Question

Question 2 You are working as a crime scene investigator and must predict the temperature of a homicide victim over a 5-hr period. You know that the room where the victim was found was at 10 "C when the body was discovered. Newton's law of cooling says that the temperature of a body changes at a rate proportional to the difference between its temperature and that of the surrounding medium (the ambient temperature), where T = the temperature of the body (°C), t time (min), k = the proportionality constant (per minute), and Ta - the ambient temperature (C) Use Euler's method to compute the victim's body temperature for the 5-hr period using values of k = 0.12/hr and At =0.5 hr. Assume that the victim's body temperature at the time of death was 37 C. Use Matlab or hand calculations to present the results in a table. Ensure columns represent values for t, Ta, T and dT/dt.

Answer 1

(a) For the constant temperature case, Newton's law of cooling is written as dT dt -0.135(T-10) The first two steps of Euler's methods are T(0.5)-7(0)--(0) x Δι = 37 + 0. 12(10-37)(0.5)-37-32400× 0.50 = 35.3800 T(1)35.38000.12 (10 35.3800) (0.5)35.3800 3.0456 x 0.5033.8572 The remaining calculations are summarized in the following table dt 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00 37.0000 35.3800 33.8572 32.4258 31.0802 29.8154 28.6265 27.5089 26.4584 25.4709 24.5426 dTldt 3.2400 3.0456 2.8629 2.6911 -2.5296 2.3778 2.2352 2.1011 1.9750 -1.8565 1.7451 10 10 10 10 10

(b) For this case, the room temperature can be represented as Ta 20-2t where t time (hrs). Newton's law of cooling is written as dT .12(T-20+21) dt The first two steps of Euler's methods are T(0.5) 37 + 0.12(20-37)(0.5) = 37-2.040 x 0.50 = 35.9800 T(1) = 35.9800 + 0.12(19-35.9800)(0.5) = 35.9800-2.0376 x 0.50 = 34.9612 The remaining calculations are summarized in the following table dTldt 37.0000 35.9800 34.9612 33.9435 32.9269 31.9113 30.8966 29.8828 28.8699 27.8577 26.8462 2.0400 2.0376 0:00 0:30 1:00 1:30 2:00 2:30 3:00 3:30 4:00 4:30 5:00 20 19 18 17 16 15 14 13 12 -2.0353 -2.0332 2.0312 2.0294 2.0276 2.0259 -2.0244 2.0229 10

Comparison with (a) indicates that the effect of the room air temperature has a significant effect on the expected temperature at the end of the 5-hr period (difference 26.8462 - 24.5426 2.3036°C). (c) The solutions for (a) Constant Ta, and (b) Cooling Ta are plotted below: 40 Constant Ta -Cooling Ta 36 32 28 24 0:00 1:002:00 3:00 4:00 5:00