Change in concentration over time

In summary, the problem involves a tank with a capacity of 400L filled with a mixture of water and chlorine with a concentration of 0.05g of chlorine per liter. Fresh water is pumped in at a rate of 4L/s and the mixture is stirred and pumped out at a rate of 10L/s. After 15 seconds, the concentration of chlorine in the tank can be calculated using the equation |x|= (|200-3t|)^(5/3)+20-200^(5/3).
  • #1
Scalper
1
0

Homework Statement


A tank with a capacity of 400L is full of a mixture of water and chlorine with a concentration of 0.05g of chlorine per liter. The chlorine concentration is to be reduced by pumping in fresh water at the rate of 4L/s. The mixture is kept stirred and pumped out at the rate of 10L/s. What is the concentration of chlorine in the tank 15 seconds later?


Homework Equations


Let x= g chlorine in the tank
let t= time in seconds

The Attempt at a Solution


dx/dt=(rate in)-(rate out)

dx/dt= [(4L/s)(0g/l)]-[(10L/s)(x/(400-6t))]

dx/dt=0-10x/(400-6t)

dx/dt=-5x/(200-3t)

(-1/5)(dx/x)=dt/(200-3t)

integrating both sides:

(-1/5)*ln(|x|)=(-1/3)*ln(|200-3t|)+C

3*ln(|x|)=5*ln(|200-3t|)+C

ln(|x|)=(5/3)*ln(|200-3t|)+C

|x|=e^[(5/3)*ln(|200-3t|)]+C

|x|=e^[ln(|200-3t|)]^(5/3)+C

e^ln cancels

|x|=(|200-3t|)^(5/3/)+C

solving for C, When t=0, x=400*.05=20

|20|=(|200-3*0|)^(5/3)+C

20-200^(5/3)=C

then solving for x when t=15

x=(|200-3*15|)^(5/3)+20-200^(5/3)

this gets a -2300ish, which I am sure is not right. what did i do wrong? lol
 
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  • #2
Scalper said:
ln(|x|)=(5/3)*ln(|200-3t|)+C

|x|=e^[(5/3)*ln(|200-3t|)]+C

"]" is at the wrong place: |x|=e^[(5/3)*ln(|200-3t|)+C]

ehild
 

1. What is meant by "change in concentration over time" in science?

Change in concentration over time refers to the rate at which the concentration of a substance changes over a given period. It is a measure of how quickly the amount of a substance in a solution changes over time.

2. How is the change in concentration over time calculated?

The change in concentration over time can be calculated by taking the difference between the initial concentration and the final concentration and dividing it by the time elapsed between the two measurements. This calculation is typically represented as ΔC/Δt.

3. What factors can cause a change in concentration over time?

There are several factors that can cause a change in concentration over time, including the addition or removal of substances from the solution, changes in temperature or pressure, and chemical reactions occurring within the solution.

4. Why is it important to measure the change in concentration over time in scientific experiments?

Measuring the change in concentration over time is important in scientific experiments because it allows researchers to track and understand the kinetics of a reaction. It also helps to determine the rate at which substances are reacting and the factors that may be influencing the reaction.

5. How can the change in concentration over time be controlled or manipulated in a laboratory setting?

The change in concentration over time can be controlled or manipulated in a laboratory setting by adjusting the initial concentration of a substance, changing the reaction temperature, using catalysts, or altering the pH of the solution. Additionally, the addition or removal of substances from the solution can also affect the change in concentration over time.

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