Question

A tank contains 100 gal of brine made by dissolving 80 Ib of salt in water. Pure water runs into the tank at the rate of 4 gal/min, and the well-stirred mixture runs out at the same rate. Find (a) the amount of salt in the tank at any time t and (b) the time required for half the salt to leave the tank.

Answer 1

Answer:

Step-by-step explanation:

Amount of brine = 100 gal

amount of salt = 80 lb

rate of water running = 4 gal / min

Let A be the amount of salt in lb in the tank at any time t.

(a) dA/dt = rate of salt in - rate of salt out

Rate of salt in = 0

rate of salt out = A x 4 / 100 = A / 25 lb/min

So, dA/dt = 0 - A/25 = - A/25 lb/min

$\frac{dA}{A}=-\frac{1}{25}dt$

$\int \frac{dA}{A}=-\frac{1}{25}\int dt$

$A(t) = C e^{-0.04t}$

When t = 0, A = 80 lb

So,

$A(t) = 80 e^{-0.04t}$

(b) Now, A = 40 lb

So,

$40 = 80 e^{-0.04t}$

$0.5 = e^{-0.04t}$

take log on both the sides

$ln 0.5 = - 0.04 t$

t = 17.3 minutes