A vertical container with base area measuring 14.0 cm by 17.0 cm is being filled with identical pieces of candy, each with a volume of 50.0 mm3 and a mass of 0.0200 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 0.250 cm/s, at what rate (kilograms per minute) does the mass of the candies in the container increase?
Solution:
First, let us find the mass density of the candy. This is equal to the mass over the volume of the candy. Therefore,
\rho =\frac{m}{V} = \frac{0.0200 g}{50.0 mm^{3}}
\rho = 4.00 \times 10^{-4} \frac{kg}{cm^{3}}
The total mass of the candies in the container when filled to a height of h, is equal to M = \rho Ah where A is the base area of the container which stays constant. In this problem, A=(14.0 cm)(17.0 cm) = 238 cm^{2} .
Now, we will calculate the rate of mass change, which is given by:
\frac{\text{d}M}{\text{d}t}=\frac{\text{d}(\rho Ah)}{\text{d}t}=\rho A\frac{\text{d}h}{\text{d}t}
We know that \frac{\text{d}h}{\text{d}t} is given to us as 0.250\frac{cm}{s}, therefore, we simply plug it into our equation.
=(4.00 \times 10^{-4} \frac{kg}{cm^{3}})(238 cm^{2})(0.250\frac{cm}{s})
=0.0238\frac{kg}{s}=1.43\frac{kg}{min}
And so, the rate at which the mass of the candies increase at is 1.43 \frac{kg}{min}
This is question can be found in Fundamentals of Physics, 10th edition, chapter 1, question 31.
