Newtons cooling law problem given in precalc

[pezgoon]

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

Homework Statement

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.

Homework 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 can't remember at this point) is

y-T=e^(kt + C)

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 into try and get the temp at the time comes out wrong. Can someone please help?? Thanks to whomever can

 

[Mark44]

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

Homework Statement

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.

Homework 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 can't remember at this point) is

y-T=e^(kt + C)

The two equations above are equivalent. IOW, they say exactly the same thing.

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 into try and get the temp at the time comes out wrong. Can someone please help?? Thanks to whomever can

Show us what you've tried. Assume that the body was found at time t = X, where X represents the number of minutes after the person was murdered. You know the temperature of the body at three different times: t = 0 (98.6 deg. F), t = M + 8 (83 deg. F), and t = M + 30 (78 deg. F). Use these times and temperatures in your equation to find k, C, and M.

 

[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 don't 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 that's 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 I am completely lost again

 

[Mark44]

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

OK, this is what I get.

so then since I don't 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 that's constant and figure 30 minutes after he died the temp would be at 93.6 so i did:

No, that's not right. You're assuming that the temperature is going to drop at a constant rate. It's not. The temperature drop is rapidest at the time when the body's temperature is largest in comparison with the ambient room temperature. In the half-hour between 9:08 and 9:38 the body's temperature dropped 5 degrees. In the half-hour right after death, the temperature drop would have been greater than this.

Because of this assumption, all your work from here on is incorrect.

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 I am completely lost again

You got C. Let's see if you can work through the rest of the problem. I'll set it up slightly different than I set it up before.

This time, let M = number of minutes between death and time 9:08
Then M + 30 = number of minutes between death and time 9:38

At 9:08, the body temp was 83 degrees, so
ln(83 - 72) = kM + C

At 9:38, the body temp was 78 degrees, so
ln(78 - 72) = k(M + 30) + C

You have C, and you now have two equations in two unknowns. Can you solve for k and M?

I get that he was killed sometime between 6am and 7am.

 

[pezgoon]

ok well see through me trying to do all of this this was the problem, i don't know how to solve for both variables, i feel like itll end up being this after subtracting c from ln(83-72)
-.883015943= k
-----------
m
or

-.883015943= m
-----------
k

(m and k are being divided out)

and if i tried the second temp itd be something similar and i have no idea how to go any further than that

 

[Mark44]

Once you have solved for k in one equation (with M on the other side), substitute for k in the other equation.

 

[pezgoon]

do I just use k=(-.883015943)/m ?

i tried doing that and this is what happened
ln(78-72)=((-.883015943)/m)*(m+30)+C

so it becomes

1.791759469=-.883015943+30+c ?

or would it be

1.791759469=(-.883015943)*(m+30)+C

or would the 30 multiply to the -.883015943?

like i can never remember when variables cancel does it just become 1 or (in this case)
1m?
doing this it turns out as either

35.66090963=m

by subtracting c+30 first then dividing by -.8830... or

-35.31004725=m
by dividing -.8830... first then subtracting C+30.

but neither of these seem right at all

after 7 hours and 30 or so minutes of working on this problem (total) this is just getting rediculous

 

[Mark44]

do I just use k=(-.883015943)/m ?

Yes.

i tried doing that and this is what happened
ln(78-72)=((-.883015943)/m)*(m+30)+C

Yes, this is fine.

so it becomes

1.791759469=-.883015943+30+c ?

No. What happened to m?

or would it be

1.791759469=(-.883015943)*(m+30)+C

No, it's not that one, either. This time you lost one of the m's.

or would the 30 multiply to the -.883015943?

like i can never remember when variables cancel does it just become 1 or (in this case)
1m?

If you can't remember when variables cancel, then you should NEVER cancel them, at least not until you understand what's really happening with this operation.

Cancelling means removing factors that are equal to 1, as in this example.
$$\frac{Ma + Mb}{M}~=~\frac{M(a + b)}{M}~=~\frac{M}{M}\frac{a + b}{1}~=~ a + b$$
But cancellation cannot be done in the following example.
$$\frac{Ma + b}{M}~\neq~a + b$$
M is not a factor of the numerator.

doing this it turns out as either

35.66090963=m

by subtracting c+30 first then dividing by -.8830... or

-35.31004725=m
by dividing -.8830... first then subtracting C+30.

but neither of these seem right at all

after 7 hours and 30 or so minutes of working on this problem (total) this is just getting rediculous

 

[bsodmike]

For those looking, the solution has been shown here:
https://www.physicsforums.com/showthread.php?p=2405613#post2405613