Newton's Law of Cooling


by maccaman
Tags: cooling, newton
maccaman
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#1
Aug7-04, 07:33 AM
P: 49
Im having a bit of trouble using Newton's Law of cooling. I have been given the formula [tex]T(t)=T_s + (T_0 - T_s) e^{-kt} [/tex]. Im trying to find the time of death of a body, and im given the time the body was found 12.00pm, its temp around 90, 2nd check, at around 1.00pm, temp is 80, and the surrounding temp is 75. How would i find the time of death, assuming that the body when it died was 100.

Note these values are prolly inacurate, im just wanting to know how to use it to find the time of death. Thanks.
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arildno
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#2
Aug7-04, 08:50 AM
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1.you already know the value [tex]T_{s}=75[/tex]
2. Let t=0 be at 12.00 pm.
Hence, you have T(0)=90, which detemines in your formula [tex]T_{0}=90[/tex]
3. At t=1, the body temperature T(1), has decreased to 80;
hence, you may determine k by the relation:
[tex]e^{-k}=\frac{T(1)-T_{s}}{T_{0}-T_{s}}[/tex]
4.
Here, you are given T(t*)=100, where t* is the time of death, measured in hours relative to 12.00 pm (t=0)
Plug and chug and solve for t* (it will be given as a negative value, since it happened prior to 12.00pm
da_willem
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#3
Aug7-04, 09:18 AM
P: 603
Quote Quote by maccaman
Im having a bit of trouble using Newton's Law of cooling. I have been given the formula [tex]T(t)=T_s + (T_0 - T_s) e^{-kt} [/tex]. Im trying to find the time of death of a body, and im given the time the body was found 12.00pm, its temp around 90, 2nd check, at around 1.00pm, temp is 80, and the surrounding temp is 75. How would i find the time of death, assuming that the body when it died was 100.

Note these values are prolly inacurate, im just wanting to know how to use it to find the time of death. Thanks.
If you set your clock at zero at the time of death and call 12pm t0 and 1pm t1 then you get two equations:

T(t0)=Ts+(T0-Ts)e^(-kt0)
T(t1)=Ts+(T0-Ts)e^(-kt1)
=Ts+(T0-Ts)e^(-k(t0+1)) t1=t0+1 (in hours!)
=Ts+(T0-Ts)e^(-k(t0))e^(-k)

so: 90 = 75 + (100-75)e^(-k*t0)
and:80 = 75 + (100-75)e^(-k*t0)e^(-k)

These are two equations with two unknowns (k and t0), so can be solved. t0 is the time of death in hours before 12pm.

Nenad
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#4
Aug7-04, 12:24 PM
P: 698

Newton's Law of Cooling


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

[tex]90=75 + (100 - 75) e^{-kt} [/tex]

[tex]15=25e^{-kt} [/tex]

[tex]ln0.6=-kt [/tex]

[tex]\frac{ln0.6}{t}=-k [/tex]

now sub this into the second equation

[tex]80=75 + (90 - 75) e^{-kt} [/tex]

[tex]5=15e^{ln0.6/t \ [60min]} [/tex]

[tex]ln\frac{5}{15}=ln0.6/t \ [60min] [/tex]

[tex]t = 28min [/tex]
Therefore the body dies 28 minutes befoure 12, or at 11:32a.m.
Nenad
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#5
Aug7-04, 12:38 PM
P: 698
sorry about my notation on the second part, Im not shure how to use this latex stuff yet.
Ruthie
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#6
Jul2-09, 04:30 PM
P: 1
I have a question that needs an answer asap!!!!

A body is found at 6.00am. . . .body temperature 25 celsius.. . .30minutes later. .body temperature is 22 celsius. Surrounding temperature is 15 celsius and normal body is 37 celsius. Estimate time of death.
ideasrule
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#7
Jul3-09, 02:27 AM
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Quote Quote by Ruthie View Post
I have a question that needs an answer asap!!!!

A body is found at 6.00am. . . .body temperature 25 celsius.. . .30minutes later. .body temperature is 22 celsius. Surrounding temperature is 15 celsius and normal body is 37 celsius. Estimate time of death.
The above posts have all the equations you need to solve this problem. Please first explain what you don't understand; then we'll know how to help.
Nusc
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#8
Nov29-09, 03:29 PM
P: 781
Quote Quote by Ruthie View Post
I have a question that needs an answer asap!!!!

A body is found at 6.00am. . . .body temperature 25 celsius.. . .30minutes later. .body temperature is 22 celsius. Surrounding temperature is 15 celsius and normal body is 37 celsius. Estimate time of death.


How do you do that problem without knowing the temperature at 6:00 AM?
angel101
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#9
Apr1-10, 09:21 PM
P: 1
what is 'e' in that formula?


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