Just before midday the body of an apparent homicide victim i

Just before midday the body of an apparent homicide victim is found in a room that is kept at a constant temperature of 70?F . At 12 noon the temperature of the body is 80?F and at 1 P. M., it is 75?F . Assume that the temperature of the body at the time of death was 98.6?F and that it has cooled in accord with Newton\'s Law of Cooling. How long before noon did the death occur?

Solution

Newton\'s Law yields dB / dt = k [ B - Ta] ,

B = body temp , Ta = ambient temp , B (0) = 80 , B(1) = 75 , Ta = 70

B(t) = c e^(kt) + Ta--->

B = 10 e^(kt) + 70--->

75 = 10 e^k + 70--> e^k = 1/2

now find t if

98.6 = 10(1/2)^t + 70

after getting value of \"t\"

t=-1.516

answer will be 12 + t..{ t < 0 }

12-1.516=10.48(death occur at 10.48 am )