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
Are you sure this is calculus? You want exponential decay to room temp?
Or is it algebra and you want a linear function?
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
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
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.
To determine when the murder took place, we can use Newton's Law of Cooling, which describes how the temperature of a body decreases or increases over time.
The formula for Newton's Law of Cooling is:
T(t) = Ts + (T0 - Ts) * e^(-kt)
Where:
- T(t) is the temperature of the body at time t.
- Ts is the temperature of the surrounding.
- T0 is the initial temperature of the body.
- k is the cooling constant.
- e is the base of the natural logarithm.
Let's break down the problem step by step:
Step 1: Identify the given information:
- Initial temperature (t = 0): T0 = 32.5 C
- Temperature after one hour (t = 1 hour): T(1) = 30.3 C
- Surrounding temperature: Ts = 20.0 C
- Normal body temperature: Ta = 37.0 C
Step 2: Calculate the cooling constant (k):
To calculate k, we can use the relation between the temperature at t = 0 and t = 1. The equation becomes:
T(1) = Ts + (T0 - Ts) * e^(-k * 1)
Substituting the known values:
30.3 = 20 + (32.5 - 20) * e^(-k)
Simplifying:
10.3 = 12.5 * e^(-k)
Divide by 12.5:
0.824 = e^(-k)
Take the natural logarithm of both sides:
ln(0.824) = -k
Using a calculator, we find:
k ≈ -0.192
Step 3: Determine the time of death (t):
To find when the murder took place, we need to find the value of t when the body temperature reaches the normal body temperature of 37.0 C. The equation becomes:
Ta = Ts + (T0 - Ts) * e^(-k * t)
Substituting the known values:
37.0 = 20 + (32.5 - 20) * e^(-0.192 * t)
Simplifying:
17.0 = 12.5 * e^(-0.192 * t)
Divide by 12.5:
1.36 = e^(-0.192 * t)
Take the natural logarithm of both sides:
ln(1.36) = -0.192 * t
Using a calculator:
t ≈ 4.597
Step 4: Convert the time to hours:
The time is given in hours, so the murder took place approximately 4.6 hours before the temperature measurement at 1:30 pm.
Step 5: Calculate the exact time of the murder:
Subtracting 4.6 hours from 1:30 pm, we can determine the exact time the murder took place:
1:30 pm - 4 hours and 36 minutes = 9:54 am
So, the murder took place around 9:54 am.