A tank contains 2 m^3 of water and 20 g of salt. Water containing a salt concentration of 2 g of salt per m^3 of water flows into the tank at a. rate of 2 m^3/min, and the mixture in the tank flows out at the same rate. We call Q(t) the quantity of salt at time t in the tank. In order to have Q(t) lessthanorequalto g, we have to wait for:

A. 6 min

B. 2 ln(6) min

C. 4 In (2) min

D. 2 In(8) min

E. In (8) min

Option E is correct.

t = In 8

Step-by-step explanation:

First of, we take the overall balance for the system,

Let V = volume of solution in the tank at any time = 2 m³ (constant)

Rate of flow into the tank = Fᵢ = 2 m³/min

Rate of flow out of the tank = F = 2 m³/min

Component balance for the concentration.

Let the initial amount of salt in the tank be Q₀ = 20g

The rate of flow of salt coming into the tank be 2 g/m³ × 2 m³/min = 4 g/min

Amount of salt in the tank, at any time = Q

Rate of flow of salt out of the tank = (Q × 2 m³/min)/V = (2Q/V) g/min

But V = 2 m³

Rate of flow of salt out of the tank = Q g/min

The balance,

Rate of Change of the amount of salt in the tank = (rate of flow of salt into the tank) - (rate of flow of salt out of the tank)

(dQ/dt) = 4 - Q

dQ/(Q - 4) = - dt

∫ dQ/(Q - 4) = ∫ - dt

Integrating the left hand side from Q₀ to Q and the right hand side from 0 to t

In [(Q - 4)/(Q₀ - 4)] = - t

In (Q - 4) - In (Q₀ - 4) = - t

In (Q - 4) = In (Q₀ - 4) - t

Q₀ = 20

In (Q - 4) = (In (16)) - t

In (Q - 4) = 2.773 - t

(Q - 4) = e⁽²•⁷⁷³ ⁻ ᵗ⁾

Q(t) = 4 + e⁽²•⁷⁷³ ⁻ ᵗ⁾

For Q to go less than or equal to 6g, we calculate the time it takes to get to 6 g of salt in the tank

In (Q - 4) = (In (16)) - t

t = In 16 - In (Q - 4)

t = In 16 - In (6 - 4)

t = In 16 - In (2)

t = In (16/2)

t = In 8

step-by-step explanation:

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