Group Members: Ryjll Morris, Jason Zhu, Aron Singh

Cooling

Newtons law of cooling states that the rate of heat loss of an object is directly proportional to the difference in the temperatures between the object and it’s medium. In other words, according to the textbook, for the temperature T(t) of an object at time (t), in a medium with a temperature Tm(t), the rate of change of T at any time (t) is proportional to T(t) – Tm(t).

This is ultimately expressed as

T = Tm + (To - Tm){e^-kt}

This formula is derived from the first order linear differential equation:

{T^'} = -k(T - Tm)

We solved this equation in the following steps:

Distributing K yields

{T^'} + kT = kTm

The complementary equation solution is in the form of T= ue^{-kt}, and {u^'}e^{-kt} = kTm

Therefore, solving for {u^'} we get

{u^'} = kTme^{-kt}

Integrating this equation to get u gives us

u = Tme^{kt} +c

Now, here we can recall that T = ue^{-kt}so substituting what we got for u into this equation we get

T = Tm +ce^{-kt}

Because c represents any constant, if the time is set to T(0) = To for the initial temperature, c would be (To – Tm), which is the inital temperature of the object minus the temperature of the medium.

So now substituting the value for c into the equation above we get

T = Tm + (To - Tm)e^{-kt}

Mixing

The mixing section deals with the same first order linear differential equations and how a solution changes over time. The application is a salt water solution with a given concentration being added at a specific rate to a tank that initially contains saltwater with a different concentration. the objective with these problems is to determine the quantity of salt in the tank as a function of time. So, lets say we have a tank that initially contains 40 lbs of salt dissolved in 600 gallons of water. At to = 0, the initial time zero, water that contains a \frac{1}{2} pound of salt per gallon is poured into the tank at the rate of 4 gallons per minute and the mixture is drained from the tank at the same rate.

Here we have to find a differential equation for the quantity Q(t) of salt in the tank at time t>0, solve the equation to determine Q(t) and then find the lim t \rightarrow \infty Q(t).

So here, Q , which is the rate of change of the quantity of salt in the tank, changes with respect to time. Therefore, if rate in represents th rate at which salt enters the tank and rate out represents the rate at which salt leaves the tank then

Q^' = rate in – rate out.

Here, rate in is (concentration) * (rate of flow)

=\frac{1}{2} (lb/ gal) * 4 (gal/min) = 2 lb/min.

Now to determine rate out, we have to do a little more work. Think about the fact that we are removing 4 gallons of mixture per minute but there is always 600 gallons in the tank. Using this as a ratio we can see that

6 : 4 = \frac{6}{4} = 1.5

1.5 : 1

so we are actually removing 1/150 of the mixture per mintue. Now because the salt is evenly distrubuted, we are also removing 1/150 of the salt per minute.

Therefore, if at time (t) there are Q(t) pounds present in the tank the rate out at any time (t) = \frac{Q(t)} {150}

So this will look like

rate out =\frac {Q(t)}{600} *4 = \frac{Q(t)} {150}

Recalling our previous formula for Q^'

{Q^'} = 2lb/min – \frac{Q(t)}{150}

Rearranging this in the Linear differential format gives us

{Q^'} + \frac{Q}{150} = 2

Now we know thate^{-t/150}is a solution of the complementary equation for this differential, which is in the form of

Q = ue^{-t/150}, where {u^'}e^{-t/150} = 2

So once again, making u^' the subject we get

{u^'} = 2e^{t/150}

Integrating this we get

u = 300e^{-t/150} +c

Recalling now that Q =ue^{-t/150} and substituting the value of u into this equation we obtain

Q = 300 +ce^{-t/150}

From the previous section, we learnt that c = (To -Tm) so, applying value to this case would make c = (Q(0) – Q(t))

From our example we know that Q(0) the inital quantity of salt is 40 lbs, and we just found that Q(t) is 300

So c = 40 – 300 = -260

Therefore Q =300 -260e^{-t/150}

Now from this we can see that the limt \rightarrow \infty Q(t) = 300 for any value of Q(0).

EXAMPLES:

Cooling Problem 1

Cooling Problem 3

Cooling Problem 5 and 7

Mixing Problems

EXERCISES:

In textbook: Section 4.2 Exercises

#8 – A tank initially contains 40 gallons of pure water. A solution with 1 gram of salt per gallon of water is added to the tank at 3 gal/min, and the resulting solution drains out at the same rate. Find the quantity Q.t / of salt in the tank at time t > 0.

#4(a) – A thermometer initially reading 212 F is placed in a room where the temperature is 70 F. After 2 minutes the thermometer reads 125 F.
(a) What does the thermometer read after 4 minutes