Newton's Law of Cooling and such (3 Viewers)

[fashionista]

hi!
this question is annoying because i keep gettin a totally different answer to the thing they want me to show.

A tank contains 100 L of brine whose conc. is 3 g/L. 3 L of brine whose conc is
2 g/L flow into the tank each minute and at the same time 3 litres of mixture flow out each minute.

(a) Show that the quantity of salt, Q grams, in the tank at any time t is given by:
Q=200+100e^(-0.03t)

i got the answer as being something like Q=36-264e^(-t/6)

PLEASE HELP ME!!

 

[cj_bridle]

this is a question from fitzpatrick page 129 Q7...

I have the solution.. ill type it up for you now.. let me find it

 

[cj_bridle]

This is by far the hardest exponential growth and decay question they can give in my opinion.

Suppose the amount of salt at any time 't' is Q.

Since there is always 100L in the tank the amount of salt = Q/100 grams per L

dQ/dt = rate of inflow - rate of outflow
= 6 - 3Q/100
= (600 - 3Q)/100

Therefore dt/dQ = 100/(600-3Q)
(There is a tricky manipulation you have to do here otherwise you will get a negative log later.)
Multiply through by -1: -dt/dQ = 100/(3Q - 600)
By integration: -t = 100/3[ln(3Q-600) + C
when t = 0, Q = 300
Therefore C = -100/3ln(300)

Thus, -t = 100/3[ln(3Q - 600) - 100/3ln(300)
t = 100/3ln(300) - 100/3[ln(3Q - 600)
t = 100/3[ln[300/(3Q-600)]

300/(3Q - 600) = e^0.03t
(3Q - 600)/300 = e^-0.03t

By simplification: 3Q = 600 + 300e^-0.03t

Therefore Q = 200 + 100e^-0.03t

 

[Jase]

wait.. but after t=100, the original concentration brine with 3 g/L runs out and is now filled with 2 g/L salt only. Which means the eqaution should be different after t=100. so why is it limited to 200?

 

[cj_bridle]

umm this is going to be hard to explain..

when i did the rate of inflow - rate of outflow bit.. i used generalised concentrations of salt ( Q/100 g/L ) at any one point. and with concentrations.. its not a matter of 'once the 3g/L moves out its only 2g/L..' they would mix together... thats why this works... (i hope that explained it)

and its limited to 200 because for any value of 't' the amount of salt wont get below 200g. This is proven in part (c) of that question in fitzpatrick.

 

[fashionista]

oh thank u thank u!!
when i waq doing it i was taking the rate of outflow of salt to be Q/6 so therefore
dQ/dt= (36-Q)/6 so that was my problem
thank u thank u!!

 

[Premus]

Hey

I really dont understand how you have done this q, cj_bridle .....


I understand....
"Suppose the amount of salt at any time 't' is Q.
Since there is always 100L in the tank the amount of salt = Q/100 grams per L"

------------------------------------------------------------------------
....but have no idea how you got.....

"dQ/dt = rate of inflow - rate of outflow
= 6 - 3Q/100 <---------------this part here - how did you do it??
= (600 - 3Q)/100 "

nor do i understand

"when t = 0, Q = 300" ??

Pls explain ?
other than that the algebra is alright haha...
it would be great if anyone could post up all the parts to this question

Thanks a lot!!!

 

[Vortex]

amount of salt flowing into the tank is given by

"3 L of brine whose conc is
2 g/L flow into the tank each minute"

therefore 6 grams of salt

amount of salt flowing out of the tank is given by

"at the same time 3 litres of mixture flow out each minute"

therefore 3 x the concentration of whatever the mixture at the time is
the concentration of the mixture at the time is the amount of salt divided by the volume i.e. Q/100
therefore amount of salt flowing out is 3Q/100

therefore dQ/dt = 6 - 3Q/100



when t = 0, Q=300
this is given by "A tank contains 100 L of brine whose conc. is 3 g/L."
so at the start, there is 3 grams of salt per litre and 100 L in the tank therefore the amount of salt is 300g

 

[cj_bridle]

I saw "-dt/dQ" and gave up trying to understand it. There has never been a rates question, from 1986 to 2003, of this calibre in the HSC. I severely hope they don't give us this sort of question for this year's HSC.

your quite correct komaticom. i doubt there will be a question this hard in the paper.

vortex: thanks for helping

 

[ngai]

i doubt there will be a question this hard in the paper.

i agree
this stuff is way too hard