# In a tank of l00 liters of brine containing dissolved salt.

In a tank of l00 liters of brine containing dissolved salt. Pure water is alloved to run into the tank at the rate of 3 li/min . Brine runs out of the tank at rate of 2 liters minute. The instantaneous concentration in the tank is kept uniform by stirting. How much salt is in the tank at the end of 1 Hr?
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Orlando Paz
Let $Q=$ amount of salt in the mixture
$100+\left(3-2\right)t=100+t=$ volume of mixture at any time t
$\frac{q}{100+t}=$ concentration of salt
$\frac{dq}{dt}=$ rate of gain-rate of loss
$\frac{dq}{dt}=0-2\left(\frac{q}{100+t}\right)$
$\int \frac{dq}{dt}=-2\int \frac{dt}{100+t}+c$
$\mathrm{ln}q=-2\mathrm{ln}\left(100+t\right)+c$
$\mathrm{ln}q=-\frac{2}{2}{\mathrm{ln}\left(100+t\right)}^{2}+c$
When $t=0:q=50kg$
$\mathrm{ln}50=-{\mathrm{ln}\left(100+1\right)}^{2}+c$
$c=13.12236$
When $t=60min,Q=?$
$\mathrm{ln}q=-{\mathrm{ln}\left(100+60\right)}^{2}+13.12236$
$\mathrm{ln}q=2.972$
$q=19.53kg$
###### Not exactly what you’re looking for?
veiga34

Suppose that the amount of salt in the mixture be Q.
The volume of mixture at time t will be as follows:
$100+\left(3-2\right)t$
Recall the fact that the concentration of salt will be given as follows:
$\text{Concentration of salt}=\frac{\text{Amount of salt in mixture}}{\text{Volume of mixture at any time t}}$
$=\frac{Q}{100+\left(3-2\right)t}$
$=\frac{Q}{100+t}$
Also, the rate of gain minus rate of loss will be $\frac{dQ}{dt}$, where Q is the amount of salt in the mixture.
Here rate of gain is 0 and rate of loss is $2\left(\frac{Q}{100+t}\right)$. That is,
$\frac{dQ}{dt}=0-2\left(\frac{Q}{100+t}\right)$
$\frac{dQ}{Q}=-2\left(\frac{dt}{100+t}\right)$
$\int \frac{dQ}{Q}=-2\int \left(\frac{dt}{100+t}\right)$
$\mathrm{ln}Q=-2\mathrm{ln}\left(100+t\right)+C$
$\mathrm{ln}Q=-2{\mathrm{ln}\left(100+t\right)}^{2}+C$
Initially the amount of salt in the mixture is 55 kg. That is, $Q=55,t=0$.
Substitute :
$\mathrm{ln}55=-{\mathrm{ln}\left(100+0\right)}^{2}+C$
$C=13.218$
By the end one 1 hours, t will be 60 minutes.
Substitute :
$\mathrm{ln}Q=-{\mathrm{ln}\left(100+60\right)}^{2}+13.218$
$\mathrm{ln}Q=3.06765$
${e}^{\mathrm{ln}Q}={e}^{3.06765}$
$Q=21.4913$
Thus, the amount of salt in tank after one hour is 21.4913 Kg.

###### Not exactly what you’re looking for?
Vasquez

$100+\left(3-2\right)t=100+t=$ volume of mixture at any time t