Please help me finish these two questions. (1 point) A tank contains 70 kg of sa

Question

Please help me finish these two questions.

(1 point) A tank contains 70 kg of salt and 1000 L of water. A solution of a concentration 0.035 kg of salf per ier cnicrs a tank a ihe raik 1 /min. The soluüion is micd and drains irom th tank ai the same rate. (a) What is the concentration of our solution in the tank initially? concentration = .07 | (kg/L) - (b) Find the amount of salt in the tank after 3.5 hours. amount = "(kg) (c) Find the concentration of salt in the solution in the tank as time approaches infinity. concentration = .035 (kg/L)

Explanation / Answer

I am allowed to answer only 1 question at a time. So I will answer 1st question. Please aks other as different question

Let S Kg be the amount of salt in the tank at any time t.
Concentration of salt at any time = S kg /1000 L = S/1000 Kg/L

Since mixture leaves at the rate of 10 L/min, salt is leaving at the rate of,
(S/1000) Kg/L *(10 L/min) = S/100 Kg/min
so,
dS/dt = - s/100
or,
dS/S = -dt/100
inetgrating both sides we get,
ln|S| = -t/100 + C
S = C*e^(-t/100)

we have that amount of salt initially was 70 Kg
so, at t= 0 , S = 70
put this in above equation to get C
S = C*e^(-t/100)
70 = C*e^(-0/100)
C= 70

So , equation for amount of salt becomes,
S = 70*e^(-t/100)

This is amount of salt
Concentration of salt will be S/1000

a)
at t= 0
S = 70
concentration = 70/1000 = 0.07 Kg/L

b)
at t = 3.5 hr = 3.5*60 min = 210 min
S = 70*e^(-t/100)
= 70*e^(-210/100)
= 8.57 Kg
concetration = 8.57/1000 = 8.57*10^-3 Kg/L
Answer: 8.57 Kg

c)
at t = infinity
S = 70*e^(-t/100)
= 70*e^(-infinity/100)
= 0
Answer: 0 Kg/L

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