A tank contains 100 gallons of water in which 40 pounds of salt are dissolved. I

Question

A tank contains 100 gallons of water in which 40 pounds of salt are dissolved. It is desired to reduce the concentration of salt to 0.1 pounds per gallon by pumping in pure water at the rate of 5 gallons per minute and allowing the mixture (which is kept uniform by stirring) to flow out at the same rate. How long will this take?

Answer: 20 Ln 4 = 27.7 min.

Help me out with a step by step solution.

The rate at which a body cools is proportional to the difference between it and the surrounding medium. Mrs. Olson put a boiling coffee (100 C) to cool. If the temperature of the coffee dropped 20 degrees in 30 minutes, what is the temperature 30 minutes later?

Answer: 64 degrees.

Please help me with a step by step solution to the answer?

Explanation / Answer

Initially, the tank contains 40 pounds of salt. Since 5 gallons of water are leaving per minute, every minute, you are losing 5/100 = 0.05 of the solution. This means that 5% of the salt leaves every minute. Therefore, the differential equation, if we let S be the total amount of salt, looks like:

dS/dt = -0.05S

We can rearrange to get all the S's on the same side and move the t over to get:

dS/S = -0.05dt

Now we integrate both sides. Note that we only have to put the constant on one side. Integrating, we get:

ln(S) = -0.05t + C

S = e^(-0.05t+C) = e^C*e^(-0.05t) = Ce^(-0.05t)

Note that since C is an arbitrary constant, we can replace e^C with a different constant C. So the general form of the solution is S = Ce^(-0.05t). To find the specific solution, we know that at t = 0, there are 40 pounds of salt. So that means that C = 40, so we get

S = 40e^(-0.05t)

We want to find when S = (0.1 pounds/gallon)(100 gallons) = 10 pounds.

This occurs when e^(-0.05t) = 10/40 = 1/4, or:

e^(0.05t) = 4

0.05t = ln(4)

t = 20ln(4)

Now, we'll solve the second problem in a similar way. Let the temperature be T. Then the differential equation is given by:

dT/dt = k(T-T0)

where T0 is the temperature of the surroundings. We can then do a variable substitution where U = (T-T0) and dU/dt = dT/dt to get that

dU/dt = kU

dU/U = kdt

Then integrate to get

ln(U) = kt + C

ln(T-T0) = kt+C

T-T0 = e^(kt+C) = e^C*e^(kt) = Ce^(kt)

T = Ce^(kt) + T0

We know that initially, the coffee was 100 degrees. This implies that

C + T0 = 100

Similarly, after half an hour, the coffee was 80 degrees, so:

Ce^(k/2) + T0 = 80

Since we have two equations and three unknowns (the rate ratio, the constant out in front, and the temperature of the room), this is not enough information to solve the problem. However, if we assume that the room temperature of T0 = 0, then we get:

C = 100

Ce^(k/2) = 80

Ce^k = 80*(80/100) = 64

So the final temperature is Ce^k + T0 = 64.

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