Answer:
Step-by-step explanation:
Assuming x(t) to be the number of salt
The maximum bottom of deposits =300 Litre
Initial Concentration of the saturated salt solution = 1 kg of salt for 3 liters of water
[tex]x_ic_i= \dfrac{1}3[/tex]
⇒x = 0
The differential equation is :
[tex]\dfrac{dx(t)}{dt}[/tex] [tex]={x_ic_i}-{x_oc_o}[/tex]
[tex]= \dfrac{1}{3} - \dfrac{x(t)}{300}[/tex]
[tex]\dfrac{dx(t)}{dt} + \dfrac{x(t)}{300} = \dfrac{1}{3}[/tex]
Using the first order differential equation of the form :
x' + p(t) x = q(t)
where
[tex]p(t) = \dfrac{1}{300}[/tex] ; [tex]q(t) = \dfrac{1}{3}[/tex]
By integrating the factor:
[tex]\mu = e^{\int\limitsP(t)dt } \\ \\ =e^{\int\limits \dfrac{1}{300}dt } \\ \\ = e^{\frac{t}{300} }[/tex]
[tex]x(t) = \dfrac{1}{\mu(t) }\int\limits \mu (t) * q(t) dt \\ \\ = e ^{- t/300} \ \ [\int\limits e ^{+ t/300} * \dfrac{1}{3}dt ][/tex]
[tex]= e ^{- t/300} \ \ [\dfrac{1}{3} * \dfrac{e^{t/300}}{1/300} + C ][/tex]
[tex]x(t) = 100 + Ce ^{-t/300[/tex]
Thus the differential equation solution is : [tex]x(t) = 100 + Ce ^{-t/300[/tex]
However; we have initial concentration x(0) = 0
SO;
[tex]0 = 100 + Ce^{-0/300}[/tex]
C = -100
[tex]x(t) = 100 -100e ^{-t/300[/tex]
when x → ±[infinity]
[tex]\infty = 100 + Ce^{- \infty /300}[/tex]
[tex]x(t) = 100 -100e ^{-t/300[/tex]
Now; the amount of salt after an hour is as follows:
1 hour = 60 minutes
t = 60 min
[tex]x(60) = 100 -100e ^{-60/300[/tex]
= 100 -100 (0.818)
= 100 - 81.8
= 18.2
x(60) = 18.2 kg
