Determination of the Time of Death
My name is M. Rizki Maulana R. usually called Rizki and number ID 07305141019. I start to collect idea at Saturday January 8, 2009. My partner is Dwi Bagus Bramantiyo and his number ID is 07305141028. We are begin activity in my room, Karangmalang A5, January 11, 2009 at 10 o’clock and finish at 13 o’clock.
I’m raise topic “Determination of the Time of Death”. Why did I choose that? Because, I was inspirited by study about Differential Equation. Profit from that topic is we can know to determine time of death in investigation of homicide or accidental death with mathematical way.
From experimental observation it is known that, to an accuracy satisfactory in many circumstances, the surface temperature of an object change at rate proportional to the difference between the temperature of the object and that of the surrounding environment. This is known as Newton’s Law of cooling. Thus, if θ(t) is the temperature of the object at time t, and T is the constant ambient temperature, then θ must satisfy the linear differential equation
dθ/dt = -k(θ–T)................(1)
where k>0 is a constant of proportionality. The minus sign in Eq.(1) is due to the fact that if the object is warmer than its surrounding, (θ–T), then it will become cooler with time. Thus dθ/dt<0. t="0" size="1">0. We assume that at the time of death td the body temperature θd had the normal value of 98.6oF or 37oC. If we assume that Eq.(1) is valid in this situation, then our task is to determine td.
The solution of Eq.(1) subject to the initial condition θ(0) = θ0 is
lanjut bacanya.....
θ(t) = T + (θ0 – T)e^(-kt)............................. (2)
However, the cooling rate k that appears in expression is yet unknown. We can determine k by making a second measurement of the body’s temperature at some later time t1, suppose that θ=θ1 when t= t1. By substituting the value in Eq.(2) we find that
θ1 - T = (θ0 – T)e^(-kt)
hence
k = -(1/t1)ln[(θ1 – T)/ (θ0 – T)]....................................... (3)
where θ0 , θ1, T and t0 are known quantities.
Finally, to determine td we substitute t = td and θ = θd in Eq.(2) and then solve for td. We obtain
td= -(1/k) ln [(θd – T)/ (θ0 – T)]......................................... (4)
In this process, I don’t have any troubles to explain that. But, I have difficulties to spell my English well. My partner could understand what I had explained. To make sure that he has understood, I gave him question like this :
For example, suppose that the temperature of the corpse is 85oF when discovered and 74oF two hours later, and that the ambient temperature is 68oF.
And he could do it well. His result is :
From Eq.(3)
k = -(1/2)ln[(74 – 68)/ (85 – 68)] = 0.5207/hr,
and from Eq.(4)
td = -(1/0.5207) ln [(98.6 – 68)/ (85 – 68)] ≈ - 1.129 hr.
Thus we conclude that the body was discovered approximately 1 hr, 8 min after death.
That was my experience of studying English 2 with the lecturer is Mr. Marsigit, MA. I’m feel so exciting.
