Question

A tank initially contains 50 gallons of brine, with 30 pounds of salt in solution. Water runs into the tank at 6 gallons per minute and the well-stirred solution runs out at 5 gallons per minute. How long will it be until there are 25 pounds of salt in the tank?The amount of time until 25 pounds of salt remain in the tank is " ? " minutes.Set up a differential equation and separate variables.

Answer 1

Answer:

1.857 minutes

Step-by-step explanation:

From the given information:

Consider the amount of salt present in the tank as x(t) at a given time (t); &

The volume of the solution = V(t)

At t = 0 i.e. (at initial conditions)  x(0) = 30; and V(0) = 50

However, the overall increase taken place for 1 gallon per minute is:

V(t) = 50 + t

The amount of salt x(t) at any given point for time (t) is;

$\dfrac{x(t)}{V(t)}= \dfrac{x(t)}{50+t}$

After 5 gallons of solution exit per minute; the concentration of the salt solution changes at:

$\dfrac{dx(t)}{dt}= -\dfrac{5x(t)}{50+t}$

Taking the integral of what we have above, we get:

In x(t) = - 5 In(t + 50) + In (C)

In x(t) = In (t+ 50)⁻⁵ + In (C)

In x(t) = In C ( t + 50)⁻⁵

x(t) = C(t + 50)⁻⁵      (General solution)

To estimate the required solution; we apply the initial conditions x(0) = 30;

Thus;

x(0) = C(50)⁻⁵ = 30

⇒ C = 30 × 50⁵

Hence; x(t) =  30 × 50⁵ × (t + 50)⁻⁵

The above expression can be re-written as:

$x(t) = 25 \implies 30 \times \bigg ( \dfrac{50}{t+50} \bigg ) ^5= 25$

i.e.

$\bigg ( \dfrac{50}{t+50} \bigg ) ^5= \dfrac{25}{30}$

$\dfrac{50}{t+50}= \bigg ( \dfrac{25}{30}\bigg) ^{\dfrac{1}{5}}$

$\dfrac{50}{t+50}= 0.964192504$

${50}= 0.964192504(t+50)$

50 = 0.964192504t + 48.2096252

50 - 48.2096252 = 0.964192504t

1.7903748 =  0.964192504t

t = 1.7903748 / 0.964192504

t ≅ 1.857 minutes

We can thereby conclude that the estimated time it will require until there are 25 pounds of salt in the tank is 1.857 minutes.