Question

A tank contains 250 liters of fluid in which 40 grams of salt is dissolved. Brine containing 1 gram of salt per liter is then pumped into the tank at a rate of 5 L/min; the well-mixed solution is pumped out at the same rate. Find the number A(t) of grams of salt in the tank at time t.

Answer 1

Answer:

$A(t)=250-210e^{-\frac{t}{50}}$

Step-by-step explanation:

We are given that

Volume,V=250 L

Mass of salt=m=40 g

Brine containing 1 g per liter

Pumped into the tank at the rate=5 L/min

We have to find the number A(t) of grams of salt in the tank at time t.

Rate of change of salt in the tank

$\frac{dA}{dt}=$Rate in-Rate out

Rate in=5 L/min

Rate out=$\frac{A}{250}\times 5=\frac{A}{50}$ L/min

$\frac{dA}{dt}=5-\frac{A}{50}=\frac{250-A}{50}$

$\int \frac{50dA}{250-A}=\int dt$

$-50ln(250-A)=t+C$

Using the formula

$\int \frac{dx}{x}=ln x$

A=40 and t=0

$-50ln(250-40)=0+C$

$-50ln(210)=C$

Substitute the value

$-50ln(250-A)=t-50ln(210)$

$-50ln(250-A)+50ln(210)=t$

$50ln\frac{210}{250-A}=t$

$t=50ln\frac{210}{250-A}$

$\frac{t}{50}=ln\frac{210}{250-A}$

$\frac{210}{250-A}=e^{\frac{t}{50}$

$\frac{250-A}{210}=e^{-\frac{t}{50}}$

$250-A=210e^{-\frac{t}{50}}$

$A(t)=250-210e^{-\frac{t}{50}}$