A tank contains 200 liters of a liquid in which 30 g of salt have been dissolved

Question

A tank contains 200 liters of a liquid in which 30 g of salt have been dissolved. Another salt mixture that has 1 gallon of salt per liter enters the tank at a rate of 4L/min. The well-mixed solution leaves the tank with the same reason. Find the amount A(t) of grams of salt in the tank at time t.

68. Un tanque contiene 200 litros de un líquido en el que se han disuelto 30 g de sal. Otra mezcla de sal que tiene 1 galón de sal por litro entra al tanque a una razón de 4L/min. La solución bien mezclada sale del tanque con la misma razón. Encuentra la cantidad A() de gramos de sal que hay en el tanque al tiempo t. a) A (r) = 200-30 b) A(t)-200-30e-m70 OA(r) = 200-1702"50 d) A (r) = 170-30c"170 e) Ninguna de las anteriores

Explanation / Answer

ANSWER:

Since the rate of pumping in and out is same , 200L of fluid in tank is always goning to be there.
At time t, the amount of salt in tank is A(t) (in grams)
So concentrations of salt at time t is A/200 g/L

Amount of brine pumped in: 4L (with salt concentration = 1 g/L)
Amount of salt pumped in: 4L * 1g/L = 4g

Amount of brine pumped out at time t: 4L (with salt concentration = A/200 g/L)
Amount of salt pumped out at time t: 4L * A/200 g/L = A/50 g

dA/dt = 4 - A/50
50 dA/dt = 200 - A
50/(200 - A) dA = dt
-50 ln|A-200| = t + C
ln|A-200| = -t/50 + C .... where C = -C/50
A - 200 = C e^(-t/50) ..... where C = e^C
A = 200 + C e^(-t/50)

Initially, brine contains 30 g of salt
A(0) = 30
200 + C e^0 = 30
C = -170

A(t) = 200 - 170 e^(-t/50)

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