Answer:
The person was murdered at 11.40 PM
Step-by-step explanation:
Given information :
Initial body temperature, [tex]T_{0}[/tex] = 90[tex]^{o}F[/tex]
Police arrive at a murder scene at 3 : 30 AM, at 5.30 AM (120 minutes), temperature of the body had dropped to 85 [tex]^{o}F[/tex].
T(120) = 85 [tex]^{o}F[/tex]
The temperature of the crime scene, C = 70[tex]^{o}F[/tex]
to determine the time of the murderer, we can use the Newton's lau of cooling:
T(t) = C + ([tex]T_{0}[/tex] - C) [tex]e^{-kt}[/tex]
where
T(t) = temperature at any given time
C = surrounding temperature
[tex]T_{0}[/tex] = initial temperature of heated object
k = cooling constant
thus
T(t) = C + ([tex]T_{0}[/tex] - C) [tex]e^{-kt}[/tex]
T(120) = 70 + (90 - 70) [tex]e^{-120k}[/tex]
85 = 70 + 20 [tex]e^{-120k}[/tex]
[tex]e^{-120k}[/tex] = (85-20) / 70
-120 k = ln (65/70)
k = - ln (65/70) / 120
= 0.00062
so we have the cooling equation
T(t) = 70 + ([tex]T_{0}[/tex] - 70) [tex]e^{-0.0062t}[/tex]
The temperature at the time of the death is 98.6[tex]^{o}F[/tex] which was the initial temperature of the person murdered, the temperature dropped to T(t) = 85 [tex]^{o}F[/tex] , now we can find the time of him being murdered.
T(t) = 70 + (98.6 - 70) [tex]e^{-0.0062t}[/tex]
85 = 70 + 28.6 [tex]e^{-0.0062t}[/tex]
[tex]e^{-0.0062t}[/tex] = (85- 28.6)/70
-0.00062t = ln ((85- 28.6)/70)
t = - ln ((85- 28.6)/70)/ 0.00062
t = 350 minutes
therefore, the preson being murdered 350 minutes (5 hours 50 minutes) before 5.30, which was 11. 40 PM
