A tank contains 100 grams of a substance dissolved in a large amount of water. The tank is filtered in such a way that water drains from the tank, leaving the substance behind in the tank. Consider the volume of the dissolved substance to be negligible. At what rate is the concentration (grams/liter) of the substance changing with respect to time in each scenario?
(a) the rate after 5 hours, if the tank contains 60 L of water initially, and drains at a constant rate of 4 L/hr?
I don't know how the liquid can drain and leave the substance behind, if it's dissolved. Maybe it's just in suspension, and the liquid drains through a filter.
Anyway, the volume of water at time t hours is
v = 60-4t
The concentration is
c(t) g/L = 100/v(t) = 100/(60-4t) = 25/(15-t)
dc/dt = 25/(15-t)^2
so, at t=5,
dc/dt = 25/(15-5)^2 = +0.25 g/L-hr
Well, well, well, looks like we have a draining dilemma! Let's calculate the rate of concentration change, shall we?
We know that the initial concentration of the substance in the tank is 100 grams divided by 60 liters of water, which is approximately 1.67 grams per liter.
Since water drains at a constant rate of 4 liters per hour, after 5 hours, a total of 20 liters of water will have drained.
Now, to find the new concentration after 5 hours, we need to find out how much substance is left in the tank. Since the volume of the dissolved substance is considered negligible, it means that the substance is not changing its mass over time. So, the remaining 80 liters of water will still contain 100 grams of the substance.
Therefore, the new concentration after 5 hours is 100 grams divided by (60 liters - 20 liters), which is 100 grams divided by 40 liters, which simplifies to be 2.5 grams per liter.
To find the rate of concentration change, we need to calculate the difference between the new concentration and the initial concentration, and divide it by the time taken. So, (2.5 - 1.67) grams per liter divided by 5 hours.
Doing the math, we get a rate of (0.83 grams per liter) divided by (5 hours), which simplifies to be 0.166 grams per liter per hour.
So, after 5 hours, the concentration of the substance is changing at a rate of 0.166 grams per liter per hour. Keep that in mind, and may your tank drain with comedic relief!
To find the rate at which the concentration of the substance is changing with respect to time, we can use the formula:
Rate of change of concentration = (Rate of drain * Initial concentration) / (Initial volume - Rate of drain * time)
Given:
Initial volume (V0) = 60 L
Rate of drain (R) = 4 L/hr
Time (t) = 5 hr
Initial concentration (C0) = 100 g
Plugging in the values into the formula:
Rate of change of concentration = (4 * 100) / (60 - 4 * 5)
= 400 / (60 - 20)
= 400 / 40
= 10 g/L
Therefore, the rate at which the concentration of the substance is changing with respect to time after 5 hours is 10 grams per liter.
To find the rate at which the concentration of the substance is changing with respect to time, we need to consider the following steps:
Step 1: Determine the amount of substance remaining in the tank after 5 hours.
Step 2: Calculate the concentration of the substance in the tank after 5 hours.
Step 3: Find the rate at which the concentration is changing with respect to time.
Let's go through each step in detail:
Step 1: Determine the amount of substance remaining in the tank after 5 hours.
We know that the tank initially contains 100 grams of the substance. Since the volume of the dissolved substance is considered negligible, we can assume that the amount of substance remaining after 5 hours is also 100 grams.
Step 2: Calculate the concentration of the substance in the tank after 5 hours.
The concentration of a substance is given by the formula:
Concentration = (Amount of substance) / (Volume of solution)
In this case, the amount of substance is 100 grams, and the volume of the solution (water) is changing with time.
Initially, the volume of water in the tank is 60 L, and it drains at a constant rate of 4 L/hr. After 5 hours, the total volume of water drained is:
5 hours * 4 L/hr = 20 L
Therefore, the remaining volume of water in the tank after 5 hours is:
Initial volume - Volume drained = 60 L - 20 L = 40 L
To calculate the concentration, we divide the amount of substance (100 grams) by the volume of the solution (40 liters):
Concentration = 100 grams / 40 liters
Step 3: Find the rate at which the concentration is changing with respect to time.
To find the rate at which the concentration is changing with respect to time, we take the derivative of the concentration with respect to time. Since the volume of water is draining at a constant rate, the derivative of the concentration is zero.
Therefore, the rate of change of the concentration after 5 hours is 0 grams per liter per hour.
Summary:
After 5 hours, if the tank initially contains 60 L of water and drains at a constant rate of 4 L/hr, the concentration of the substance remains constant at 100 grams per 40 liters, and the rate of change of the concentration is 0 grams per liter per hour.