Solution: Sherlock Holmes

Apply Newton’s Law of Cooling:
dT/dt = k(T-T_s)
T_s =21 (Room Temperature)
dT/(T-T_s)= k.dt
\int \frac {dT}{T-T_s}=\int k.\mathrm{d}t
ln (T-T_s)=kt+C
e^{kt+C}= T-T_s
T= Ae^{kt} +T_s
A=e^{C}
T=Ae^{kt}+21

At t=0, (normal body temperature)
T_o= A + 21 =37
=> A=16

Let t =x hour 1st temperature taken
T(x)= 16e^{kx}+21=29
16e^{kx} = 8
e^{kx}= \frac {1}{2} …[1]

t = x+1 hour later

T(x+1) =16e^{k(x+1)} + 21 = 27
e^{k(x+1)} = 3/8
e^{kx}e^{k}= 3/8
e^{k} = 3/4
k = ln(3/4)

[1]: kx = ln(1/2)
x = ln(1/2) / ln(3/4)
x=2.41 hr = 2 hr 25m
Murder Time= 2am -2h25m =11:35 p.m. [QED]

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