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