(1 point) A tank contains 100 kg of salt and 2000 L of water. Pure water enters

Question

(1 point) A tank contains 100 kg of salt and 2000 L of water. Pure water enters a tank at the rate 10 L/min. The solution is mixed and drains from the tank at the rate 5 L/min.

(a) What is the amount of salt in the tank initially?
amount =  (kg)

(b) Find the amount of salt in the tank after 4.5 hours.
amount =   (kg)

(c) Find the concentration of salt in the solution in the tank as time approaches infinity. (Assume your tank is large enough to hold all the solution.)
concentration =   (kg/L)

Note: please show clearly the final answer

Explanation / Answer

let s(t) be the amount of salt at time t and w(t) the amount of water at time t
then we have
s(0)=100 ,w(0)=2000 and
s' = -5s/w
w' = 5
thus w(t) = 5t + k since w(0) = 2000 then k=2000 thus we have
w(t) = 5t + 2000

thus we have
s' = -5s / (5t + 2000) = -s / (t + 400)
ds/dt = -s / (t+ 400)
-1/s ds = 1/(t+400) dt integrating both sides we get
-ln(s) = ln(t+400) + k
s(0) = 100 thus
-ln(100) = ln(400) + k
k = - ln(40000)
thus
-ln(s) = ln(t+400) - ln(40000)
ln(s) = ln(40000) - ln(t+400)
ln(s) = ln( 16000 / (t+200) )
s = 40000/(t+400)

thus we have
s(t) = 40000/(t+400)
w(t) = 5t+2000

thus the answers are
1) s(0) = 100 kg
2) s(4*60+5) = s(245) = 40000/645 = ~ 62.01550 ( given time 4.5 hours means 4hours 5minuts )
3) concentration is given as
c(t) = s(t)/w(t) = 40000/[ (t+400) * (5t+2000) ]
and it is easy to see that as t goes to infinity the denominator goes to infinity and thus the concentration goes to zero

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