Question

body was found at 10 a.m. outdoors on a day when the temperature was 40oF. The medical examiner found the temperature of the body to be 80oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.

Answer 1

have no idea what i have to do...

Answer 2

soooooo the thing i dont understand about this problems would be what is what? like what Ta / To / T(t) ...

Answer 3

i think knowing what to input in the equation would lead me to solve the equation

Answer 4

@jim_thompson5910 @Hero @mathslover @mathmale @zepdrix Helloon_n i was wondering if you can help be verify the variables on the problem

Answer 5

this is what i believe - Ta is 40 and To is 80 BUT then again whats T(t)

Answer 6

ohohohohoh i forgot to mention even if i have what each of them are i still dont get what numbers they are

T(t)=temp of the obj at time t
To= initial time
Ta = ambient temp
k= positive constant

Answer 7

I'd suggest you look up "Newton's Law of Cooling" for some references. Here's one result I obtained by Googling that:
http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

I'd suggest you think of T as the temperature of the body, and \[T _{a}\] as the ambient temperature. After all, your goal here is to find the time at which this person died, given that the ambient temp is 40 deg F and the temp of the body when found was 80 deg F (down from about 98.6 deg F).

Answer 8

http://www.ugrad.math.ubc.ca/coursedoc/math100/notes/diffeqs/cool.html

Answer 9

they tell you what \(k\) is so you do not have to find it

Answer 10

Once you have a formula for T as a function of t, let T=80 and Ta=40. This leaves only time t to solve for.

Answer 11

Newton's Law of Cooling can be calculated based on \(\theta(t\)) using


\(\dfrac{d\theta}{dt} = -k(\theta - S)\)

where \(\theta\) is the intial temp \(S\) is the body temp and \(t\) is the time. k is a given constant.

Answer 12

the first question is, What is the rate? it will help us define the answer, when the person died.

Answer 13

okkkkkkkkk sooooo.... the equation would look like this ??? \[T(t)= 40+ (80-40) e ^{-.1947t}\]

Answer 14

Actually, the rate of change of the temperature depends upon the difference between the ambient temperature and the temperature of the object in question. This can be represented by a differential equation, which can then, in turn, be solved. Hero has presented one workable form (of several) of this d. e. I propose that your first step is to solve this separable d. e.

Answer 15

close \[T(t)= 40+ (98.6-40) e ^{-.1947t}\]

Answer 16

I think more important is HOW one goes about obtaining such a formula for the object temperature, T(t).

Answer 17

actually i can solve the equation with e because we turn it into ln eventually ...

Answer 18

i will hazard a guess that this is not a difeq class, just exponentials

Answer 19

for example this as a previous problem we solve ...

Answer 20

@satellite73 where did 98.6 come from ??? and why use that and not 80?