Question

(1 point) A tank contains 2640 L of pure water. A solution that contains 0.09 kg of sugar per liter enters the tank at the rate 3 L/min. The solution is mixed and drains from the tank at the same rate. (a) How much sugar is in the tank at the beginning

Answer 1

Answer:

t=0      Sugar = 0 Kg

t=1min Sugar=0.27 Kg

Step-by-step explanation:

Data

Tank = 2640 L (pure water)

Sol=0.09kg Sugar per liter

Vin = Vout = 3L/min

Sugar in the beginning = ?

if beginning is t = 0min Amount of sugar = 0, this is due to the fact that at the moment of entering the tank the content is only water , but if beginning is t= 1min then;

$\frac{3L}{min}*\frac{0.09Kg}{L}=0.27kg/min$

Answer 2

Answer: y(t)= 237.6(1-e^(-t/880))

Step-by-step explanation:

In the attachment