A tank contains 2920 L of pure water. A solution that contains 001 kg of sugar per liter enters a tank at the rate 4 L/min The solution is mixed and drains from the tank at the same rate.
(a) How much sugar is in the tank initially?
(b) Find the amount of sugar in the tank after t minutes. (Answer should be a function of t)
(c) Find the concentration of sugar in the solution in the tank after 39 minutes.
For a), I put 0.
For b), I have so far: Q'=R_in*C_in - R_out*C_out= (4)(0.01)-(4)*(Q(t)/V)
Not sure what to do next. Help would be greatly appreciated.
To find the amount of sugar in the tank initially (a), we can multiply the density of the solution by the volume of water in the tank. Since the concentration of sugar is given in kg per liter, we need to convert the volume of water from liters to kg.
Density of pure water = 1 kg/L
Therefore, the amount of sugar in the tank initially is:
(a) Amount of sugar = Density of solution * Volume of water
Amount of sugar = 0.001 kg/L * 2920 L
Amount of sugar = 2.92 kg
For b) to find the amount of sugar in the tank after t minutes, you are on the right track with the equation Q' = R_in * C_in - R_out * C_out. Let's use that equation to solve for the amount of sugar in the tank as a function of t.
Q' = Rate of sugar entering the tank - Rate of sugar leaving the tank
Rate of sugar entering the tank = R_in * C_in
Rate of sugar entering the tank = (4 L/min) * (0.001 kg/L)
Rate of sugar entering the tank = 0.004 kg/min
Rate of sugar leaving the tank = R_out * (Q(t) / V)
Rate of sugar leaving the tank = (4 L/min) * (Q(t) / 2920 L)
Now we can substitute these values back into the equation:
Q' = 0.004 - (4 * Q(t) / 2920)
To find the amount of sugar in the tank after t minutes, we need to solve this differential equation. Pulling the derivative of Q(t) and substituting back into the equation:
Q'(t) = dQ(t) / dt
dQ(t) / dt = 0.004 - (4 * Q(t) / 2920)
Separating variables:
dQ(t) = (0.004 - (4 * Q(t) / 2920)) * dt
Now, we can integrate both sides of the equation:
∫ dQ(t) / (0.004 - (4 * Q(t) / 2920)) = ∫ dt
After integrating both sides, we get:
ln|0.004 - (4 * Q(t) / 2920)| = t + C
where C is the constant of integration.
To find the particular solution, you can use the initial condition from part (a) where Q(0) = 2.92 kg.
At t = 0, we have:
ln|0.004 - (4 * 2.92 / 2920)| = 0 + C
ln|0.004 - 0.004| = C
ln(0) = C
Since ln(0) is undefined, we can say that C = -∞
Now the equation becomes:
ln|0.004 - (4 * Q(t) / 2920)| = t - ∞
Taking the exponential of both sides:
|0.004 - (4 * Q(t) / 2920)| = e^(t - ∞)
Since e^(-∞) approaches 0, this simplifies to:
0.004 - (4 * Q(t) / 2920) = 0
Now, solve for Q(t):
4 * Q(t) = 0.004 * 2920
Q(t) = 0.004 * 2920 / 4
Q(t) = 2.92 kg
Therefore, the amount of sugar in the tank after t minutes remains constant at 2.92 kg.
For part (c), we can use the initial equation:
Q' = 0.004 - (4 * Q(t) / 2920)
Substituting t = 39 into this equation:
Q'(39) = 0.004 - (4 * Q(39) / 2920)
Since we know Q(t) remains constant at 2.92 kg, substitute this value into the equation:
Q'(39) = 0.004 - (4 * 2.92 / 2920)
Q'(39) = 0.004 - 0.004
Q'(39) = 0
Therefore, the concentration of sugar in the solution in the tank after 39 minutes is 0 kg/L.
To solve this problem, let's break it down step by step:
(a) To find the amount of sugar in the tank initially, we are given that the tank contains 2920 L of pure water, so the initial amount of sugar is 0 kg.
(b) To find the amount of sugar in the tank after t minutes, let's set up an equation using the principle of conservation of mass. The rate of solution entering the tank is 4 L/min and the concentration of sugar in the entering solution is 0.01 kg/L. The rate of solution leaving the tank is also 4 L/min. Let's denote the quantity of sugar in the tank at time t as Q(t) (in kg).
The equation becomes:
dQ(t)/dt = (4 L/min) * (0.01 kg/L) - (4 L/min) * (Q(t) / 2920 L)
To solve this differential equation, let's rewrite it in the form of dQ(t)/dt = K - (Q(t) / T), where K = (4 L/min) * (0.01 kg/L) and T = 2920 L.
Rearranging the variables, we get:
dQ(t)/dt + (Q(t) / T) = K
This is a first-order linear ordinary differential equation. We can solve it using an integrating factor. The integrating factor is e^(1/T * t) since the coefficient of Q(t) is 1/T.
Multiplying the equation by the integrating factor, we get:
e^(1/T * t) * dQ(t)/dt + (e^(1/T * t) / T) * Q(t) = K * e^(1/T * t)
Now, let's integrate both sides of the equation:
∫[e^(1/T * t) * dQ(t)/dt] dt + ∫[(e^(1/T * t) / T) * Q(t)] dt = ∫[K * e^(1/T * t)] dt
The first integral on the left side is simply e^(1/T * t) * Q(t).
The second integral can be rewritten as:
∫[(e^(1/T * t) / T) * Q(t)] dt = (1 / T) ∫[e^(1/T * t) * Q(t)] dt
Now, we have:
e^(1/T * t) * Q(t) + (1 / T) ∫[e^(1/T * t) * Q(t)] dt = ∫[K * e^(1/T * t)] dt
The integral on the right side is a simple integration of a constant, which gives us:
K * ∫[e^(1/T * t)] dt = K * (T * e^(1/T * t) + C)
Let's substitute the value of K back in and simplify further:
e^(1/T * t) * Q(t) + (1 / T) ∫[e^(1/T * t) * Q(t)] dt = K * (T * e^(1/T * t) + C)
e^(1/T * t) * Q(t) + (1 / T) ∫[e^(1/T * t) * Q(t)] dt = (4 L/min) * (0.01 kg/L) * (2920 L * e^(1/T * t) + C)
To solve for the integral, we can use the method of integration by parts by setting u = Q(t) and dv = e^(1/T * t) dt. This will give us ∫[e^(1/T * t) * Q(t)] dt.
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