Question

Police arrive at a murder scene at 2:30 AM and immediately record the body's temperature which was 93°F. At 3:30 AM, after thoroughly inspecting and fingerprinting the area, they again took the temperature of the body which had dropped to 84°F. The temperature of the crime scene has remained at a constant 80°F. Determine when the person was murdered. (Assume that the victim was healthy at the time of death. That is, assume that the temperature of the body at the time of death was 98.6°F.) The person was murdered at (Do not round until the final answer. Then round to the nearest minute as needed.)

Answer 1

solution- imore los temperature of By Newton's lan of cooling we have. die-k (0-0) the surroundings. O= temperature attimet. In our situatuation, K-Constant. do = -K (0-80) =f-ket 0-80 do tene. en 20-80) = -ktylne e=93°F . -kt 0-80 = ce O= soteekt. we compute the time in minutes. when t= 2:30 hrs a 150 minutes, when ta 3.30 uro = 210 minutes, 8=84°F. - kx150 - 150k 93=80+ ce = 13 84=80+ cé ce -210k - ce-210k divide by 2, =13 4 = 13 e-150k +210k e +60k 60k = ln (12) ka then (13) = 0.02 cxe-21080 02 4 4.2 = qe = 266.75 e-4.2 c = 4 e-210x0.02 0 - 80 +26675 -0.025 Now. when. D=9806, then. -0.02t 98.6=80+266175ē e-0.02t = -0.07. 18.6 266.75 -0.02t en (0.07) en (0.07) ta - 132.96 0.02 The person was murdered at [2:113 AM (Aus)