Question

A body was found at 10 a.m. in a warehouse where the temperature was 40°F. The medical examiner found the temperature of the body to be 80°F. What was the approximate time of death?

Use Newton's law of cooling with k=0.1947.

Answer 1

Approximate time of death is `8:02:24` am.

Important to note that this is the skin temperature of the body. The medical examiner would be measuring internal temperature which would decrease much slower.

Explanation:

Newton's law of cooling states that the rate of change of temperature is proportional to the difference to the ambient temperature. Ie

`(dT)/(dt) prop T - T_0`

If `T > T_0` then the body should cool so the derivative should be negative, hence we insert the proportionality constant and arrive at

`(dT)/(dt) = -k(T - T_0)`

Multiplying out the bracket and shifting stuff about gets us:

`(dT)/(dt) + kT = kT_0`

Can now use the integrating factor method of solving ODEs.

`I(x) = e^(intkdt) = e^(kt)`

Multiply both sides by `I(x)` to get

`e^(kt)(dT)/(dt) + e^(kt)kT = e^(kt)kT_0`

Notice that by using the product rule we can rewrite the LHS, leaving:

`d/(dt)[Te^(kt)] = e^(kt)kT_0`

Integrate both sides wrt to `t`.

`Te^(kt) = kT_0 int e^(kt) dt`

`Te^(kt) = T_0e^(kt) + C`

Divide by `e^(kt)`

`T(t) = T_0 + Ce^(-kt)`

Average human body temperature is `98.6°"F"`.

`implies T(0) = 98.6`

`98.6 = 40 + Ce^0`

`implies C = 58.6`

Let `t_f` be the time at which body is found.

`T(t_f) = 80`

`80 = 40 + 58.6e^(-kt_f)`

`40/(58.6) = e^(-kt_f)`

`ln(40/(58.6)) = -kt_f`

`t_f =- ln(40/(58.6))/k`

`t_f = - ln(40/(58.6))/(0.1947)`

`t_f = 1.96 hr`

So from time of death, assuming body immediately started to cool, it took 1.96 hours to reach 80°F at which point it was found.

`1.96hr = 117.6min`

Approximate time of death is `8:02:24` am