# Help with a Newtons Cooling Law Problem

1. Sep 17, 2009

### ZeromusX99

1. The problem statement, all variables and given/known data
A thermometer reading 70o F is placed in an oven preheated to a constant temperature. Through a glass window in the over door, an observer records that the thermometer reads 110o F after 1/2 minute and 145o F after 1 minute. How hot is the oven.

2. Relevant equations
T(t)=(To-TR)ekt+TR

3. The attempt at a solution
Ok so I let said
TR= Oven Temperature
To=70
To(.5)=110
To(1)=145

I then set up the general equation as
T(t) = (70-TR)ekt+TR

Then using the two times and temperatures I was given I created these two equations.
145=(70-TR)ek+TR
110=(70-TR)e.5k+TR

I then moved over the TR on the end over to the left side giving me these two equations
145-TR=(70-TR)ek
110-TR=(70-TR)e.5k

Dividing the top equation by the bottom i get
(145-TR)/(110-TR)=e.5k

Taking the natural log of both sides and then dividing by .5 gets me
k= 2*ln((145-TR)/(110-TR))

To me this seems like I may have overcomplicated something or made a mistake as plugging that value of k back into the equation seems like it would make solving for TR incredibly difficult, not even sure how I would be able to solve for TR using that value for k.

So what I'm asking is if the steps I took were the right ones to do and if I should go ahead and try to solve for TR using that value for k or if there is a simpler way. Thanks in advance.

Last edited: Sep 18, 2009
2. Sep 18, 2009

### Billy Bob

Welcome to PF, ZeromusX99.

You did all the hard work already!

From this last equation, you can substitute your expression for e.5k into 110-TR=(70-TR)e.5k and solve for TR.

3. Sep 18, 2009

### D H

Staff Emeritus
Whoa! This would say the final temperature is either plus or minus infinity.

You have a sign error.

Using your result, k= 2*ln((145-TR)/(110-TR)), what is e.5k? (It isn't all that complex.)

Note well: This is a valid approach but it will of course yield the wrong answer because of your sign error.

Aside: There is a much easier approach. What happened in the second half minute?

EDIT
Your approach is fine. It will just happen to yield a negative value for k.
The comment about an easier approach is still valid.

4. Sep 19, 2009

### ZeromusX99

Thanks for the suggestion Billy Bob substituting for e.5k worked, guess I was just over thinking things.

Also, thanks to your suggestion D H of looking at the second half minute, where temperature increased by 35, I found the simpler way of doing the calculation TR, I'll go ahead and post it here

Instead of dividing I just subtract the equations I initially listed to get
35 = (70-TR)(e-k-e-.5k

Then do another subtraction of equations using my T(.5) and T(0)
110 = (70-TR)(e-.5k+TR
70 = (70-TR)+TR

I get
40 = (70-TR)(e-.5k-1)

dividing my
35 = (70-TR)(e-k-e-.5k)
by
40 = (70-TR)(e-.5k-1)
then gets me
7/8 = (e-k-e-.5k)/(e-.5k-1)

pulling an e-.5k from the numerator then gives me
7/8 = e-.5k(e-.5k-1)/(e-.5k-1)
(e-.5k-1) cancels leaving me with
7/8 = e-.5k

natural of both sides gives me ln(7/8) = -.5k so
k= -2ln(7/8)

Plugging this into my equation
40 = (70-TR)(e-.5k-1)
gives me
40 = (70-TR)(eln(7/8)-1)
which is just
40 = (70-TR)((7/8)-1)
leaving me with
40 = (70-TR)(-1/8)
multipling both sides by -1/8 gives me
-320 = 70 - TR[/SUB
Add TR and 320 to both sides then gives me
TR[/SUB = 390

Hehe next time I'll remember to make it -kt as well

5. Sep 19, 2009

### D H

Staff Emeritus
Correct.

You didn't quite get the gist of my hint.
Write Newton's law of cooling as

$$T(t)-T_R = (T_0-T_R)e^{-kt}$$

During the first half minute the temperature rose from 70o F to 110o F:

$$110-T_R = (70-T_R)e^{-kt/2}$$

from which

$$e^{-kt/2} = \frac{110-T_R}{70-T_R}$$

During the next half minute the temperature rose from 110o F to 145oo F. Thus

$$145-T_R = (110-T_R)e^{-kt/2}$$

Substituting the expression for $\exp(-kt/2)$ in the above,

$$(145-T_R)(70-T_R) = (110-T_R)^2$$

Simplifying,

\aligned 10150-215T_R+T_R^2 &= 12100-220T_R+T_R^2 \\ (220-215)T_R &= (12100-10150) \\ T_R &=390 \endaligned

6. Sep 19, 2009

### ZeromusX99

Ah I understand now thanks for all the help D H