Section 2.3 Modeling with First Order: Problem 8 Previous Problem ListNext (1 po

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Section 2.3 Modeling with First Order: Problem 8 Previous Problem ListNext (1 point) A tank contains 100 kg of salt and 1000 L of water. Pure water enters a tank at the rate 12 L min. The solution is mixed and drains from the tank at the rate 6 L/min. (a) What is the amount of salt in the tank initially? amount = E(kg) (b) Find the amount of salt in the tank after 3.5 hours. (kg) amount = (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)

Explanation / Answer

Let s(t) be the amount of salt in (kgs) after t minutes and w(t) is the amount of water at t time.

We have given that

s(0)=100kg , pure water enters a tank at the rate 12L/min and The solution is mixed and drains from the tank at the rate 6 L/min

ds/dt=(rate in) -(rate out) since ds/dt is the rate of change of the amount of salte

rate in =(100kg/1000L)*12L/min=1.2kg/min where (rate in) is the rate at which salt enters the tank

The tank always contains 1000L of Liquid,so the concentration of time t is s(t)/1000 (measured in kg per liter)

rate out =[(s(t)kg)/1000L]*5L/min where (rate out) is the rate at which salt leaves the tank

rate out =(s(t)/200)(kg/min)

ds/dt=(rate in) -(rate out)=1.2kg/min -(s(t)/200) (kg/min)

ds/dt =1.2-s(t)/200 =(240-s(t))/200

separating above equation

ds/(240-s(t)) =dt/200

integrating the above equation

ln(240-s(t)) =t/200 +C --------1

When t=0

ln(240-s(0))=0+C

ln(140)=C since we have given s(0) =100kg

C=ln(140) is substituting into equation 1

ln(240-s(t)) =t/200 +ln(140)

ln(240-s(t))-ln(140)=t/200

ln((240-s(t))/140)=t/200 since by using logarithmic formula

(240-s(t))/140=e^(t/200) which implies 240-s(t)=140*e^(t/200)

s(t)=240-140*e^(t/200)

a) When t=0 s(0)=240-140*e^(0/200)=240-140=100kg

the amount of salt in the tank initially is s(0)=100kg

b) When t=3.5 hours

we are converting hours into minutes 1 hour = 60 minutes so 3.5 *60 =210 minutes

s(210)=[240-140*e^(210/200)] =-160.071kg

the amount of salt in the tank after 3.5 hours is -160.071 kg

c) Let c(t) be the concentration and w(t) be the amount of the water at time t.

We have given that w'=12L-6L=6L

w'=6

dw=6dt

integrating the above equation

w(t)=6t+C

When t=0 and w(0)=1000

1000=0+C then C=1000

w(t)=6t+1000

c(t)=s(t)/w(t)

c(t)=[240-140*e^(t/200)]/(6t+1000)

When t goes to infinity The denominators goes to infinity

Therefore The concentration goes to zero

lim( t--> infinity) c(t)=lim( t--> infinity) [240-140*e^(t/200)]/(6t+1000)=[240-140*e^(infinity/200)]/(6 * infinity+1000)=0

c(t)=0 when t=infinity

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