kuhse4461a

2021-12-30

Brine containing 3lb/gal of salt enters a tank at the rate of 4gal/min and the solution well-stirred leaves at 2gal/min. The tank initially contains 20 gal of water. What is the amount of salt in the tank after 5 minutes?

Pansdorfp6

Solution:
At a rate, salt enters the tube.

$R=12$ lbs/min
Due to the tenk's 2 gal/min loss of liquid

after to time: (minutes). The number of gallons of Brine in the tenk
$=20+2t$
So after 5 min
$=20+2×5$
$=30$ gallons
and 2 gallons $=.3454lbs$
So amount of salt $=30×8.3454$
$=250.362$

Heather Fulton

Given that the initial amount of salt is zero( because of pure water)
Rate in 4gal/min
Rate out 2gal/min initial volume is 20 gal
Brine containing 3lb/gal of salt
The inteving rate 4gal/min
So intering concentration
$=3×4$
$=12$
Volume $=20gal$
Let the Amount of Brine is y(+)
Then rate change
${y}^{1}=rat{e}_{\mathrm{ln}}-rat{e}_{out}$
$=12-\frac{2×y}{20+\left(4-2\right)t}$
${y}^{1}=12-\frac{y}{10+t},y\left(0\right)=0$
Salving

$y×\left(10+t\right)=\int 12×\left(10+t\right)dt+c=6{\left(10+t\right)}^{2}+c$
$y\left(+\right)=6\left(10+t\right)+\frac{c}{10+t}$
By $y\left(0\right)=0⇒0=60+\frac{c}{10}⇒c=-600$
$y\left(t\right)=6\left(10+t\right)-\frac{600}{10+t}$
The amount after 5 min
$y\left(5\right)=6×\left(10+5\right)-\frac{600}{10+5}=90-\frac{600}{15}$
$y\left(5\right)=90-40=50lb$

karton

xleb123

Given information:
- Brine enters the tank at a rate of 4 gallons per minute, with a concentration of 3 pounds of salt per gallon.
- The solution is well-stirred, meaning the concentration remains uniform throughout the tank.
- The solution leaves the tank at a rate of 2 gallons per minute.
- The tank initially contains 20 gallons of water.
To determine the amount of salt in the tank after 5 minutes, we need to find $S\left(5\right)$.
Let's break down the problem into two parts: the amount of salt entering the tank and the amount of salt leaving the tank.
1. Amount of salt entering the tank:
The rate of brine entering the tank is 4 gallons per minute, with a concentration of 3 pounds of salt per gallon. Therefore, the amount of salt entering the tank per minute is $4×3=12$ pounds. Thus, the total amount of salt entering the tank in the first $t$ minutes is $12t$ pounds.
2. Amount of salt leaving the tank:
The rate of solution leaving the tank is 2 gallons per minute. Since the solution is well-stirred, the concentration of salt in the tank is uniform. Therefore, the concentration of salt leaving the tank is also 3 pounds per gallon. Thus, the amount of salt leaving the tank per minute is $2×3=6$ pounds. Therefore, the total amount of salt leaving the tank in the first $t$ minutes is $6t$ pounds.
Now, let's calculate the amount of salt in the tank after 5 minutes.
At time $t$, the amount of salt in the tank is given by:

The initial amount of salt in the tank is 0 pounds since it contains only water initially.
Using the formulas we derived earlier, we have:
$S\left(t\right)=0+12t-6t$
Simplifying the expression:
$S\left(t\right)=6t$
Finally, we can calculate $S\left(5\right)$ to find the amount of salt in the tank after 5 minutes:
$S\left(5\right)=6×5=30$ pounds.
Therefore, the amount of salt in the tank after 5 minutes is 30 pounds.

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