Question

4.4 Applications: Decay Practice Exercises 4.4.5 - Forensics A coroner arrives at a murder scene at 2 A.M. He takes the temperature of the body and finds it to be 61.6º. He waits 1 hr, takes the temperature again, and finds it to be 57.2º. The body is in a freeer, where the temperature is 10°. When was the murder committed?

Answer 1

From Newton's law of cooling

dT/dt = k [ T(t) - Ta(t) ] [ where dT/dt is change of temperature, k is temperature decay constant, T(t) body temperature at examination time, Ta is environement temperature ]

solving the above equation we get

T(t) = Ta + (T₀ - Ta)e ^-kt_c [ where T0 initial body temperature which is 98.6 fahrenhite, t_c number of hours spent ]

so we can write

61.6 = 10 + ( 98.6 - 10 )e ^-kt_c  

61.6 = 10 + ( 88.6)e ^-kt_c

51.6 = ( 88.6)e ^-kt_c

e ^-kt_c = 51.6 / 88.6 ----------------- 1

57.2 = 10 + (98.6 - 10 )e ^-k(t_c+1)  

47.2 = 88.6x e ^-k(t_c+1)

e ^-k(t_c+1) = 47.2 / 88.6 --------------------2

So 1 divide by 2 :

e ^{-k ( t_c - t_c + 1)} = 51.6 / 47.2

e^-k = 1.09

log (1.09 ) = -k

-k = 0.0387

So for -k = 0.0387 in the equation 1

e^0.0387(t_c) = 51.6 / 88.6

log ( 51.6/88.6) = 0.0387t_c

t_c $\approx$ -6 hours

So the muder happened 2AM - 6 = 8pm approximately.