A Thermometer Reading 70 F Is Placed in an Oven Preheated to a Constant Temperature

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Help with a Newtons Cooling Law Problem

• Thread starter ZeromusX99
• Start date Sep 17, 2009

Homework Statement

A thermometer reading 70^o F is placed in an oven preheated to a constant temperature. Through a drinking glass window in the over door, an observer records that the thermometer reads 110^o F later on 1/two minute and 145^o F after i minute. How hot is the oven.

Homework Equations

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

The Try at a Solution

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

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

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

I so moved over the T_R on the terminate over to the left side giving me these two equations
145-T_R=(lxx-T_R)due east¹⁰⁰⁰
110-T_R=(lxx-T_R)e^.5k

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

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

To me this seems similar I may accept overcomplicated something or made a error equally plugging that value of m back into the equation seems like it would make solving for T_R incredibly difficult, not fifty-fifty certain how I would exist able to solve for T_R using that value for one thousand.

So what I'm asking is if the steps I took were the right ones to do and if I should go alee and try to solve for T_R using that value for one thousand or if in that location is a simpler manner. Thanks in accelerate.

Last edited: Sep eighteen, 2009

Answers and Replies

Welcome to PF, ZeromusX99.

You did all the difficult work already!

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

Dividing the pinnacle equation past the lesser 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.

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, boring adding elided>To me this seems similar I may have overcomplicated something ...

Using your result, m= 2*ln((145-T_R)/(110-T_R)), what is e^.5k? (Information technology isn't all that complex.)

Note well: This is a valid arroyo merely 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 simply happen to yield a negative value for 1000.
The comment about an easier approach is however valid.

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

Also, thank you to your proposition D H of looking at the 2nd 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 go
35 = (70-T_R)(eastward^-k-east^-.5k

Then do some other subtraction of equations using my T(.v) and T(0)
110 = (seventy-T_R)(due east^-.5k+T_R
70 = (70-T_R)+T_R

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

dividing my
35 = (seventy-T_R)(e^-k-east^-.5k)
by
forty = (70-T_R)(e^-.5k-i)
then gets me
seven/eight = (e^-k-e^-.5k)/(e^-.5k-1)

pulling an e^-.5k from the numerator then gives me
7/viii = e^-.5k(east^-.5k-i)/(due east^-.5k-1)
(e^-.5k-one) cancels leaving me with
7/eight = e^-.5k

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

Plugging this into my equation
twoscore = (lxx-T_R)(e^-.5k-1)
gives me
forty = (70-T_R)(e^ln(7/eight)-1)
which is but
40 = (lxx-T_R)((7/8)-1)
leaving me with
xl = (70-T_R)(-1/viii)
multipling both sides by -1/8 gives me
-320 = 70 - T_{R[/SUB
Add TR and 320 to both sides then gives me
TR[/SUB = 390}

Hehe next fourth dimension I'll recollect to make it -kt as well

Correct.

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

[tex]T(t)-T_R = (T_0-T_R)e^{-kt}[/tex]

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

[tex]110-T_R = (70-T_R)e^{-kt/2}[/tex]

from which

[tex]due east^{-kt/2} = \frac{110-T_R}{70-T_R}[/tex]

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

[tex]145-T_R = (110-T_R)e^{-kt/2}[/tex]

Substituting the expression for [itex]\exp(-kt/2)[/itex] in the in a higher place,

[tex](145-T_R)(seventy-T_R) = (110-T_R)^2[/tex]

Simplifying,

[tex]\aligned
10150-215T_R+T_R^2 &= 12100-220T_R+T_R^two \\
(220-215)T_R &= (12100-10150) \\
T_R &=390
\endaligned[/tex]

Ah I sympathize now cheers for all the help D H

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