For obvious reasons, the dissecting-room of a certain coroner is kept very cool

Question

For obvious reasons, the dissecting-room of a certain coroner is kept very cool at a constant temperature of 5 degree C (= 41 degree F). Early one morning, while doing an autopsy on a murder victim, the coroner himself is killed and the victim's body is stolen. At 10 a.m. the coroner's assistant discovers his chiefs body and finds its temperature to be 23 degree C, and at noon the body's temperature is down to 18.5 degree C. Assuming the coroner had a normal temperature of 37 degree C (= 98.6 degree F) when he was alive, when was he murdered

Explanation / Answer

here we will apply Newton's Law of Cooling:

T(t) = T + (T - T)e^(-kt)

Where:
T(t) = temperature at time "t"
t = hours past base time
T = temperature at base time
T = temperature of room
k = constant
given-
at 10AM, the temp is 23 c and at 12PM, the temp is 18.5. c

So we take To = 23, T(t)= 18.5 after 2 hours.solve for k knowing the room temperature is 5c

T(t) = T + (T - T)e^(-kt)
18.5 = 5 + (23 - 5)e^(-2k)
18.5 = 5 + 18e^(-2k)
13.5 = 18e^(-2k)
0.75 = e^(-2k)

take Natural log,

ln(0.75) = -2k
k = - ln(0.75) / 2

Now use same constant to solve for "t" using the T0 =37c T(t)= 23 c.we will get t.

T(t) = T + (T - T)e^(-kt)
23 = 5 + (37 - 5)e^(ln(0.75)t/2)
0.5625 = e^(ln(0.75)t/2)
Natural log of both sides:
ln(0.5625) = ln(0.75)t / 2
solve for t:
2 ln(0.5625) = t ln(0.75)
t = 2 ln(0.5625) / ln(0.75)
t = 4

, the time of death was 10-4 =6AM

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