Question

A tank contains 2840 L of pure water. A solution that contains 0.07 kg of sugar per liter enters a tank at the rate 4 L/min The solution is mixed and drains from the tank at the same rate. (a) How much sugar is in the tank initially? A(0)= 0 (kg) (b) State the rate at which the sugar is entering the tank. (kg/min) (c) State the concentration of sugar in the tank at time t, using the letter A to represent the amount of sugar in the tank at time t. (kg/L) (d) State the rate at which the sugar is leaving the tank, Rout, using the letter A to represent the amount of sugar in the tank at time t. (kg/min) (e) State the differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

Answer 1

Answer:

(a) A(0)= 0 (kg)

(b) $R_{in}=0.28\dfrac{kg}{min}$

(c) $C(t)=\dfrac{A(t)}{2840}$

(d) $R_{out}=\dfrac{A(t)}{710}$

(e) $\dfrac{dA}{dt}=0.28-\dfrac{A(t)}{710}$

Step-by-step explanation:

A tank contains 2840L of pure water.

A solution that contains 0.07 kg of sugar per liter enters a tank at the rate 4 L/min. The solution is mixed and drains from the tank at the same rate.

(a) Amount of sugar initially in the tank.

Since the tank initially contains pure water, the amount of sugar in the tank

A(0)= 0 (kg)

(b) Rate at which the sugar is entering the tank. (kg/min)

$R_{in}$=(concentration of sugar in inflow)(input rate of the solution)

$=(0.07\dfrac{kg}{liter}) (4\dfrac{liter}{min})\\R_{in}=0.28\dfrac{kg}{min}$

(c) Concentration of sugar in the tank at time t

Volume of the tank =2840 Liter

Concentration c(t) of the sugar in the tank at time t

Concentration, C(t)= $\dfrac{Amount}{Volume}$

$C(t)=\dfrac{A(t)}{2840}$

(d) Rate at which the sugar is leaving the tank

$R_{out}$=(concentration of sugar in outflow)(output rate of solution)

$=\dfrac{A(t)}{2840})( 4\dfrac{Liter}{min})=\dfrac{A}{710}\\R_{out}=\dfrac{A(t)}{710}$

(e) Differential equation representing the rate at which the amount of sugar in the tank is changing at time t.

$\dfrac{dA}{dt}=R_{in}-R_{out}\\\dfrac{dA}{dt}=0.28-\dfrac{A(t)}{710}$