A body was found at midnight and it was 80 degrees. 2 hours later, it was 75 degrees. The room is 60 degrees. What time did the body die

Question

I know to use the cooling formula, but i dont know how to solve

Answer 1

Lauren,

 

Let   t = time in hours since the body died.  And t = t_m is hours after death at midnight.

   T(t) = The body temperature as a function of time after death (in degrees F)

      T_o = Body temperature at instant of death or at time zero (= 98.6^oF)

      T_s = Constant temperature of the surroundings (the morgue) = 60^oF

       K = A constant in the exponential decay of the body's temperature (to be determined)

 

Then Newton's formula is :  T(t) = T_s + (T_o - T_s) e ^(-Kt)  =  60 + (98.6 - 60) e ^(-Kt)    = 60 + 38.6 e ^(-Kt) .

 

Apply the known data:

 1.   T(t_m) = 80 = 60 + 38.6 e ^(-Kt_m)  which simplified becomes  e ^(-Kt_m)  = 20/38.6 = 0.51813,

             so -Kt_m = ln_e (0.51813) = - 0.65752  or     Kt_m = 0.65752.

 2.   T(t_m + 2) = 75 = 60 + 38.6 e ^{(-K[t_m + 2])},  which simplified becomes  e ^{(-K[t_m + 2])} = 15/38.6 = 0.38860,

             so (-K[t_m + 2]) = ln_e (0.38860) = - 0.94520  or K[t_m + 2] = 0.94520  

 

Two equations with two unknowns, so solve for K and t_m.

Rewrite the second one as  2K = 0.94520 - Ktm = 0.94520 - 0.65752 = 0.28768 , K = 0.14384

And from the first equation  t_m = 0.65752/K = 0.65752/0.14384 = 4.57 hours

 

By Newton's model of temperature loss, the body died about 4.57 hours BEFORE midnight since by midnight its temperature had already dropped from a living person at 98.6^oF to 80^oF.

 

NOTE: In reality, the body was not lying in the morgue with its ambient temperature 60, but was found in some other environment where it had lay for 4 or 5 hours (where K would have a different value). Then it was brought to the morgue and measured at 80.  So our answer is really an imprecise estimate, and one would probably say something like the time of death was between 4 & 6 hours before midnight.