Question

A tank contains initially 2500 liters of 50% solution. Water enters the tank at the rate of 25 iters per minute and the solution flows out at the rate of 50 liters per minute. Find the percentage of salt after 20 minutes

Answer 1

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

$\dfrac{dA}{dt}=rate\ in\ solution -rate\ out\ solution$

$\dfrac{dA}{dt}=concentration\times rate\ in\ of\ water -concentration\times rate\ out\ of\ water$

$\dfrac{dA}{dt}=0\times 25-\dfrac{A}{2500}\times 50$

$\dfrac{dA}{dt}=-\dfrac{A}{50}$

Now by integration

$\ln A=-\dfrac{1}{50}t+C$

Where C is the constant

Given that at t=0 ,A=1250

So $C=\ln 1250$

$\ln A=-\dfrac{1}{50}t+\ln 1250$

When t=20 min

$\ln A=-\dfrac{1}{50}\times 20+\ln 1250$

A=837.90

So percentage of solution after 20 min

$=\dfrac{1250-837.9}{1250}\times 100$

Percentage of solution=32. 96%