Find the amount of salt in the tank after 4.5 hours.

Question

A tank contains 70 kg of salt and 2000 L of water. A solution of a concentration of 0.0175 kg of salt per liter enters a tank at the rate of 7 L/min. The solution is mixed and drains from the tank at the same rate.

I know the following:

The initial concentration of the solution is .035 kg/L.

As the time approaches infinity, the concentration of salt in the solution will be 0.0175 kg/L.

Answer 1

This is a first-order linear differential equation. I will solve for S(t), the salt in the tank at time t in minutes.

We can write the differential equation for dS / dt:

dS / dt = (- S(t) / 2000 L ) * (7 L / 1 min) + (7 L / min ) * (0.0175 kg / L)

The first term is the rate of change from the mixed solution leaving the tank, and the second term is the rate of change from a differently mixed solution going into the tank.

This is a separable equation: We divide the left side by the right, and multiply by dt:

dS / (0.1225 - (7/2000) * S(t) ) = dt

Rearrange:

dS / ( S(t) - 35 ) = -0.0035 * dt

Integrate:

ln[ S(t) - 35 ] = -0.0035 * t + C

Exponentiate:

S(t) - 35 = A * e^( -0.0035 * t)

(Here, we have defined A = e^C )

Finish up, keeping in mind that S(0) = 70

S(t) = 35 * e^( -0.0035 * t) + 35

This gives the initial result of 70kg and the asymptotic limit of 0.0175 kg / L.

Enjoy!