The tidy toilet bowl has been used with a blue disinfectant in the tank. The sou

Question

The tidy toilet bowl has been used with a blue disinfectant in the tank. The source of the blue antibacterial mixture is removed from the tank, but the five gallons inside the tank still have the blue dye dissolved in the concentration of c0 = 4.18 ppm. It is not known how much water one flush removes from the tank but after each flush the blue water is replaced by clear tap water, i.e. water with no dye in it. Measurement of the concentration of dye in the tank after one flush gave c1=3.344 ppm. Assuming perfect mixing between flushes, what is the concentration after the second flush? c2= What is the concentration of the dye after the tenth flush? c10= What is the volume of water removed from the tank by one flush? What is the concentration of the dye in the bowl after very many flushes?

Explanation / Answer

The amount of mixture in the tank is 5 galloons. Initial concentration of blue dye is 4.18 ppm. Let wiith every flush, x galloons of mixture is replaced by clear tap water. (x<=5)

c0 = 4.18 ppm

Amount of blue dye inside the tank initially = 4.18 * 10^(-6) * 5 galloons

After the first flush, amount of blue dye inside the tank will be = 4.18 * 10^(-6) * (5-x) galloons

c1 = 3.344 ppm

According to this data amount of blue dye inside the tank after first flush= 3.344 * 10^(-6)*5 galloons. So,

3.344 * 10^(-6)*5 = 4.18 * 10^(-6) * (5-x)

(5-x) = 3.344*5/4.18 = 4

x = 1 galloon.

The volume of water removed from the tank by one flush is 1 galloon.

So, c1 * 5 = c0* (5-x) => c1 = c0 * (5-1)/5 = c0*4/5

c2 = c1*4/5 = c0* ((4/5)^2) = 4.18 * 0.64= 2.6752

The concentration of dye after 2nd flush will be 2.6752.

The concentration of dye after nth flush will be

cn = c0 * ((4/5)^n)

So concentration of dye after 10th flush is c10 = c0 * ((4/5)^10) = 4.18 *0.107374 = 0.44882332

The concentration of dye after very many flushes is limit(n->infinity) cn = limit(n->infinity) 4.18 *((4/5)^n)

= 0 ppm

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