Question

pure water is poured at the rate of 3 gal/min to a tank containing 300 lbs. of salt dissolved in 100 gals. of water and solution kept while stirring, forced out at 2 gals/min. Find the amount of salt at the end of 1 hr. and 30 minutes?

Answer 1

Tank loses 2 gal/min while adding 3 gal/min for a net of 1 gal/min
\[Volume = 100 + t\]
Salt content in the tank starts at 300 and loses 2C lbs every min
where C(t) is the concentration of salt at time t
C(t) = Salt content/Volume
\[C(t) = \frac{300-2t*C(t)}{100+t}\]
solving for C(t)
\[\rightarrow C(t) = \frac{300}{100+3t}\]
Let S(t) be salt content at time t
S(t) = C(t)*Volume
\[\rightarrow \ s(t) = \frac{300(100+t)}{100+3t}\]
when t=90
\[s(90) = \frac{300(190)}{100+270} = 203.57\]

Answer 2

thanks for the effort dumb cow, but i already got the answer. . it's 83.10 lbs.

Answer 3

Can you please explain to me how you got that answer?
Thanks

Answer 4

this problem is actually related to mixture problem related to differential equation application. . have you encountered this one?

Answer 5

no

Answer 6

here,

let me show you how i got it. .

Answer 7

we got 100 +t as denominator for the because the rate of going in minus the rate in going out (3t - 2t) resulted to t. .