question

Grains of fine California beach sand are approximately spheres with an average radius of 50 μm and are made of silicon dioxide, which has a density of 2.4 × 10^3 kg/m3. 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 1.2 m on an edge?

Answer 1

Answer on Question #51352 – Math – Geometry

Grains of fine California beach sand are approximately spheres with an average radius of $50\,\mu\mathrm{m}$ and are made of silicon dioxide, which has a density of $2.4*10^{3}\,\mathrm{kg/m^{3}}$. 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 cube $1.2\,\mathrm{m}$ on an edge.

Solution

$$r = 50\,\mu\mathrm{m} = 5*10^{-5}\,\mathrm{m}; \quad \rho = 2.4*10^{3}\,\mathrm{kg/m^{3}}; \quad a = 1.2\,\mathrm{m}; \quad m - ?$$

One grain has the surface area: $S_g = 4\pi r^2$.

The surface area of cube: $S_c = 6a^2$.

Number of grains, that have total surface area equal to the surface area of cube, is given by

$$N = S_c / S_g = \frac{3a^2}{2\pi r^2}.$$

Volume of one grain: $V_0 = 4\pi r^3 / 3$.

$$m = \rho * V_0 * N = \rho \frac{4\pi r^3 * 3a^2}{3 * 2\pi r^2} = 2\rho r a^2 = 2*2.4*10^{3}*5*10^{-5}*1.44 = 0.3456\,\mathrm{kg}$$

Answer: $m = 0.3456\,\mathrm{kg}$

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