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Zepler [3.9K]
3 years ago
9

fish tank initially contains 35 liters of pure water. Brine of constant, but unknown, concentration of salt is flowing in at 5 l

iters per minute. The solution is mixed well and drained at 5 liters per minute. Let xx be the amount of salt, in grams, in the fish tank after tt minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dtdx/dt, in terms of the amount of salt in the solution
Chemistry
1 answer:
weeeeeb [17]3 years ago
7 0

Answer:

Therefore, the rate of change in the amount of salt is \frac{dx}{dt} =( 5c}{ - \frac{x }{20})

\frac{grams }{min}

Explanation:

Given:

Initial volume of water V = 35 lit

Flowing rate = 5 \frac{Lit}{min}

The rate of change in the amount of salt is given by,

   \frac{dx}{dt} = ( Rate of salt enters tank - rate of sat leaves tank )

Since tank is initially filled with water so we write that,

x(0) = 0

Let amount of salt in the solution is c,

  \frac{dx}{dt} = \frac{5c}{1 } - \frac{x(t) \times 5}{100}

  \frac{dx}{dt} =( 5c}{ - \frac{x }{20}) \frac{grams}{min}

Therefore, the rate of change in the amount of salt is \frac{dx}{dt} =( 5c}{ - \frac{x }{20})

\frac{grams }{min}

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