Question 1116711: A coroner found that the temperature of a murdered body was
90 F at 11:00 pm and 88.5 F
an hour later. The temperature of the room was 70 F. According to Newton’s law of Cooling, the temperature, in F, of the body at time T, in hours, is given by T(t)=70+ce^kt, where
c and k are constants. Normal body temperature is
98.6 F. Find the values of c and k .
Determine when the murder was committed.
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! 98.6=70+c*e^0
c=28.6 ANSWER
90=70+28.6*e^kt
88.5=70+28.6e^k(t+1)
From the first:
0.6993=e^kt
ln both sides
kt=-0.3577
From the second:
18.5=28.6*(e^kt)*e^k
0.6469=e^k(t+1)
ln both sides
-0.4356=k(t+1)=kt + k
-0.4356+0.3577=-0.0779=k, by substitution ANSWER
kt=-0.3577
-0.0779*t=-0.3577
t=4.59 hours earlier than 11 pm or 6.41 hours time or 6:25 pm to the nearest minute ANSWER
check
t(4.59)=70+28.6*e*(4.59*(-0.0779)
=70+20=90
at 5.59 hours
t(5.59=70+28.6(e^(5.59*(-0.0779))=88.5
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