Police arrive at a murder scene at 3 : 30 am and immediately record the body's temperature which was 90degrees°f. at 5: 30 am, after thoroughly inspecting and fingerprinting the area, they again took the temperature of the body which had dropped to 85degrees°f. the temperature of the crime scene has remained at a constant 70degrees°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 98.6degrees°f.) the person was murdered at ? . (do not round until the final answer. then round to the nearest minute asneeded.)

The person was murdered at 11.40 PM

Step-by-step explanation:

Given information :

Initial body temperature, = 90

Police arrive at a murder scene at 3 : 30 AM, at  5.30 AM (120 minutes), temperature of the body had dropped to 85 .

T(120) = 85

The temperature of the crime scene, C = 70

to determine the time of the murderer, we can use the Newton's lau of cooling:

T(t) = C + ( - C)

where

T(t) = temperature at any given time

C = surrounding temperature

= initial temperature of heated object

k = cooling constant

thus

T(t) = C + ( - C)

T(120) = 70 + (90 - 70)

85 = 70 + 20

= (85-20) / 70

-120 k = ln (65/70)

k = - ln (65/70) / 120

  = 0.00062

so we have the cooling equation

T(t) = 70 + ( - 70)

The temperature at the time of the death is 98.6  which was the initial temperature of the person murdered, the temperature dropped to T(t) =  85 , now we can find the time of him being murdered.

T(t) = 70 + (98.6 - 70)

85 = 70 + 28.6

= (85- 28.6)/70

-0.00062t = ln ((85- 28.6)/70)

t = - ln ((85- 28.6)/70)/ 0.00062

t = 350 minutes

therefore, the preson being murdered 350 minutes (5 hours 50 minutes) before 5.30, which was 11. 40 PM

7/2

(decimal: 3.5)

welcome ; )

step-by-step explanation: