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7am one morning detectives find a murder victim in a closed meat locker. The temperature of the victim measures 88. Assume the meat locker is always kept at 40, and at the time of death the victim's temperature was 98.6. When the body is finally removed at 8am, its temperature is 86. How big an error in the time of death would result if the live body temp was known only to be between 98.2 and 101.4?

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wizard123
Using Newtons law of cooling:
[tex]T = (T_0 -40)e^{kt} + 40[/tex]
Solving for time with T = 88 which is temperature at 7am:
[tex]t = \frac{ln(\frac{88-40}{T_0 - 40})}{k} [/tex]

Next solve for k using given information that after 1 hour at 8am the temperature decreased to 86. Let T0 = 88, T = 86.
[tex]k = ln(\frac{86-40}{88-40}) = ln(\frac{23}{24})[/tex]

Finally sub in values of 98.2 and 101.4 for T0 to get different times of death.
[tex]t = \frac{ln(\frac{88-40}{98.2-40})} { ln(\frac{23}{24})} = 4.527 = 4 hr 32 min[/tex]
This puts time of death about 2:28 am.

[tex]t = \frac{ln(\frac{88-40}{101.4-40})} { ln(\frac{23}{24})} = 5.785 = 5 hr 47 min[/tex]
This puts time of death about 1:13 am

The difference is 1 hr 15 min , this is maximum error in reporting time of death.

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