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This exercise uses Newton's Law of Cooling.
Newton's Law of Cooling is used in homicide investigations to determine the time of death. The normal body temperature is 98.6°F. Immediately following death, the body begins to
cool. It has been determined experimentally that the constant in Newton's Law of Cooling is approximately k - 0.1947, assuming time is measured in hours. Suppose that the
temperature of the surroundings is 60°F.
(a) Find a function TCC) that models the temperature t hours after death.
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(6) If the temperature of the body is now 78°F, how long ago was the time of death? (Round your answer to the nearest whole number.)
Xhr

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By applying the Newton's law of cooling and performing the neccessary calculations, the time of death of the body is about 4 hours ago.

Newton's law of cooling

According to the Newton's law of cooling;

T(t) = T(s) + (To - Ts)e^-kt

T(t) = temperature at time t

T(s) = temperature of the surroundings

To = Initial temperature

k = cooling constant

t = time taken

Substituting the values;

T(t) = 60 + (98.6 - 60)e^- 0.1947t

To find the time taken to attian a temperature of 78°F;

78 = 60  + (98.6 - 60)e^- 0.1947t

78= 60 + 38.6e^- 0.1947t

78 - 60 = 38.6e^- 0.1947t

18/38.6 = e^- 0.1947t

0.466 = e^- 0.1947t

ln0.466 = lne^- 0.1947t

ln0.466 =- 0.1947t

t = ln0.466 /- 0.1947

t = 4 hours

Learn more about Newton's law of cooling: https://brainly.com/question/14643865

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