1

The victim, a middle-aged male potato, was found dead in his

home. Using crime-scene body- temperature data, can you

determine the time of death?

What Time Did the Potato Die?

Prepared for SSAC byRebecca SundermanEvergreen State College, Olympia WA© The Washington Center for Improving the Quality of Undergraduate Education. All rights reserved. 2005

Quantitative concepts and skillsUnit conversionsTrendlinesLogarithmsGraphs, linear Graphs, semilogarithmic

2

Forensic scientists use multiple techniques to approximate the time of death. Data include body temperature, chemical composition of eye vitreous, stomach contents, presence of rigor mortis, fixed or unfixed lividity, and the developmental stage of insects. It is often a combination of these data that ultimately provides investigators with a time frame for the window of death.

Upon death, a body begins to cool to the temperature of its surrounding environment. In this module you will use temperature data taken over a period of time to determine the time of death of a vegetable victim, a potato.

Useful Conversion Factors and Constants:

Living Potato Temp = 325°F

°C = (°F –32)*(5/9)

Preview

3

Slide 3: •Spells out the problem that requires solving.

Slides 4-5:•Provide the data needed to solve the problem and build the necessary spreadsheet.

Slides 6-9: •Construct the graphs for solving the problem.

Slide 10:•Examines the results.

Slide 11:•Provides the end-of-module assignment.

Overview of Module

4

The CSI unit arrived at Pete Potato’s house shortly after 2 AM. Katy Carrot, from across the street, called 911 at 1:28 AM and reported seeing both the front door on the Potato residence open and a large gourd running away from the potato home.

Investigators found the deceased body of a middle-aged, male potato on the kitchen floor. Body temperature and time data were collected at the scene.

PROBLEM:

What time did the victim die?

Problem

5

Recreate this spreadsheet

Procedure: Setting up a Spreadsheet

Be sure to catch this two-minute jump.

To expand the column width, bring the cursor

between the columns to activate the arrows

shown here. Left-click and drag to get the

desired width.Portions of Column A contain

sequential numbers. An alternative to typing in each Column A number

would be to:

(1) Type in the first three numbers of a sequence. (2) Highlight these three numbers and bring the cursor to the black square on the lower right corner. A black plus symbol will appear. (4) Left-click on the plus symbol and drag—with the button still depressed—to row 29. When you release the mouse your column should be filled.

You may have noticed that Cells B8, B20, and B29 on your sheet do

not have three digits. To establish a rule for a

given set of numbers, highlight all of the

numbers and right-click. Select “Format Cells”

and then the “Number” tab. Highlight “Number” from the category list and set the “decimal

places” to 1.

6

1. Insert a column between “Clock Time” and “Potato Temp” labeled “Investigation Time”.

Although “Clock Time” is useful to the investigator at the crime scene, a single unit time reference must be established

to analyze these data.

Right-click on the column letter of “Potato Temp” and select “Insert” from the menu.

2. Establish the first temperature reading to be at time 1.0 minutes and fill in the remaining cells accordingly.

Don’t forget th

e two-m

inute jump!

Procedure: Setting up a Spreadsheet

Note that this selection means that 2:11 and 2:10 correspond to investigation

times of 0.0 and -1.0 min, respectively.

7

3. Select “next”.

5. Under the “Titles” tab provide a name for your chart and titles, including units, for both axes. Under the “Legend” tab uncheck the box “show legend”.

Procedure: Creating a Graph

Create a graph of the victim’s temperature as a function of time.

1. Highlight Cells B3 to C29 and click on the “Chart Wizard”.

2. Select “XY (Scatter)” from the “Standard Types”, “chart type” menu, then click on the “chart sub-type” menu option without lines.

The majority of scientific work does not involve “connect-the-dot” style graphing!

6. Select “as a new sheet”, then select “finish”.

4. You have already selected your data range, so select “next” again.

Victim Temperature Values

0

10

20

30

40

50

60

70

0 10 20 30

Time (minutes)

Te

mp

era

ture

(d

eg

C)

Potato Temp

8

1. Enlarge the font sizes for all graph titles by double clicking on the title, selecting the “Font” tab, and changing the font size.

2. Remove the chart background color by selecting “Clear” from the menu that appears when you right-click the mouse, while the cursor is over the chart area.

Procedure: Polishing the Graph

3. Remove the gridlines from your graph by selecting “Clear” from the menu that appears when you right-click the mouse, while the cursor is over a gridline.

4. Adjust the y-axis range to run from 45°C to 70 °C by right-clicking on the y-axis and selecting “Format Axis”.

Victim Temperature Values

0

10

20

30

40

50

60

70

0 10 20 30

Time (minutes)

Te

mp

era

ture

(d

eg

C)

Potato Temp

9

1. Use the equation from Slide 2 and calculate the living potato temperature in degrees Celsius.

2. Create a new column on your spreadsheet containing this temperature.

3. You will need to add another data series to your previous graph.

The entire column (D) will have the same number.

4. Right-click on the chart area and select “Source Data”. Under the Series tab select “Add”.

5. Put the cursor in the “x values” box and click on the icon in the far right of this box. Click and highlight the “Investigation Time” column (C5 to C31).

6. Select the “Living Temp” column data for the y-axis.

Procedure: Adding the Known

Victim Temperature Values

0

50

100

150

200

0 10 20 30

Time (minutes)

Te

mp

era

ture

(d

eg

C)

Potato Temp

Living Temp

10

A trendline provides a possible best-fit curve to a set of data points. The “fit” of the curve can be monitored by examining the R2 value. (R2 values range from 0.0 to 1.0 with 1.0 being a perfect fit, which occurs when all of the experimental data points fall directly on the curve.) The equation of

the trendline allows for mathematical modeling.

Procedure: Adding the Trendline

1. Right-click on any data point in the “Living Temperature” data series and select “Add Trendline”. Under the type tab select “linear”.

2. Right-click on any data point in the “Potato Temp” data series and select “Add Trendline”. Under the type tab select “linear” and check the boxes for “Display equation” and “Display R2” under the options tab. Move the equation box above the data points.

3. Right-click on any data point in the “Potato Temp” data series and select “Add Trendline” (be sure to try to click on a data point and not the first trendline, or it won’t let you add another trendline). Under the type tab select “exponential” and check the boxes for “Display equation” and “Display R2” under the options tab. Move this equation box below the data points and make the text blue. You can delete the new additions to the legend as they are not necessary.

Victim Temperature Values

y = -0.5308x + 63.116R2 = 0.9843

y = 63.449e-0.0095x

R2 = 0.99150

50

100

150

200

0 10 20 30

Time (minutes)

Te

mp

era

ture

(d

eg

C)

Potato Temp

Living Temp

11

Solving the Problem

Step One: ExtrapolationRight-click the two Potato Temp trendlines and select “clear”. Now add two new trendlines (a linear and exponential) but this time enter a numerical value into the “Forecast Backward” box under the “options” tab. Thus, your data points are actual observations, and your trendlines are interpretations, or back extrapolations. Keep changing your numerical values until your exponential trendline intersects the Living Temp trendline.

The values being extended correspond to the x-axis. Using your general knowledge of baked potatoes, how long do you expect a potato to be warm to the touch after removing it from an oven? Enter this time estimate as your forecast number.

Step Two: InterpretationYou have two trendlines and thus two possible solutions. You also have two methods for reaching a solution value: graphical analysis and algebraic analysis.

Time of Death = 12: 32 AM

Solution:The exponential trendline makes more sense. It has a larger R2 value and its prediction is logical. (A potato can stay warm only for so long.)

Graphical…The best-fit line crosses the living temp line at –100 minutes or 12:31 AM (2:12 less 101 minutes).

Algebraic…The best fit line crosses the living temp line where y = 162.8 deg C in the best-fit equation, when you solve for x. Then x = -99.2 minutes or 12:32 AM.

Victim Temperature Values

y = -0.5308x + 63.116R2 = 0.9843

y = 162.78R2 = #N/A

y = 63.449e-0.0095x

R2 = 0.9915

0

50

100

150

200

-140 -120 -100 -80 -60 -40 -20 0 20 40

Time (minutes)

Te

mp

era

ture

(d

eg

C)

Potato Temp

Living Temp

12

End-of-Module Assignment

1. When we were adding trendlines, why was it not necessary to include the R2 value or trendline equation for the living potato temperature series?

2. Why is the algebraic solution more reliable than the graphical solution of this problem? (slide 11)

3. Why is the solution the “Recorded Time” minus “algebraic solution” plus one minute? (slide 11)

4. A second victim, a Zack Zucchini, was found in the house. Given the data below, determine this second victim’s time of death.

A Living Zucchini has a temperature of 300°F.