Answer:
a. 36.9876kg of salt
b. As time reaches infinity, the concentration of salt will equal the amount of salt going in (0.0125kg/L)
Step-by-step explanation:
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a) There is a mass of 25 kilograms of salt in the tank after 3.5 hours.
b) The concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute.
Salt concentration (c(t)), in kilograms per liter, is usually modelled by first order non-homogeneous linear differential equation, whose form is derived from principle of mass conservation and described below:
[tex]\frac{dc(t)}{dt} + \frac{\dot V}{V} \cdot c(t) = \frac{\dot V\cdot c_{o}}{V}[/tex] (1)
Where:
The solution of this differential equation is:
[tex]c(t) = \left(c_{i}-c_{o}\right)\cdot e^{-\frac{\dot V}{V}\cdot t }+c_{o}[/tex] (2)
Where:
[tex]c_{i} = \frac{m_{s}}{V}[/tex] (3)
Where [tex]m_{s}[/tex] is the initial salt mass in the tank, in kilograms and t is the time, in minutes.
a) The amount of salt is found by multiplying the volume occupied by the water and the salt concentration in the tank. If we know that [tex]m_{s} = 50\,kg[/tex], [tex]V = 2000\,L[/tex], [tex]\dot V = 7\,\frac{L}{min}[/tex], [tex]c_{o} = 0.0125\,\frac{kg}{L}[/tex] and [tex]t = 3.5\,h[/tex], then the amount of salt in the tank:
[tex]c_{i} = \frac{50\,kg}{2000\,L}[/tex]
[tex]c_{i} = 0.025\,\frac{kg}{L}[/tex]
[tex]c(12600) = (0.025-0.0125)\cdot e^{-\frac{7}{2000}\cdot (12600)}+0.0125[/tex]
[tex]c(12600) = 0.0125[/tex]
And the salt mass in the tank is:
[tex]m = \left(0.0125\,\frac{kg}{L} \right)\cdot \left(2000\,L\right)[/tex]
[tex]m = 25\,kg[/tex]
There is a mass of 25 kilograms of salt in the tank after 3.5 hours. [tex]\blacksquare[/tex]
b) For [tex]t \to +\infty[/tex], [tex]c(t) \to c_{o}[/tex]. Therefore, the concentration of salt in the solution in the tank as time approaches infinity is 0.0125 kilograms per minute. [tex]\blacksquare[/tex]
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