Estimating Time of Death

by
Tom Woodson
Clayton Tyndall
Chris Stepheson

Abstract

An experiment in estimating the time of death was performed where a potato was used instead of a human body. The time of death for the potato was considered to be the time it was taken out of the oven. The temperature of the potato at the time of death was 194 F. Temperature readings of the potato were taken every fifteen minutes for three hours. Estimates for the time of death at the corresponding temperatures were calculated with an average percent error of 7.29%. The ambient temperature was 75.2 F.

Introduction

Newton's Law of Cooling states that the rate of cooling of an object is proportional to the temperature difference between the object and its surroundings, provided that this difference is not too large.1 When estimating the time of death, one needs to know the temperature of the surroundings and the temperature of the body at two different times in order to make an accurate estimate.

It is assumed that the temperature of the body T(t) is governed by Newton's Law of Cooling, (1)

where k is a negative constant, is the ambient temperature, and time t is the number of hours since the time of death. Also, the temperature of a human body at the time of death is considered to be 98.6 F, T(0) = 98.6 .

Consider a crime scene in Wilmington in which a man is killed in his apartment. The body was discovered in the apartment early in the morning and at 7:00 A.M. The coroner measured its temperature, at that time. One hour later another temperature, , was taken. The coroner noted that the temperature of the murder victim's apartment was maintained at a constant temperature of 70 F. The ambient temperature in this case remained constant, but keep in mind this is not always the case.

Let T(t) be the temperature t hours after the body was 98.6 F. The ambient temperature was a constant 70 F after the person's death. Newton's Law of Cooling states that . (2)

Therefore, (2) can be solved to obtain (3)

which for our example is (4)

Using the two temperature measurements taken by the coroner, the following equations can be established. (5)
and , (6)

where is the number of hours after death. The values for k and can be determined from the equations above. If 70 is subtracted from both equations and equation (5) is divided by equation (6), the following ratio is obtained (7)

Therefore, hours. This value for k can be substituted into (5) to obtain (8)

Therefore, the resulting value for is calculated to be 10.92 hours. From this calculated value it is believed that the man died about 10.92 hours before 7:00 A.M., which would be around 8:00 P.M. the previous evening. (Figure 1) Figure 1 represents the cooling of the victim's body. If the body is in a room where the temperature is a constant 70 F, then the temperature of the body at t1 corresponds to equation (4). Therefore, if t1 and T(t) are known originally, the error in the calculation of t1 can be calculated using the following equation: (9)

Using the examples from the previous crime scenario, let's simulate another crime where a potato is the murder victim. Assume that the temperature of the potato is 194 F at the time of death so that T(0) = 194 F. Make sure the temperature of the apartment is kept constant during the experiment. Take multiple temperature readings of the potato after it "dies". Estimate the time of death of the potato for every corresponding temperature and compare these estimates with the true values. Instead of using equation (7) to determine k, construct a semi-log plot to obtain a better estimate for this value.

Procedure

A potato will be used in the place of a human body. The potato will be baked in a conventional oven and then taken out of the oven once a temperature of 194 F is obtained. Once the potato is taken out of the oven, it will be allowed to cool in a room where the temperature of the room is 75.2o F. Temperature readings of the potato will be taken every fifteen minutes in order to determine how well the baked potato obeys Newton's Law of cooling.

Results and Discussion
 ```Experimental Data ``` ```Estimated ``` ```% error ``` ```Time(min) ``` ```Temp( C ) ``` ```Temp( F ) ``` ```T-Ta ``` ```LN(T-Ta) ``` ```Time (min) ``` ```of t ``` ```Avg. error ``` ```0 ``` ```90 ``` ```194.0 ``` ```118.8 ``` ```4.8 ``` ```0.0 ``` ```0.00% ``` ```7.29% ``` ```15 ``` ```84.2 ``` ```183.6 ``` ```108.4 ``` ```4.7 ``` ```7.9 ``` ```47.35% ``` ```30 ``` ```72.5 ``` ```162.5 ``` ```87.3 ``` ```4.5 ``` ```26.6 ``` ```11.47% ``` ```45 ``` ```62.4 ``` ```144.3 ``` ```69.1 ``` ```4.2 ``` ```46.7 ``` ```3.81% ``` ```60 ``` ```56 ``` ```132.8 ``` ```57.6 ``` ```4.1 ``` ```62.4 ``` ```4.01% ``` ```75 ``` ```50.1 ``` ```122.2 ``` ```47.0 ``` ```3.8 ``` ```80.0 ``` ```6.63% ``` ```90 ``` ```46.6 ``` ```115.9 ``` ```40.7 ``` ```3.7 ``` ```92.4 ``` ```2.65% ``` ```105 ``` ```42.9 ``` ```109.2 ``` ```34.0 ``` ```3.5 ``` ```107.8 ``` ```2.67% ``` ```120 ``` ```39.8 ``` ```103.6 ``` ```28.4 ``` ```3.3 ``` ```123.2 ``` ```2.70% ``` ```135 ``` ```37.2 ``` ```99.0 ``` ```23.8 ``` ```3.2 ``` ```138.7 ``` ```2.77% ``` ```150 ``` ```36.5 ``` ```97.7 ``` ```22.5 ``` ```3.1 ``` ```143.4 ``` ```4.37% ``` ```165 ``` ```34.6 ``` ```94.3 ``` ```19.1 ``` ```2.9 ``` ```157.7 ``` ```4.45% ``` ```180 ``` ```32.5 ``` ```90.5 ``` ```15.3 ``` ```2.7 ``` ```176.7 ``` ```1.84% ```

 ```exp(k*t) ``` ```T(true) ``` ```slope error ``` ```1 ``` ```194.0 ``` ```2.5% ``` ```1.190056 ``` ```175.0 ``` ```1.416232 ``` ```159.1 ``` ```1.685395 ``` ```145.7 ``` ```2.005714 ``` ```134.4 ``` ```2.386911 ``` ```125.0 ``` ```2.840557 ``` ```117.0 ``` ```3.38042 ``` ```110.3 ``` ```4.022888 ``` ```104.7 ``` ```4.78746 ``` ```100.0 ``` ```5.697343 ``` ```96.1 ``` ```6.780155 ``` ```92.7 ``` ```8.068761 ``` ```89.9 ``` Figure 2 shows the cooling of the potato over time. Notice the potato has a faster cooling rate at higher temperatures, and how the cooling rate starts to level off at lower temperatures. Also note how the data agree with the actual function .(10)

The reason that some of the data points deviate from the function could be associated with error in reading the thermometer. If there were no errors in the calculations for the estimated times of death, the data would lie perfectly on the curve of the actual function. Figure 3 represents a semi-log plot of the temperatures versus time. The semi-log plot was constructed by plotting the natural logarithm of the temperature difference between the body temperature of the potato and the ambient temperature versus time. The semi-log plot opposed to equation (7) was used to calculate the constant k, because a better estimate for this value can be obtained from many data points than just two. The slope of the line in the semi-log plot corresponds to the constant k. A value of -0.0116 minutes-1 was calculated for k, which has a correlation coefficient of 0.9929. The error in the slope of the line was calculated as 2.5%, which was determined from the following equation: , (11)2

where R2 is the correlation coefficient, and n is the number of samples. The y-intercept corresponds to the natural logarithm of the temperature difference between the body temperature of the potato and the ambient temperature at t=0 minutes.

The estimated time of death of the potato at the corresponding temperatures was calculated by using equation (3). The percentage of error in the estimated time of death was calculated for all of the corresponding temperatures using equation (9). An overall average of the percent error was calculated as 7.29% respectively.

Conclusion

Overall, the experiment was successful. The data that was taken and fitted to equation (3) obeyed Newton's Law of Cooling fairly well. The value for the k constant was calculated as -0.0116 minutes-1 with a correlation coefficient of 0.9929. An error of 2.5% was calculated for the slope of the line in the semi-log plot, which corresponds to the k constant. The estimated time of death at the corresponding temperatures agreed fairly well with the actual time of death, with the exception of t=7.9 minutes. There is a 47.35% error in this estimated time of death, which could possibly result in the misreading of the thermometer. Instead of using a thermometer, a temperature probe with a digital readout would reduce the error in the temperature readings. A temperature probe would be more accurate than the thermometer. An overall average percent error in the estimated time of death was calculated as 7.29%. This is good, considering that the 47.35% error at t=7.9 minutes is included in the overall average.

References

1. Lomen, David and David Lovelock. Exploring Differential Equations via Graphics and Data. New York: John Willey & Sons, 1996 731-735.

2. Heald, MA., American Journal of Physics. 1977 p11.