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.

1 Answer
Aug 9, 2016

Answer:

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