Question

A tank contains 5,000 L of brine with 11 kg of dissolved salt. Pure water enters the tank at a rate of 50 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? y = ___________kg (b) How much salt is in the tank after 20 minutes? () y = ________ kg

Answer 1

Answer:

a

  $y(t) = 11e^{-\frac{t}{100} }$

b

  $y(20) = 9 \ kg$

Step-by-step explanation:

From the question we are told that  

    The volume of brine in the tank is  $V_L = 5000 \ L$

     The mass of the salt is $y(0)= 11 \ kg$

     The rate at which water enters that tank is $\r V = 50 \ L / min$

Generally the rate at which salt enters the tank is mathematically represented as

           $z = \frac{0}{V_L} * \r V$

=>        $z = \frac{0}{5000} * 50$

Here the zero implies that there is no salt entering the tank

Generally the rate at which salt leaves  the tank is mathematically represented as

           $u = \frac{y(t)}{V_L} * \r V$

=>        $u = \frac{y(t)}{5000} * 50$        

=>        $u = \frac{y(t)}{100}$

Generally the rate of salt flow in and out of the tank is              

       $\frac{dy}{dt} = z - u$

=>    $\frac{dy}{dt} = - \frac{y(t)}{100}$

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

=>    $ln(y(t)) = -\frac{t}{100} + c$

at time  t = 0  y(0) =  11

=>  $ln(11) = -\frac{0}{100} + c$

=>   $c = ln (11)$

So

    $ln(y(t)) = -\frac{t}{100} + ln (11)$

taking exponent of both sides

    $y(t)=e^{ -\frac{t}{100} + ln (11)}$

=>  $y(t) = e^{-\frac{t}{100} } * e^{ln(11)}$

=>  $y(t) = 11e^{-\frac{t}{100} }$

Generally at t = 20 minutes

 The amount of salt in the tank is mathematically represented as

         $y(20) = 11e^{-\frac{20 }{100} }$

=>      $y(20) = 9 \ kg$