
#1
Aug704, 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. 



#2
Aug704, 08:50 AM

Sci Advisor
HW Helper
PF Gold
P: 12,016

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 



#3
Aug704, 09:18 AM

P: 603

T(t0)=Ts+(T0Ts)e^(kt0) T(t1)=Ts+(T0Ts)e^(kt1) =Ts+(T0Ts)e^(k(t0+1)) t1=t0+1 (in hours!) =Ts+(T0Ts)e^(k(t0))e^(k) so: 90 = 75 + (10075)e^(k*t0) and:80 = 75 + (10075)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. 



#4
Aug704, 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. 



#5
Aug704, 12:38 PM

P: 698

sorry about my notation on the second part, Im not shure how to use this latex stuff yet.




#6
Jul209, 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. 



#7
Jul309, 02:27 AM

HW Helper
P: 2,324





#8
Nov2909, 03:29 PM

P: 781

How do you do that problem without knowing the temperature at 6:00 AM? 



#9
Apr110, 09:21 PM

P: 1

what is 'e' in that formula?



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