Question

A vertical container with base area measuring 14 cm by 16 cm is being filled with identical pieces of candy, each with a volume of 50.0 mm^3 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 0.21 cm/s, at what rate does the mass of the candies in the container increase?

Answer 1

Rate of change of mass is given by,

`(dm)/(dt)=18.816\,g/s`

Step-by-step explanation:

In the question,

We have the Base Area of the vertical container = 14 cm x 16 cm

Now,

Let us take the height of the container = h

Rate of change of height with time, dh/dt = 0.21 cm/s = 2.1 mm/s

So,

Volume of the container = Base Area x Height

So,

V = 14 x 16 x h

V = 140 x 160 x h (because, 1 cm = 10 mm)

V = (22400)h

Now,

Volume of one of the candy = 50 mm³

Mass of the candy = 0.0200 g

So,

Density of the candy = Mass/ Volume

So,

`Density=(0.0200)/(50)\\Density=0.0004`

Now,

V = (22400)h

On differentiating with respect to time, t, we get,

`(dV)/(dt)=(22400h)/(dt)\\(d)/(dt)((mass)/(density))=22400.(dh)/(dt)\\(1)/(density).(dm)/(dt)=22400.(dh)/(dt)\\`

Therefore, on putting the value of density in the equation and also the value of rate of change of height with time, we get,

`(1)/(density).(dm)/(dt)=22400.(dh)/(dt)\\(1)/(0.0004).(dm)/(dt)=22400(2.1)\\(dm)/(dt)=18.816\,g/s`

Therefore, the rate of change of mass with respect to time is given by,

`(dm)/(dt)=18.816\,g/s`