Question

A tank contains 1,000 L of brine with 12 kg of dissolved salt. Pure water enters the tank at a rate of 10 L/min. The solution is kept thoroughly mixed and drains from the tank at the same rate. (a) How much salt is in the tank after t minutes

Answer 1

To find the amount of salt in the tank after t minutes, use the formula: Amount of salt = Initial concentration x Volume of solution. The initial concentration is 0.012 kg/L and the volume of the solution is 1,000 L.

Step-by-step explanation:

To find the amount of salt in the tank after t minutes, we need to consider the rate at which pure water enters and leaves the tank. Since pure water enters the tank at a rate of 10 L/min and the solution is thoroughly mixed, the volume of the solution remains constant at 1,000 L. The concentration of the solution can be calculated as the mass of salt divided by the volume of the solution.

Initially, the concentration of the solution is 12 kg/1,000 L = 0.012 kg/L.

After t minutes, the amount of salt in the tank can be calculated using the formula:

Amount of salt = Initial concentration x Volume of solution

= 0.012 kg/L x 1,000 L

= 12 kg.

Answer 2

y = 12*e^[-t/100]

Step-by-step explanation:

We have to mix problems we use:

dy / dt = (rate in) - (rate out)

The water entering the tank at a speed of 10 L / min but it has no salt, therefore

rate in = 0

The tank has 1000 liters of brine with 12 kg of salt initially, therefore the concentration of salt at time "t" is:

y (t) / 1000

then rate out would be:

rate out = y (t) / 1000 * 10 = y (t) / 100

The difference equation would then be:

dy / dt = 0 - y (t) / 100

1 / y (t) * dy = 1/100 * dt

We integrate from both sides and we have:

ln y = - (1/100) * t + C1

y = e ^ [- (1/100) * t + C1]

y = e ^ [- (1/100) * t] * e ^ [C1]

We assume that C = e ^ [C1]

Thus:

y = C * e ^ [- (1/100) * t]

now y = 12 to t = 0, replacing:

12 = C * e ^ [- (1/100) * 0]

12 = C, therefore we would have:

y = 12 * e ^ [- t / 100]