Question

A tank contains 400 liters of fresh water. By mistake, 100 kg of salt are poured into the tank instead of 90 kg. To correct this condition, 10 liters of brine are allowed to flow out per minute while 10 liters of fresh water per minute are pumped into the tank. If the mixture is kept uniform by stirring, find the time required for the tank to contain the desired amount of salt. 

Answer 1

Let s(t) = amount, in kg of salt at time t.

ds/dt= (rate of salt into tank) − (rate of salt out of tank)

$\frac{ds}{dt}=100-\frac{10 s}{400}=\frac{40000-10s}{400}$

$\frac{ds}{40000-10s}=\frac{dt}{400}$

$\intop\frac{ds}{40000-10s}=\int \frac{dt}{400}$

$-\frac{1}{10} ln(40000-10s)=\frac{t}{400}+C$

$ln(40000-10s)=-\frac{t}{40}+C_2$

$40000-10s=C_3 e^{-t/40}$

$s=-\frac{C_3e^{-t/40}-40000}{10}$

S(0)=0

$0=-\frac{C_3e^{-0/40}-40000}{10}$

$C_3=40000$

$s=-\frac{40000e^{-t/40}-40000}{10}=-4000(e^{-t/40}-1)$

If s=90

$90=-4000(e^{-t/40}-1)$

$e^{-t/40}=90/(-4000)+1=0.9775$

$-\frac{t}{40}=ln(0.9775)=-0.0228$

t=0.91 min.