At 7am one morning detectives find a murder victim in a closed meat locker. The temperature of the victim measures 88degrees. Assume the meat locker is always kept at 40 degrees, and at the time of death the victim's temperature was 98.6 degrees. When the body is finally removed at 8am, its temperature is 86 degrees.
a) when did the murder occur?
b) How big an error in the time of death would result if the live body temperature was known only to be between 98.2 degrees and 101.4 degrees?
Please, help

Question

At 7am one morning detectives find a murder victim in a closed meat locker. The temperature of the victim measures 88degrees. Assume the meat locker is always kept at 40 degrees, and at the time of death the victim's temperature was 98.6 degrees. When the body is finally removed at 8am, its temperature is 86 degrees.
a) when did the murder occur?
b) How big an error in the time of death would result if the live body temperature was known only to be between 98.2 degrees and 101.4 degrees?
Please, help

loser66 · Answer

@dan815

freckles · Answer

i'm thinking we will use newton's law of cooling

loser66 · Answer

Yes, it is. But I got negative value which is impossible. That is why I posted it here.

freckles · Answer

for question A?

loser66 · Answer

Let 7:00am be the initial time, we set T(0) = 88.
And the Newton's law gives us $T(t) = 40 -\alpha e^{-kt}$
I put t =0 to solve for alpha
After 1hour (8:00 am) , T(60) = 86 to solve for k
Hence, I got $T(t) = 40 +48e^{-0.000709t}$ 
But, if I calculate the time T(t) = 98.6, I got t = -281.... :(

freckles · Answer

well -281 would just mean 281 mins before the initial time 7am

loser66 · Answer

No, it is not that since they remove the body at 8:00, So it is impossible its death happened after 8

freckles · Answer

what?

freckles · Answer

281 min before 7am is definitely before 7am or 8am

loser66 · Answer

And if it is before 7:00, how can the degree increase from 88 to 98.6?

freckles · Answer

I'm going to do the problem...
but from your answer it looks like you have 281 mins before 7am which is ...2:19 am

freckles · Answer

give me a sec to check your other work

freckles · Answer

$$T(t) \ \text{ where } t \text{ is \in minutes } \ T(0)=T_0=88^\circ \text{ ---this is at } 7am \ T(60)=T_{60}=86^o \text{--- 60 minutes later which is } 8am \ \ T_a =40^\circ \text{ ---constant temp of the refrigerator } \ \frac{dT}{dt}=-k(T-40^ \circ) \ \frac{dT}{T-40^\circ}=-k dt \ \ln|T-40^\circ|=-kt+C \ T-40^\circ=a e^{-kt} \ T=40^\circ+ae^{-kt}  \ T_0=40^\circ+a=88^\circ \implies a= 88^\circ-40^\circ=48^ \circ \ \text{ so now we have } \ T=40^\circ+48^\circ e^{-k t} \ T_{60}=40^\circ+48^\circ e^{-k \cdot 60} =86^\circ  \implies \ e^{- k 60}=\frac{46}{48} \ -k 60 =\ln(\frac{46}{48}) \ k=\frac{-1}{60} \ln(\frac{46}{48}) \ k=0.00070933$$
so far all of your stuff is right up to this point

freckles · Answer

$$T=40^\circ+48^ \circ e^{-0.00070933 t}$$
now we are given T at the time of death and asked to find t

freckles · Answer

$$98.6^\circ=40^\circ+48^\circ e^{-0.00070933 t}$$
solving this will give us the answer in terms of the initial time being 7am
so if the t is negative that means it happened before 7am

freckles · Answer

which would make sense because how can the death happen after them finding the body :p

freckles · Answer

and I also get t is approximately -281 when T is 98.6 

which means 281 minutes before 7am is when the death happened

loser66 · Answer

I am thinking of the case $a =\pm e^C$

freckles · Answer

$$281 \text{ minutes }=4 \text{ hours } \text{ and } 41 \text{ minutes } \ 7 am -(4 \text{ hours  and }  41 \text{ minutes })=3 am -(41 \text{minutes }) \ =(2 \text{ hours} \text{ and } 60 \text{ minutes } ) -(41 \text{ minutes } ) \ =2 \text{ hours } \text{ and } (60 \text{ minutes } -41 \text{ minutes }) \ =2 \text{ hours } \text{ and } 19 \text{ minutes }  \ =2:19 am$$

loser66 · Answer

We already worked on a = e^C, (a>0), but it still has the case a<0 since |T-40| has 2 values of the right hand side.

loser66 · Answer

I meant from $ln|T-40|= -kt+C$, we have $|T-40|= e^{-kt +C}=e^Ce^{-kt}$\) 
Hence if we let a = e^C, then a must be $a =\pm e^C$

freckles · Answer

which question are you working on?

freckles · Answer

e^c is positive

loser66 · Answer

a) I am still doubt when the result is -281

freckles · Answer

why?

freckles · Answer

the negative just means before the initial time which was 7am

freckles · Answer

-281= 281 minutes before 7am=2:19am

loser66 · Answer

because the Newton's of cooling applied to heating also, right?

freckles · Answer

well the body is cooling over time not heating ...

freckles · Answer

so before 7am the body should be hotter

freckles · Answer

which it is

loser66 · Answer

oh, I got what you meant

freckles · Answer

http://www.sosmath.com/diffeq/first/application/newton/newton.html here is a very similar example

freckles · Answer

end of page

freckles · Answer

the - just means before the initial time you used

freckles · Answer

we should be able to use 8am as the initial time if you wanted
but 7am would have to be at t=-60

loser66 · Answer

:) 
I used their computer to see the answer, it gave me 2:22:36

freckles · Answer

2:19am is pretty close 
-281 was an approximated answered 
just like our value for k was too

loser66 · Answer

Ooooooooooooook
how about part b?

freckles · Answer

$$\text{ so } T=40^\circ+48^\circ e^{- .00070933t} \ \text{ sounds like it wants us to solve for } t \ \text{ given } 98.2^\circ<T<101.4^\circ$$...
maybe

freckles · Answer

and then once you solve for t
you can subtract -281 on all sides just to show how close to the actual time the numbers were

loser66 · Answer

Yes, I think so. If it is so, we just calculate t for the limits only and then subtract to the previous result, right?

loser66 · Answer

I got what you mean. Thanks for the help. :)

freckles · Answer

np
and yes you solve both T=98.2 and T=101.4 separately 
because we know this is a decreasing function (there are no weird increases to worry about)

loser66 · Answer

:)