Newton's Law of Cooling - Differential Equation

Reuel · 2011-02-02 01:41:49

A heating a cooling differential equation word problem:

"Blood plasma is stored at 40 degrees F. Before it can be used, it must be at 90 degrees F. When the plasma is placed in an oven at 120 degrees F, it takes 45 minutes for the plasma to warm to 90 degrees F. Assume Newton's Law of cooling applies. How long will it take the plasma to warm to 90 degrees F if the oven temperature is set at 100 degrees F?"

Looking up Newton's Law of Cooling, I believe a good starting place is to state the equation as

where T is the temperature at time t and that change is directly proportional to the difference between the surrounding temperature of the chamber and that of the object.

Listing the initial values...

(for a 120F oven)

And so, what the problem is asking for is for us to find the time it will take to warm the plasma all the way up to 90 F if the oven is only 100. Just working off of the definition of Newton's Law of Cooling, here is a first attempt at setting up the problem:

That isn't right but I am not sure how to specify two different temperatures for two different objects, the oven and the plasma, t 0 and time t.

Reuel · 2011-02-02 01:50:16

Wait, the temperature of the oven, in this model, is constant. Could I just write

?

bobbym · 2011-02-02 02:01:45

Hi Reule;

I have it as -k in my notes?!

Reuel · 2011-02-02 02:05:46

Hi bobbmy.

I am just going by what my teacher said.

"The rate of cooling of an object is directly proportional to the difference in temperature of the surroundings and the temperature of the object."

bobbym · 2011-02-02 02:08:33

Why not solve the DE you were given and then just plug in.

Where R is the temperature of the surrounding environment.

Reuel · 2011-02-02 02:08:59

What DE was I given? Other than the cooling law on the right side which I just based on my notes.

Last edited by Reuel (2011-02-02 02:10:32)

bobbym · 2011-02-02 02:15:07

Hi;

I mean the one you found:

Reuel · 2011-02-02 02:17:27

Because yours

and mine

have the order of the temperature of the object and the temperature of the surrounding area reversed. Which one is the proper order?

bobbym · 2011-02-02 02:22:37

Hi Reule;

I do not think it will matter. Give me a minute to complete my work and let us see if we can agree on a solution.

bobbym · 2011-02-02 02:57:42

Hi

Solve the DE.

Solve for c using the initial conditions.

Use the next value at T(45) to solve for k the constant of proportionality.

Where R is the surrounding temperature.

Reuel · 2011-02-02 03:58:01

I think solving for k was throwing me off... I wasn't sure what conditions to use for it. A lot of my trouble with these problems is knowing how to use the information given. I suppose that's true for most people.

Here is my work doing it the other way. I am not sure if our answers are the same or not.

Solving for C with t = 0,

Solving for k by following your example,

So the function would then be

If I have done my math correctly, that is.

I don't know. What do you think?

bobbym · 2011-02-02 04:07:04

Try to answer the next part of the question with your formula, I get around 95 minutes.

Reuel · 2011-02-02 04:16:16

t = 2? That isn't right. I must have messed something up.

bobbym · 2011-02-02 04:20:25

Yes, that is not correct. Your k is not right. If 120 took 45 minutes at 100 it has to take longer, a lot longer.

Reuel · 2011-02-02 04:24:05

So, apparently, it does matter whatever order the difference is put in? Or did I make some little error? It's a good thing to know.

bobbym · 2011-02-02 04:27:33

Hi Reuel;

Yes, I think so now.
Check out this page here and then see if you can do the example they have.

http://www.math.wpi.edu/Course_Material … node5.html

It is what I use for these cooling problems.

Reuel · 2011-02-02 04:30:07

Awesome! Thanks! That is going into my notebook for sure.

Let me the problem that way. Hang on.

bobbym · 2011-02-02 04:32:31

Okay, if you do not get her answer than we can look at it. Is that okay with you?

Reuel · 2011-02-02 04:42:15

Yes.

In post #10 how did you get 8/3? I keep coming up with 3/8.

bobbym · 2011-02-02 04:45:44

They are the same mine has a - sign in front.

Reuel · 2011-02-02 04:51:52

Less than two years ago I hardly knew much more than how to solve 3x = 6 for x. In that time I have gotten as far as I have and somehow, someway, in all that time I never realized a negative sign made equal the reciprocal of the inside of a logarithm. How did that ever escape me? And for so long?

You learn new stuff every day.

Finishing up the problem now...

Reuel · 2011-02-02 05:01:02

I got 95.4038 minutes.

bobbym · 2011-02-02 05:03:31

That I is correct, I think.

Reuel · 2011-02-02 05:56:43

Horray. Here is my work:

Solving first for C with the initial condition T(0) = 45 which is where T_s is 120 and where the plasma is 40 F...

Therefore, the equation is now

Solving for the proportionality constant k the initial conditions are used again:

So the new function solved for C and k is

The new function is in terms of T_s and t and so to calculate how long it will take in time (t) for the plasma to warm to 90 degrees (T(t)) the equation can be solved for t:

And by that t comes out to be approximately 95 minutes.

Thanks for the help. I now see how solving for c and then k using the initial conditions allows for the establishment of a function with which to calculate new values for time.

bobbym · 2011-02-02 06:04:04

Hi Reuel;

Glad to help! Thanks for working with me on them.

Math Is Fun Forum

#1 2011-02-02 01:41:49

Newton's Law of Cooling - Differential Equation

#2 2011-02-02 01:50:16

Re: Newton's Law of Cooling - Differential Equation

#3 2011-02-02 02:01:45

Re: Newton's Law of Cooling - Differential Equation

#4 2011-02-02 02:05:46

Re: Newton's Law of Cooling - Differential Equation

#5 2011-02-02 02:08:33

Re: Newton's Law of Cooling - Differential Equation

#6 2011-02-02 02:08:59

Re: Newton's Law of Cooling - Differential Equation

#7 2011-02-02 02:15:07

Re: Newton's Law of Cooling - Differential Equation

#8 2011-02-02 02:17:27

Re: Newton's Law of Cooling - Differential Equation

#9 2011-02-02 02:22:37

Re: Newton's Law of Cooling - Differential Equation

#10 2011-02-02 02:57:42

Re: Newton's Law of Cooling - Differential Equation

#11 2011-02-02 03:58:01

Re: Newton's Law of Cooling - Differential Equation

#12 2011-02-02 04:07:04

Re: Newton's Law of Cooling - Differential Equation

#13 2011-02-02 04:16:16

Re: Newton's Law of Cooling - Differential Equation

#14 2011-02-02 04:20:25

Re: Newton's Law of Cooling - Differential Equation

#15 2011-02-02 04:24:05

Re: Newton's Law of Cooling - Differential Equation

#16 2011-02-02 04:27:33

Re: Newton's Law of Cooling - Differential Equation

#17 2011-02-02 04:30:07

Re: Newton's Law of Cooling - Differential Equation

#18 2011-02-02 04:32:31

Re: Newton's Law of Cooling - Differential Equation

#19 2011-02-02 04:42:15

Re: Newton's Law of Cooling - Differential Equation

#20 2011-02-02 04:45:44

Re: Newton's Law of Cooling - Differential Equation

#21 2011-02-02 04:51:52

Re: Newton's Law of Cooling - Differential Equation

#22 2011-02-02 05:01:02

Re: Newton's Law of Cooling - Differential Equation

#23 2011-02-02 05:03:31

Re: Newton's Law of Cooling - Differential Equation

#24 2011-02-02 05:56:43

Re: Newton's Law of Cooling - Differential Equation

#25 2011-02-02 06:04:04

Re: Newton's Law of Cooling - Differential Equation

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