In a murder investigation, the temperature of a corpse was 32.5 degrees C at 1:30 pm and 30.3 degrees C an hour later. Normal body temperature is 37.0 degrees C and the temperature of the surroundings was 20.0 degrees C. When did the murder take place?
Accepted Solution
A:
Let [tex]T(t)[/tex] be the temperature of the body [tex]t[/tex] hours after 1:30 PM. Then [tex]T(0) = 32.5[/tex] and [tex]T(1) = 30.3[/tex].
Using Newton's Law of Cooling, [tex]\dfrac{dT}{dt} = k(T - T_s)[/tex], we have [tex]\dfrac{dT}{dt} = k(T-20)[/tex]. Now let [tex]y = T - 20[/tex], so [tex]y(0) = T(0) - 20 = 32.5 - 20 = 12.5[/tex], so [tex]y[/tex] is a solution to the initial value problem [tex]dy/dt = ky[/tex] with [tex]y(0) = 12.5[/tex]. By separating and integrating, we have [tex]y(t) = y(0)e^{kt} = 12.5e^{kt}[/tex].