Mathematics: A tank contains 9,000 L of brine with 16 kg of dissolved salt. Pure water enters the tank at a rate of 90 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

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A tank contains 9,000 L of brine with 16 kg of - id9000165871788, jjjones2233

tdahna0403

23.01.2022

$16e^{-\frac{t}{100}}$ kg

Step-by-step explanation:

We are given that

Volume,V=9000 L

$\frac{dV}{dt}=90L/min$

We have to find the salt in the tank after t minutes.

Let y(t) be the amount of salt dissolved in the tank

y(0)=16

$\frac{dy}{dt}=$ rate in-rate out

Rate in=0

Rate out= $\frac{y}{9000}\times \frac{dV}{dt}=\frac{y(t)}{9000}\times 90=\frac{y}{100}$

$\frac{dy}{dt}=0-\frac{y}{100}$

$\int dy=-\int \frac{y}{100}dt$

$\int \frac{dy}{y}=-\int\frac{1}{100}dt$

$ln y=-\frac{t}{100}+C$

$y=e^{-\frac{t}{100}+C}=e^{-\frac{t}{100}}\cdot e^C=Ke^{-\frac{t}{100}}$

Where $k=e^{C}$

Substitute t=0 and y=16

16=C

$y(t)=16e^{-\frac{t}{100}}$ kg

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