Answer: Around 0:35 Pm or 12:35 Am
Explanation:
The equation that describes the cooling of objects can be written as:
T(t) = Ta + (Ti - Ta)*e^(k*t)
Where Ta is the ambient temperature, here Ta = 25°C.
Ti is the initial temperature of the body, we have Ti = 37°C.
t is the time.
k is a constant.
So our equation is:
T(t) = 25°C +12°C*e^(k*t)
at 2pm, the temperature was 33°C
at 3pm, the temperature was 31°C.
we want to find the hour where we have our t = 0, suppose this hour is X.
then we can write our times as:
2pm ---> 2 - X
3pm ----> 3 - X
and our equations are:
33°C = 25°C + 12°C*e^(k2 - k*X)
31° = 25°C + 12°C*e^(k3 - k*X)
So we have two equations and two variables, let's solve the system.
first, simplify it a bit, for the first eq:
33 - 25 = 12*e^(k2 - k*X)
8/12 = e^(k2 - k*X)
ln(8/12) = k*2 - k*X
for the second equation we have:
31 - 25 = 12*e^(k3 - k*X)
6/12 = e^(k3 - k*X)
ln(6/12) = k*3 - k*X
So our equations are:
1) ln(2/3) = 2*k - X*k
2) ln(1/2) = 3*k - X*k
First, let's isolate one of the variables in one of the equations. let's isolate k in the first equation.
ln(2/3)/(2-X) = k
now we can replace it in the second equation:
ln(1/2) = 3*ln(2/3)/(2 - X) - X*ln(2/3)/(2-X)
now let's solve it for X, i will take a = ln(1/2) and b = ln(2/3) so it is easier to read.
a = 3*b/(2 - X) - X*b/(2 - X)
a*(2 - X) = 3*b - X*b
2a - aX = 3b - Xb
X(a - b) = 2a - 3b
X = (2*ln(1/2) - 3*ln(2/3))/(ln(1/2) - ln(2/3)) = 0.590
now, knowing that one hour has 60 minutes, then this is:
0.59*60m = 35 minutes
So the hour of death is 0:35 Pm or 12:35 Am
