Initially 15 grams of salt are dissolved into 25 liters of water. Brine with concentration of salt 4 grams per liter is added at a rate of 6 liters per minute. The tank is well mixed and drained at 6 liters per minute. a) Let x be the amount of salt, in grams, in the solution after t minutes have elapsed. Find a formula for the rate of change in the amount of salt, dx/dt, in terms of the amount of salt in the solution x. dxdt= grams/minute b) Find a formula for the amount of salt, in grams, after t minutes have elapsed. x(t)= grams c) How long must the process continue until there are exactly 20 grams of salt in the tank?

a) dx/dt = 600 - 6x

b) x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) The mass of salt in the tank attains the value of 20 g at time, t = 0.227 min = 13.62 s

Explanation:

Taking the overall balance, since the total Volume of the setup is constant, then flowrate in = flowrate out

Let the concentration of salt in the tank at anytime be C

Let the concentration of salt entering the tank be Cᵢ

Let the concentration of salt leaving the tank be C₀ = C (Since it's a well stirred tank)

Let the flowrate in be represented by Fᵢ

Let the flowrate out = F₀ = F

Fᵢ = F₀ = F = 6 L/min

a) Then the component balance for the salt

Rate of accumulation = rate of flow into the tank - rate of flow out of the tank

dx/dt = Fᵢxᵢ - Fx

Fᵢ = 6 L/min, C = 4 g/L, F = 6 L/min

dC/dt = 24 - 6C

dx/dt = 25 (dC/dt), (dC/dt) = (1/25) (dx/dt) and C = x/25

(1/25)(dx/dt) = 24 - (6/25)x

dx/dt = 600 - 6x

b) dC/dt = 24 - 6C

dC/(24 - 6C) = dt

∫ dC/(24 - 6C) = ∫ dt

(-1/6) In (24 - C) = t + k (k = constant of integration)

In (24 - 6C) = -6t - 6k

-6k = K

In (24 - 6C) = K - 6t

At t = 0, C = 15 g/25 L = 0.6 g/L

In (24 - 6(0.6)) = K

In 20.4 = K

K = 3.02

So, the equation describing concentration of salt at anytime in the tank is

In (24 - 6C) = K - 6t

In (24 - 6C) = 3.02 - 6t

24 - 6C = e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾

6C = 24 - (⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)

C = 4 - ((e⁻⁽⁶ᵗ ⁻ ³•⁰²⁾)/6

C = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

x/25 = 4 - (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)/6)

x = 100 - 4.12((e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

c) when x = 20 g

20 = 100 - 4.12(e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

80 = (e⁻ ⁽⁶ᵗ ⁻ ³•⁰²⁾)

- (6t - 3.02) = In 80

- (6t - 3.02) = 4.382

(6t - 3.02) = -4.382

6t = -4.382 + 3.02

t = - 1.362/6 = 0.227 min = 13.62 s

answer

see attachment

explanation

the given function is

this function has y-intercept at

-4 and a slope of 2.

note that, the graph has a positive slope.

the graph that represents this function is the straight line that moves upward from left to right.

the graph touches the y-axis at -4.

the correct choice is the graph shown in the attachment.