Help with a Newtons Cooling Law Problem

[ZeromusX99]

Homework Statement

A thermometer reading 70^o 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 110^o F after 1/2 minute and 145^o F after 1 minute. How hot is the oven.

Homework Equations

T(t)=(T_o-T_R)e^kt+T_R

The Attempt at a Solution

Ok so I let said
T_R= Oven Temperature
T_o=70
T_o(.5)=110
T_o(1)=145

I then set up the general equation as
T(t) = (70-T_R)e^kt+T_R

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

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

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

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

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 T_R incredibly difficult, not even sure how I would be able to solve for T_R 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 T_R using that value for k or if there is a simpler way. Thanks in advance.

 

[Billy Bob]

Welcome to PF, ZeromusX99.

You did all the hard work already!

giving me these two equations
145-T_R=(70-T_R)e^k
110-T_R=(70-T_R)e^.5k

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

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

 

[D H]

Homework Equations

T(t)=(T_o-T_R)e^kt+T_R

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

You have a sign error.

<Long, tedious calculation elided>To me this seems like I may have overcomplicated something ...

Using your result, k= 2*ln((145-T_R)/(110-T_R)), 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.

 

[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 T_R, I'll go ahead and post it here

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

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

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

dividing my
35 = (70-T_R)(e^-k-e^-.5k)
by
40 = (70-T_R)(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-T_R)(e^-.5k-1)
gives me
40 = (70-T_R)(e^ln(7/8)-1)
which is just
40 = (70-T_R)((7/8)-1)
leaving me with
40 = (70-T_R)(-1/8)
multipling both sides by -1/8 gives me
-320 = 70 - T<sub>R[/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</sub>

 

[D H]

T_R = 390

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 70^o F to 110^o 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 110^o F to 145o^o 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,

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

 

[ZeromusX99]

Ah I understand now thanks for all the help D H