CSI MATH LAB 1. A detective is called to the scene of a crime where a dead body
ID: 3018880 • Letter: C
Question
CSI MATH LAB
1. A detective is called to the scene of a crime where a dead body has just been found. She arrives on the scene at 10:23 pm and begins her investigation. Immediately, the temperature of the body is taken and is found to be 80 F. The detective checks the programmable thermostat and finds that the room has been kept at a constant 68 F. After evidence from the crime scene is collected, the temperature of the body is taken once more and found to be 78.5 F. This last temperature reading was taken exactly one hour after the first one. The next day the detective is asked by another investigator,“What time did our victim die?” Assuming that the victim’s body temperature was normal (98.6 F) prior to death, what is her answer to this question? Newton’s Law of Cooling can be used to determine a victim’s time of death.
Newton’s Law of Cooling describes the cooling of a warmer object to the cooler temperature of the environment. Specifically we write this law as,
u(t) = T + (u0 T )e^kt, k < 0
where u(t) is the temperature of the object at time t, T is the constant tem- perature of the environment, u0 is the initial temperature of the object, and k is a constant that depends on the material properties of the object. In particular, we know the investigator arrived on the scene at 10:23 ( t hours after death), the temperature of the body was found to be 80 F. One hour later, t+1 hours after death, the body was found to be 78.5 F. Our known constants for this problem are, T= 68 F and u0= 98.6 F.
Explanation / Answer
Let person dies at time x
we know the investigator arrived on the scene at 10:23 ( t hours after death), the temperature of the body was found to be 80 F
=> u(t) = T + (u0 T )e^kt
=> 80 = 68 + ( 98.6 - 68 )ekt
=> 80-68= 30.6ekt
=> 12 / 30.6 = ekt -------- 1st equation
One hour later, t+1 hours after death, the body was found to be 78.5 F.
=> u(t) = T + (u0 T )e^kt
=> 78.5 = 68 + ( 98.6 - 68 )ekt+k
=> 78.5-68= 30.6ekt+k
=> 10.5 / 30.6 = ekt+k --------- 2nd equation
dividing 2nd by 1st equation
10.5/12 = ek
=> k = ln(0.875)
=> k= - 0.13353
Using value of k in 1st equation
12/30.6 = e-0.13353t
=> ln(0.39215)= -0.13353t
=> -0.93611= -0.13353t
=>t =7.01 hours
Detective arrived 7 hours after the oerson died at 10:23 pm
=> Person died at time = 10:23 pm - 7 = 3:23 pm