A tank contains 3000 L of pure water. Brine that contains 40 g of salt per liter

Question

A tank contains 3000 L of pure water. Brine that contains 40 g of salt per liter of water is pumped into the tank at a rate of 25 L/min. Show that the concentration of salt tminutes later (in grams per liter) is

C(t) =

.

After t minutes 25t liters of brine with 40 g of salt per liter has been pumped into the tank, so it contains........

liters of water and.........

grams of salt. Therefore, the salt concentration at time t will be the following.
C(t) =

=

(b) What happens to the concentration as

t

?

lim t C(t) =  g/L

Explanation / Answer

initial amount of salt y(0)=0

rate of salt input = 25*40 =1000 g/min

rate of salt output =0

net rate =rate of salt input- rate of salt output

y'(t)=1000

=>y(t) =1000t +c

y(0)=0

=>1000*0 +c=0

=>c=0

=>y(t) =1000t

volume of total liquid after time t is v(t)=3000+ 25t

concentration of salt t minutes later ,C(t)=(y(t))/v(t)

C(t)=(1000t)/(3000+ 25t)

C(t)=(40t)/(120+t) grams per litre

After t minutes 25t liters of brine with 40 g of salt per liter has been pumped into the tank, so it contains 3000+ 25t  litres of water and 1000t grams of salt

(b)

limt C(t)

=limt ((40t)/(120+t))

=limt (40/((120/t)+1))

=(40/(0+1))

=40

as t concentration becomes 40 g/L

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