Newton´s Law of Cooling

1. What is the Newton’s Law of Cooling?
States that the rate of change of the temperature of an object is proportional to the difference between its own temperature and the ambient temperature

2. What variables in your problem correspond with the variables in the Newton’s Law of Cooling?
a) The actual body temperature at the time of death; and
b) Post mortem temperature

3. According to this Law, at what time approximately the death happened?
Between 14 to 16 minutes before 9:10PM

4. How does the room temperature affect the time of death?
The death can decrease his temperature below the room temperature. Heavy clothing and other coverings, by retaining body heat, will speed up putrefaction. Rapid putrefactive changes may been seen in corpses left in a room which is well heated, or in a bed with an electric blanket.

5. How does an illness (e.g. fever) affect the time of death?
In the temperature, because he had more temperature and more heat, that also affect the time of death, while the temperature is going down.

6. How exact is the approximation of the Newton’s Law of Cooling for predicting the time of death?
They are common, because if we consider the variables in body’s heat, the reliability of algor mortis permits an approximation to the time of death. There are many facts that modify the results.

Post Mortem changes and time of death

1. What is forensic science?
The application of scientific knowledge to questions of civil and criminal law.

2. What kind of evidence can be collected in a murder case? Things that are aaroud the body,clues

3. Why is important to determine the time of death? To know the time of death of the body, and to join things of the body with the witnesses.

4. What methods are commonly used to find the time of death? Mathematician problems, taking the temperature of the body, and to start the case.

5. What factors are considered in Algor Mortis?
Is the most useful single indicator of the time of death during the first 24 hours post mortem. It is of some importance to note that the use of body temperature estimations to assess

6. How does the environment affect the time of death?
In the temperature of the body

7. What can you tell about the different methods of temperature reading of the body? Do they make a difference in the results?
There are different methods to read temperature, and also they make difference in the results, because if you take a temperature an instant latter or before, everything would change.
The best way is to take the body temperature as early as possible. As theenvironmental temperature , all these at the time the body was discovered and any subsequent variation in these conditions. If a method of sequential measurement of body temperature is use then the thermometer should be left in situ during this time period. This latter method is much easier to undertake when using a thermo-couple with an attached print-out device. Temperature readings of the body and observations made at the scene by one physician are always available for evaluation by an expert at a later time.

BITACORA

First we started by guessing using this formula:

Ln / y - yt / = Kt + C



FIRST PART
On these part, we have to find if the crime was at the stated time, and we change the temperature from 0F to 0C, because is easier and we are used to work with these units.


Data:
Place temperature: 8:00 PM -- 300
y1 = 35.50 C t1= O ( 9:10 PM )
y2= 350C t2= 6 ( 9:16 PM )
y3= 37.50 C ( not sick person’s temperature) t3= X


First step:
We solve for C:

Ln/ 35.50 – 300 / = C

Because (K)(t) is equal to 0, so C is :
C = 1.7047

Second step:
We solve for K: (using the new value of C, and using the time difference of 6)

Ln/ 35.50 – 300 / = K (6) + 1.7047

1.6094 -1.7047 = 6K

K = - .015877

Third Step:
Now with the values of K and C and 37.50 as the initial temperature, we have to solve for the value of “t”, that would be the time.

Ln / 37.50 - 300 / = (-.015877) t + 1.7047

- 2.0149 – 1.7047 = t
.015877

t = 19.5378 min.




SECOND PART:
On this final part, we have to find again for the time with the same process used in the first part, but using new values of temperature, to confirm our investigations and find more possible answers.

Data:
Place temperature: 8:00 PM -- 300
y1 = 35.50 C t1= O ( 9:10 PM )
y2= 350C t2= 6 ( 9:16 PM )
y3= 37.50 C ( not sick person’s temperature) t3= X

Time—Temperature:

8:50 -- 30.80 C
9:00 -- 300C
9:10 -- 29.170C
9:16 -- 28.670C


First Step:
Solve for C:

Ln/ 35.50 -29.170/ = Kt + C

C=1.8453

Second Step:
Solve for K using value of C:

Ln/ 350C -29.170/= K(6) + 1..8453

1.84530 – 1.8453 = K
6
K= -.01371

Third Step:
Solve for the time using the values of C and K:

Ln/ 37.50C -29.170/= (-.01371 ) t + 1..8453

2.11986 – 1.8453 = t
.01371

t= 20.02 min

= 8:50 PM





Other temperatures, other possible answers:

with 37.80 = 22.60 min

with 370 15.50 min

with 380 24.27 min




Conclusion:

In our conclusion first we think that the problem obviously didnt have an exact answer, so we did different forms to solve and with the aproximations that we got. We can conclude that the murdered didn’t occur before the 8:50 because some approximations that we consider with the most appropriated temperature, showed us that the crime wasn’t committed before the 8:50.