Question

In a murder investigation, the temperature of the corpse was 32.5 C at 1:30pm and 30.3 C an hour later. Normal body temperature is 37.0 C and the temperature of the surrounding was 20.0 C. When did the murder take place?


PLEASE SHOW STEP BY STEP

Answer 1

Are you sure this is calculus? You want exponential decay to room temp?
Or is it algebra and you want a linear function?

Answer 2

I am going to assume it is algebra first

take t = 0 at 1:30

T = Ti - k t
Ti = 32.5
so
T = 32.5 - k t
30.3 = 32.5 - k (1 hour)
k = 2.2
so
T = 32.5 - 2.2 t
37 = 32.5 - 2.2 t
t = - 2.04 call it 2 hours
so by that linear model 11:30 am
now I will work on the more realistic calculus version

Answer 3

rate of change of temp proportional to temp above room temp which is 20

to make it easy on the arithmetic define

T' = real T - 20

dT'/dt = k (T')

dT/T' = k dt

ln T' = k t

T' = C e^kt
let's call t = 0 at 1:30
T = 32.5 so T' = 12.5
12.5 = C e^0 = C
so
T' = 12.5 e^kt
now when t = 1 hour T = 30.3 so T' = 10.3
10.3 = 12.5 e^k
ln .824 = k
k = - .194
so
T' = 12.5 e^-.194 t
so what was t when T= 37 so T'= 17 ?

17 = 12.5 e^-.194 t
1.36 = e^-.194 t
.307 = -.194 t
t = - 1.87 hours
so 1.87 hours before 1:30 pm
11:38 am

Answer 4

According to Newton's Law of Cooling
T(t) = roomtemp + (37 - 20)e^(-kt) , where t is the time in hours and k is a constant

so we get two equations:
32.5 = 20 + 17e^(-kt) ---> 12.5 = 17e^(-kt)

and
30.3 = 20 + 17e^(-k(t+1)) ---> 10.3 = 17e^(-kt - k)

divide them:
115/103 = e^k , remember to divide powers of the same base, you keep the base and subtract the exponents.
k = ln115 - ln103

then in 12.5 =17e^(-kt)
125/170 = e^(-kt)
-kt = ln125 - ln170
t = (ln125 - ln170)/-k
= (ln125 - ln170)/(ln103 - ln115) , .... only now will I do any actual calculation
= 2.79 hrs

so death occurred 2.79 hrs or 2 hrs and 47.4 minutes before 1:30 pm

I will leave it up to you to figure out that time.

Answer 5

One of us made an arithmetic mistake. It is up to Tanya to get it right :)