1. A tank initially holds 100 L of solution and the solution contains 20 kg of d
ID: 2847257 • Letter: 1
Question
1. A tank initially holds 100 L of solution and the solution contains 20 kg of
deadly chemical X. Pure solution is then poured into the tank at a rate of 5 L per minute.
The tank is also drained at this rate. Assuming uniform dispersion of the deadly chemical
X, how long will it be in minutes before there is less than 1 kg of Chemical X in the tank?
2. A thermometer sits on a kitchen counter all night where the temperature
is a steady 70F. Outside, the temperature is 10F. After being outside for 3 minutes, the
thermometer reads 25F. Find the explicit function for the temperature of the thermometer
at time t minutes.
3. A Turkey comes out of the fridge at a temperature of 35 degrees Fahren-
heit. It must be cooked to an internal termperature of 180 degrees in order to be safe to eat.
After 1 hour, you check the turkey and it measures 100 degrees. If you put it in an oven set
to 200 degrees, using Newton's Law of Heating, how long will it take to cook the Turkey in
a pre-heated oven?
Explanation / Answer
1. Let the amount (in kg) of chemical X in the tank be represented by the letter A. Then dA/dt = (rate of chemical X in) - (rate of chemical X out) = (5L/min)*(0kg/min) - (5L/min)*(Akg/100L). The 0kg/min is because the solution coming in contains 0 chemical X. The reason for the Akg/100L term is that that's the concentration of the flow leaving the tank. So we have dA/dt = -A/20. Separating variables gives us dA/A = -dt/20. Then integrating gives ln|A| = -t/20 + C. Raising e to both sides gives A = e^(-t/20 + C) or A = Ce^(-t/20). We know that there are 20 kgs in the tank to start, so A(0) = 20, which implies C is 20, giving us A = 20e^(-t/20). We want to know when chemical X reaches 1kg, so let A = 1 and solve:
1 = 20e^(-t/20) --> 1/20 = e^(-t/20) --> ln(1/20) = -t/20 --> t = -20ln(1/20) or t = 20ln(20).
2. Newton's Law of Heating/Cooling says Temperature = (Initial - Ambient)e^(rt) + Ambient. With this in mind, our equation for the thermometer is T = (70 - 10)e^(rt) + 10. We're told that after three minutes, the temperature goes down from 70 to 25, so we plug 25 in for T. This gives 25 = 60e^(3r) + 10, or 0.25 = e^(3r). Solving for r tells us that r = (1/3)ln(0.25). Therefore, our explicit function is T = 60e^(t*ln(0.25)/3) + 10.
3. Again, use Newton's Law of Heating/Cooling. T = (35 - 200)e^(rt) + 200 or T = -165e^(rt) + 200. After 1 hour, T = 100, so this gives us 100 = -165e^(1*r) + 200 --> r = ln(100/165). Now we want to know when the turkey reaches 180 degrees, so we say 180 = -165e^(t*ln(100/165)) + 200. Solving for t:
-20 = -165e^(tln(100/165)) --> ln(20/165) = t * ln(100/165) --> t = [ln(20/165)]/ln(100/165) = 4.21 hours.
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