Answer:
Percentage of solution=32. 96%
Explanation:
Given that initially tank have 50% solution.It means that amount of solution=0.5 x 2500 =1250 lts
Lets take the amount of solution at time t = A
So
[tex]\dfrac{dA}{dt}=rate\ in\ solution -rate\ out\ solution[/tex]
[tex]\dfrac{dA}{dt}=concentration\times rate\ in\ of\ water -concentration\times rate\ out\ of\ water[/tex]
[tex]\dfrac{dA}{dt}=0\times 25-\dfrac{A}{2500}\times 50[/tex]
[tex]\dfrac{dA}{dt}=-\dfrac{A}{50}[/tex]
Now by integration
[tex]\ln A=-\dfrac{1}{50}t+C[/tex]
Where C is the constant
Given that at t=0 ,A=1250
So [tex]C=\ln 1250[/tex]
[tex]\ln A=-\dfrac{1}{50}t+\ln 1250[/tex]
When t=20 min
[tex]\ln A=-\dfrac{1}{50}\times 20+\ln 1250[/tex]
A=837.90
So percentage of solution after 20 min
[tex]=\dfrac{1250-837.9}{1250}\times 100[/tex]
Percentage of solution=32. 96%
According to the question,
Amount of solution,
= [tex]0.5\times 2500[/tex]
= [tex]1250[/tex]
Let,
Now,
→ [tex]\frac{dA}{dt} = Rate \ in \ solution - Rate \ out \ solution[/tex]
or,
→[tex]\frac{dA}{dt} = Concentration\times Rate \ in \ of \ water- Concentration\times Rate \ out \ of \ water[/tex]
By substituting the values,
[tex]= 0\times 25-\frac{A}{2500}\times 50[/tex]
[tex]= - \frac{A}{50}[/tex]
By applying integration, we get
→ [tex]ln \ A = - \frac{1}{50} t+C[/tex]
We know, t = 0
A = 1250
By substituting the values, we get
→ [tex]ln \ A = - \frac{1}{50}\times 20+ln \ 1250[/tex]
[tex]A = 837.90[/tex]
hence,
The percentage will be:
= [tex]\frac{1250-837.9}{1250}\times 100[/tex]
= [tex]32.96[/tex] %
Thus the answer above is right.
Learn more about percentage here:
https://brainly.com/question/14959031
