Question

Problem 7. (1 point) (1 pt) Coroners estimate time of death using the rule of thumb that a body cools about 2 degrees F during the first hour after death and about 1 degree F for each additional hour. Assuming an air temperature of 66 degrees F and a living body temperature of 98.6 degrees F. the temperature Tt) in degrees F of a body at a time t hours since death is given by T(t) = 66 + 32.6e-kt 1. For what value of k will the body cool by 2 degrees F in the first hour? k= 2. Using the value of ke found above, after how many hours will the temperature of the body be decreasing at a rate of 1 degree F per hour? after hours 3. Using the value of k: found above, show by calculating both values that, 24 hours after death, the coroner's rule of thumb gives approximately the same temperature as the formula T(24) = degrees F rule of thumb gives T = degrees F

Answer 1

1. Given, Tlt) = 66 +32.6 e-kt Time derivative of temperature o T' =-32.6ke-kt For initial rate, substitute t=0, T' = -32.6K For initial rate of 2°F a we get a 2 = + 32.6K. => 2 = K ook - 10.0613 let the time be to oo the temperature of the body a at the rate of 1°F. dee decreasing (2) 0.0613 to = -1 - 32.6 % 0.0613 e- €-0.0613 to 5 326X0-0613 -0.0613 to = ln (22's konca) 0.0613 to = ln (32-64000 613) o to = ln (82:6X0.0613) 0.0613 c 11.3 hours. after 1103 hours.

(C) using our formula and values, The temperature is, T(24) = 66 +22.6e-0.0613X24 = 773.49°F Now by rule of thumba Temperature drop after four hours = 2x4 -8°F Additional temperature drop 24 hours = 1420 = 20°F Total temperature drop = 2 895 se rule of thumb ginen, T = 98.6–28 - 17006°F.