Question

Suppose there are two lakes located on a stream. Clean water flows into the first lake, then the water from the first lake flows into the second lake, and then water from the second lake flows further downstream. The in and out flow from each lake is 500 liters per hour. The first lake contains 100 thousand liters of water and the second lake contains 200 thousand liters of water. A truck with 500 kg of toxic substances crashes into the first lake. Assume that the water is being continually mixed perfectly by the stream.
a. Find the concentration of toxic substances as a function of time in both lakes.
b. When will the concentration in the first lake be below 0.001 kg per liter?
c. When will the concentration in the second lake be maximal?

Answer 1

Answer:

a.  For the first lake;

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m·t/200 - m·t²/80000

b. t ≈ 1.00505 hours

c. 200 hours

Step-by-step explanation:

The flow rate of water in and out of the lakes = 500 liters/hour

The volume of water in the first lake = 100 thousand liters

The volume of water in the second lake = 200 thousand liters

The mass of toxic substances that entered into the first lake =  500 kg

The concentration of toxic substance in the first lake = m₁/(100000)

Therefore, we have;

The quantity of fresh water supplied at t hours = 500 × t

The change

The change in the mass of the toxic substance with time is given as follows

dm/dt = (m - m/100000 × 500 × t)/100000

c = (m·t - 0.005·m·t²)/100000

For the second tank, we have;

c = m/100000 × 500 × t - (m/100000 × 500 × t)/200000 × 500 × t

Using an online tool, we have;

c = m·t/200 - m·t²/80000

b. When c < 0.001 kg per liter, we have m < 0.001 × 100000, which gives m < 100

Substituting gives;

0.001 = (100·t - 0.005·100·t²)/100000, solving with an online tool, gives;

t ≈ 1.00505 hours

c. For maximum concentration, we have;

c = m·t/200 - m·t²/80000

m/200000 = m·t/200 - m·t²/80000

1/200000 = t/200 - t²/80000

dc/dt = d(t/200 - t²/80000)/dt = 0

Solving with an online tool gives t = 200 hours