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
