Math CentralQuandaries & Queries


Question from Nicholas, a student:

A barrel initially has two kg of salt dissolved in twenty liters of water. If water flows in the rate of 0.4 liters per minute and the well-mixed salt water solution flow out at the same rate, how much salt is present after 8 minutes?
I tried working backwards given the answer but I can't seen to get their answer of ~1.7kg. Any help would be great! Thanks

Hi Nicholas,

The water is flowing in and out of the barrel at a constant rate but the amount of salt leaving the barrel is not constant, the rate depends on the amount of salt remaining in the solution. If S = S(t) is number of kilograms in the barrel at time t minutes then I think a reasonable assumption is that the rate of change of S is proportional to S. That is there is a positive constant k so that

dS/dt = -k S

This means that

S(t) = A e-kt

for some constant A.

You know S(0) and this will allow you to evaluate A. Also the problem has enough information to evaluate dS/dt at time t = 0 minutes. This will allow you to find k.

Let me know if you have difficulties completing the problem.


Nicholas wrote back

After going through the explanation given by Harley, I am able to get the answer but I don't really actually follow what I'm doing.
For example, am I doing the following right?
I was thinking that dS/dt = -.4L/min and that S was 20L, my originally amount of water... solving would give me k, but from that, the units for my k is just 1/min. Am I doing something wrong?


The water is flowing at -0.4 L/min not the salt. The amount of salt S(t) is in kilograms. What I know is that at time t = 0 minutes there are 2 kilograms of salt in 20 liters of water and the water is flowing at -0.2 L/min. In 20 liters of water there are 2 kilograms of salt and hence 0.2 liters of water will contain 0.02 kilograms of salt. Thus at time t = 0 minutes dS/dt = -0.02 Kg/L.

Does this help?

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