Question

A tank originally contains 100 gal of fresh water. Then water containing to of salt per gallon is poured into the tank at a rate of 2 gamin, and the mixture is allowed to leave at the same rate. After 10 min the process is stopped, and fresh water is poured into the tank at a rate of 9 gal/min, with the mixture again leaving at the same rate. Find the amount of salt in the tank at the end of an additional 10 min. (Round your answer to two decimal places.) lbs

Answer 1

If u have any doubt, please comment below. TQ

for first 10 min Let act) denotes amount of salt to tank at t m) do zrcate lin-roete ocet at do 3x2-x2 50 SQlo) co (because of fresh wateret begfring ) with 8(0)=0 appley Laplece transforma L{8} + 1/ 1 {a} = L{1} let l{@} - 05} 50 300) + 100) 965) [s+5) - Ś -> 005 50 s(50st) 2500 1505+1 - 28 r 50(5+ states Apply enverse laplace transform Llocs 21 - 25001 Posts = P6+) = 502-65-2 ta PCT) – 50C1) – 501's -t/50 Qat) - 50-50 e W.KT ***}=15 11šta] =eat -10150 When t=10mis paid = 50.500 = 8(10) = 5 Cr. = 9.0634 -115)

Second 10 thin Now = ratelin-rate out ST da dt oxa - 2 *9 with 8(o)=9.0634 100 apply Laplace transfortos Libe} + {2:0 I{ '} 180 2{}=0 SOCs) – 0(0) + 80 °C) -0 sQ6) - 9,0634+008) 0 > qcs) ( 5+ soo) - 9.0634 a pcs)(loos, 19) - 9.0634 -) QC) (941005) = 906.34 906-34 100S + 9 apply Everse transform L-t{Q Gl} 906.34 S+ 1 -athoo 9.0634 e for next lo con -901100 OCIO) - 9.0634 e 3.684 16 - 100 roo - 06+) Aug
TQ.

Please like the answer, if it is helpful. TQ