Answer
$Q(t) = 120\gamma \left(1 - e^{-t/60}\right)$; $Q_{\infty} = 120\gamma$
Work Step by Step
$\mathbf {Step \text{ }1}$ : Define the Variables and the Differential Equation
Let:
$\bullet$ $V = 120$ L (volume of the tank, constant since inflow = outflow)
$\bullet$ $Q(t)$: amount of salt (g) in the tank at time $t$ (min)
$\bullet$ Initial condition: $Q(0) = 0$ (since pure water)
$\bullet$ Inflow: salt mixture at $2$ L/min with concentration $\gamma$ g/L, so rate of salt in = $2\gamma$ g/min
$\bullet$ Outflow: well-stirred mixture at $2$ L/min, so concentration in outflow is $\frac{Q(t)}{V}$ g/L, and rate of salt out = $2 \cdot \frac{Q(t)}{120} = \frac{Q(t)}{60}$ g/min
The rate of change of salt in the tank is:
\[
\frac{dQ}{dt} = \text{rate in} - \text{rate out} = 2\gamma - \frac{Q(t)}{60}
\]
$\mathbf {Step \text{ }2}$ : Solve the Differential Equation
The equation is:
\[
\frac{dQ}{dt} = 2\gamma - \frac{Q}{60}
\]
Rewriting:
\[
\frac{dQ}{dt} + \frac{1}{60} Q = 2\gamma
\]
This is a first-order linear differential equation. The integrating factor is:
\[
\mu(t) = e^{\int \frac{1}{60} dt} = e^{t/60}
\]
Multiply both sides by $\mu(t)$:
\[
e^{t/60} \frac{dQ}{dt} + \frac{1}{60} e^{t/60} Q = 2\gamma e^{t/60}
\]
The left side is the derivative of $Q e^{t/60}$:
\[
\frac{d}{dt} \left( Q e^{t/60} \right) = 2\gamma e^{t/60}
\]
Integrate both sides with respect to $t$:
\[
Q e^{t/60} = 2\gamma \int e^{t/60} dt = 2\gamma \cdot 60 e^{t/60} + C
\]
\[
Q e^{t/60} = 120\gamma e^{t/60} + C
\]
Solve for $Q(t)$:
\[
Q(t) = 120\gamma + C e^{-t/60}
\]
Use the initial condition $Q(0) = 0$:
\[
0 = 120\gamma + C \implies C = -120\gamma
\]
Thus:
\[
Q(t) = 120\gamma - 120\gamma e^{-t/60} = 120\gamma (1 - e^{-t/60})
\]
$\mathbf {Step \text{ }3}$ : Find the Limiting Amount as $t \to \infty$
As $t \to \infty$, $e^{-t/60} \to 0$, so:
\[
Q_{\infty} = \lim_{t \to \infty} Q(t) = 120\gamma (1 - 0) = 120\gamma
\]
$\mathbf {Step \text{ }4}$ : Final Answers
$\bullet$ Expression for the amount of salt at any time $t$:
\[
\boxed{Q(t) = 120\gamma \left(1 - e^{-t/60}\right)}
\]
$\bullet$ Limiting amount as $t \to \infty$:
\[
\boxed{Q_{\infty} = 120\gamma}
\]
