Answer:
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
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