Case: At 9:00 PM a coroner arrived at a hotel room of a murder victim. The temp. of the room was 70 degrees F. It was

Question

Case: At 9:00 PM a coroner arrived at a hotel room of a murder victim. The temp. of the room was 70 degrees F. It was assumed that the victim had a body temp. of 98.6 degrees F (AT THE TIME OF DEATH)(not at 9:00PM). The coroner took the victim's temp. at 9:15 PM at which it was 83.6 degrees F and again at 10:00 PM at which it was 80.3 degrees F. A. At what time did the victim die? I solved that question to be at 7:15 PM (165 min before 10:00PM) B. If the assumption of the victim's body temp. at the time of death was found to be incorrect due to a major outbreak of influenza at the hotel during the victim's stay and the usual fever a flu case will run is 103 degrees F. What is the time of death then? I solved it to be 6:37 PM(203 min before 10:00PM). C. If the victim had no influenza at the time of death, but the room temp was really 75 degrees F what was the time of death then? answer was 7:47PM(133 min before 10:00PM) D. Assuming the victim was murdered 3 minutes of the time the

Help In Homework · Accepted Answer

Case A: In this case, the logarithmic rate of cooling per minute is rateA = log( (83.6-70)/(80.3-70) ) / 45 = 0.0062 (1/min), where "log" denotes the natural logarithm, The time needed to cool from 96.6F to 80.3F is TimeCoolA = log( (96.6-70)/(80.3-70) ) / rateA = 154 min The murder time is timeMurderA = 10:00 pm - 154 min = 7:26 PM Answer to D: if the murdered got off the elevator within 3 minutes of the murder, the 5 min sequence between 7:21 and 7:26 needs to be reviewed Case B: If the starting body temperature was 103F, but the room temperature was 70F as assumed in A, the logarithmic rate of cooling is the same as in A, and the time needed to cool from 103F to 80.3F is TimeCoolB = log( (103-70)/(80.3-70) ) / rateA = 189 min The murder time is timeMurderB = 10:00 pm - 189 min = 6:51 PM Answer to D: the sequence between 6:46 and 6:51 to be reviewed Case C: If the room temperature is 75F and the body temperature at death was 98.6F, than the logarithmic cooling rate per minute is rateC = log( (83.6-75)/(80.3-75) ) / 45 = 0.0108 (1/min) The time needed to cool from 96.6F to 80.3F is TimeCoolC = log( (96.6-75)/(80.3-75) ) / rateC = 131 min The murder time is timeMurderB = 10:00 pm - 131 min = 7:49 PM Answer to D: the sequence between 7:44 and 7:49 to be reviewed D: the answers are written in cases A, B, C

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Case: At 9:00 PM a coroner arrived at a hotel room of a murder victim

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