Question

A vertical container with base area measuring $\mathbf{14}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}\mathbf{by}\mathbf{}\mathbf{17}\mathbf{.}\mathbf{0}\mathbf{}\mathbf{cm}\mathbf{}$is being filled with identical pieces of candy, each with a volume of $\mathbf{50}\mathbf{.}\mathbf{0}\mathbf{}{\mathbf{mm}}^{\mathbf{3}}$and a mass of $\mathbf{0}\mathbf{.}\mathbf{0200}\mathbf{}\mathbf{g}$. Assume that the volume of the empty spaces between the candies is negligible. If the height of the candies in the container increases at the rate of $\mathbf{0}\mathbf{.}\mathbf{250}\mathbf{}\mathbf{cm}\mathbf{/}\mathbf{s}$, at what rate (kilograms per minute) does the mass of the candies in the container increase?

Short answer

The rate of increase in mass of candies is $1.43\mathrm{kg}/\min$.

Step-by-step solution

Given data

The dimensions of the base area are $14.0\mathrm{cm}\times 17.0\mathrm{cm}$

The volume of each candy, $\mathrm{v}=0.50{\mathrm{mm}}^3$.

The mass of each candy, $\mathrm{m}=0.0200\mathrm{g}$.

Understanding the density of a material

The density of a material is given by mass per unit volume. The rate of increase in mass indicates the amount of increase in mass per unit time.

The expression for the density is given as:

$\mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}}$ … (i)

Here, $\mathrm{p}$is the density, m is the mass and V is the volume.

Determination of the density of candy

Using equation (i), the density is calculated as:

$$\begin{gathered} \mathrm{p}=\frac{\mathrm{m}}{\mathrm{v}} \\ =\frac{0.0200\mathrm{g}}{50.0{\mathrm{mm}}^3} \\ =4\times {10}^{‐4}\frac{\mathrm{g}}{{\mathrm{mm}}^3} \end{gathered}$$

Now, convert the density into $\mathrm{kg}/{\mathrm{cm}}^3$

$$\begin{gathered} \mathrm{p}=4\times {10}^{‐4}\frac{\mathrm{g}}{{\mathrm{mm}}^3}\left(\frac{1\mathrm{kg}}{1000\mathrm{g}}\right)‐\left(\frac{1000{\mathrm{mm}}^3}{1{\mathrm{cm}}^3}\right) \\ =4\times {10}^{‐4}\frac{\mathrm{kg}}{{\mathrm{cm}}^3} \end{gathered}$$

Thus, the density of candy is $4\times {10}^{‐4}\mathrm{kg}/{\mathrm{cm}}^3$

Determination of the rate of increase in mass of the candy

As the volume of the empty spaces between the candies is negligible, the mass of the candies in the container with the heighthwill be

$\mathrm{M}=\mathrm{p}\times \mathrm{A}\times \mathrm{h}$ … (ii)

The area A is calculated as:

$$\begin{gathered} \mathrm{A}=14.0\mathrm{cm}\times 17.0\mathrm{cm} \\ =238{\mathrm{cm}}^3 \end{gathered}$$

With this, the rate of change of mass will be given by differentiating equation (ii) with respect to time. This gives,

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=\frac{\mathrm{g}\left(\mathrm{p}\times \mathrm{A}\times \mathrm{h}\right)}{\mathrm{dt}} \\ =\mathrm{p}\times \mathrm{A}\times \frac{\mathrm{dh}}{\mathrm{dt}} \end{gathered}$$

Substitute the values in the above equation.

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=4\times {10}^{‐4}\frac{\mathrm{kg}}{{\mathrm{cm}}^3}\times 238{\mathrm{cm}}^2\times 0.250\frac{\mathrm{cm}}{\mathrm{s}} \\ =0.0238\frac{\mathrm{kg}}{\mathrm{s}} \end{gathered}$$

Convert this rate from kg/s to kg/min.

$$\begin{gathered} \frac{\mathrm{dM}}{\mathrm{dt}}=0.0238\frac{\mathrm{kg}}{\mathrm{s}}‐\left(\frac{1\mathrm{s}}{0.01667\min }\right) \\ =1.43\frac{\mathrm{kg}}{\min } \end{gathered}$$

Thus, the rate of increase in mass is $1.43\mathrm{kg}/\min$