# Newtons cooling law not making sense

1. Oct 21, 2009

### pezgoon

I'm going to write this up the same way my teacher gave me the sheet

1. The problem statement, all variables and given/known data
At 9am on october 19, 2009 a body was found in room 327 at University Center. The room is kept at a constant temperature of 72 degrees. The medical examiner was called and he arrived in eight minutes. The first thing he did was to take the temperature of the body. It was 83 degrees. Thirty minutes later the temperature of the body was taken again and it was now 78 degrees. Help the police by telling them when the person was murdered.

Variables

t= time
y= temp. in degrees Fahrenheit of an object in a room
T= temperature in degrees Fahrenheit of the room
T= 72
yinitial(0)= 98.6 (assumed)
yfirst(t+8?)= 83
ysecond(t+38?)= 78
I put the question marks cause I'm not sure whether that is the right time.

2. Relevant equations
This is the equation as it was given to me

ln|y-T|= kt + C

but everywhere else I look has a different formula than this one the closest i could find/come up with (i cant remember at this point) is

y-T=e^(kt + C)

3. The attempt at a solution

I have tried the equation (and just about every other one I could find or think of) so many times that I cannot figure them out on the paper anymore and I am basically lost as to where to start because every way I've tried by plugging the numbers back in to try and get the temp at the time comes out wrong. Can someone please help?? Thanks to whomever can
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 21, 2009

### mgb_phys

At 9:08 the temperature was 11 F (above background), at 9:38 it was 6 F you need to extrapolate back to when it was 26.6 F

Hint - log scale

3. Oct 21, 2009

### pezgoon

actually i forgot to put it into the question but she wants us to use the formula and show all the work, so i cant do it the easy way lol

4. Oct 22, 2009

### pezgoon

This still makes no sense this is what I tried and got stuck at:

ln|98.6-72|=k*0+C

so k and 0 cancel and i get

3.280911216=C

so then since I dont know M (which i think you originally used x) but from going from 908 to 938 and there being a 5 degree drop I assumed thats constant and figure 30 minutes after he died the temp would be at 93.6 so i did:

ln|93.6-72|=k*30+3.280911216

and get

3.072693315=k*30+3.280911216

so I tried both dividing first and subtracting first and when i checked both dividing first gave me wild numbers and subtracting actually gave me the right answer so:

3.072693315=k*30+3.280911216
-3.280911216 -3.280911216
gives me
-.208217901/(30)=k*30/(30)
gives me
k=-.0069405967
so then to check it i did
y-72=e^(-.0069405967*30+3.280911216)
and got something around 93.59999333 or so rounding up to 93.6 so I thought it had worked so then i tried this to check it

ln|83-72|=-.0069405967*t+3.280911216
so then
2.397895273=-.0069405967*t+3.280911216
-3.280911216 -3.280911216
getting this
-.883015943/(-.0069405967)=-.0069405967/(-.0069405967)*t
equaling
127.2247879=t
so to check it i divide 127 ish by 30 to see how many halfs there are (4.240826263) then multiply that by 5 (21.20413132) then I add 83 to see if it gets back to 98.6 since that was the original, and end up with 104.2041313.... sooo im completely lost again

5. Oct 22, 2009

### mgb_phys

ln|y-T|= kt + C
Is the equation of a straight line ( eg y = mx + c) on the graph I suggested you plot.
The Y axis is the log of the temperature difference and the X axis is time.

You know 2 points,
y=ln(11), x=8 (when the body is 11F above background and 8 mins after 9:00)
y=ln(6) x=38 (30mins later)

You are trying to solve y = mx + c
ln(11) = 8m + C
ln(6) = 38m + C

In order to find the x (ie time) at which y = ln(98.6-72).
Note we took time=0 to be 9:00am so a negtive answer is minutes before 9:00am

Last edited: Oct 22, 2009
6. Oct 22, 2009

### bsodmike

08:25 am?

7. Oct 22, 2009

### mgb_phys

Sounds roughly correct - a quick graph plot would check

8. Oct 22, 2009

### bsodmike

You can see a quick plot here,
http://www.wolframalpha.com/input/?i=98.6%3D72%2B11e^%28-0.0202t%29

The equation I finally obtained was,

$T(t)=72+11e^{-0.0202t}$

The solution is obtained by solving for $t$, such that,

$98.6=72+11e^{-0.0202t}\rightarrow ln(\dfrac{11}{26.6})=0.0202t$

$t = -43.7$

Since I took T(0) = 83 °F, this meant my time was based at 09:08 am. Hence subtracting 43 minutes takes me to ~08:25 am.

Last edited: Oct 22, 2009
9. Oct 22, 2009

### Staff: Mentor

This problem was also cross-posted to the Precalculus Math section under Homework and Coursework questions.

pezgoon, that's not a good thing to do.

10. Oct 22, 2009

### bsodmike

That is certainly not a good thing to do. Reading through his responses I'll hold back the complete solution as I believe a bit more effort will easily yield the result. Rather solving the linear set of equations I found it far easier to approach from the solution for the differential solution itself. At least there is more than one way of going about it.

11. Oct 22, 2009

### bsodmike

@Mark: I just tried this out. When I took the linear equations solution method I obtained,

$t=\dfrac{ln\left(\dfrac{11}{26.6}\right)\cdot 30}{ln\left(\dfrac{\left(\dfrac{6}{26.6}\right)}{\left(\dfrac{11}{26.6}\right)}\right)}=\dfrac{ln\left(\dfrac{11}{26.6}\right)\cdot 30}{ln\left(\dfrac{6}{11}\right)}=43.7$

I've [STRIKE]probably[/STRIKE] not made a mistake [STRIKE]as k needs to be negative[/STRIKE]; for some reason when I worked through the example, t, came out as positive.

Why?!?!? Am pulling my hair out over this one hah :p

Update: argh...same result!

$\begin{eqnarray}\begin{split} ln(6)&=\left[\left(\dfrac{1}{t}\right)ln\left(\dfrac{11}{26.6}\right)\right](t+30)+ln\left(26.6\right)\\ ln(6)&=ln\left(\dfrac{11}{26.6}\right)+\left(\dfrac{30}{t}\right)ln\left(\dfrac{11}{26.6}\right)+ln\left(26.6\right)\\ ln\left(\dfrac{6}{26.6}\right)&=ln\left(\dfrac{11}{26.6}\right)+\left(\dfrac{30}{t}\right)ln\left(\dfrac{11}{26.6}\right)\\ ln\left(\dfrac{\left(\dfrac{6}{26.6}\right)}{\left(\dfrac{11}{26.6}\right)}\right)&=ln\left(\dfrac{6}{11}\right)=\left(\dfrac{30}{t}\right)ln\left(\dfrac{11}{26.6}\right)\\ t&=\dfrac{ln\left(\dfrac{11}{26.6}\right)\cdot 30}{ln\left(\dfrac{6}{11}\right)}=43.7 \label{eq:} \end{split}\end{eqnarray}$

Well here's an even odder thing...

$k=\left(\dfrac{1}{t}\right)ln\left(\dfrac{11}{26.6}\right)$

Now try,

$k=\left(\dfrac{1}{43.7}\right)ln\left(\dfrac{11}{26.6}\right) = -0.0202$

So, k is correct with t = (+ve) 43.7 s. But why positive ?!?!

Solution...