Question

A body was found at 10 a.m. outdoors on a day when the temperature was 40oF. The medical examiner found the temperature of the body to be 80oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.

\[T(t)=T _{A}+(T _{o}-T _{A)}e^{-kt}\]

A. 5 am
B. 8 am
C. 9:45 am
D. 9:30 am

Answer 1

work with the differences between the outdoor temperatures and the body temp
at 10 the difference is \(80-40=40\) and presumably at the time of death it was \(98.6\) so the difference was \(98.6-40=58.6\)
set
\[\large 58.6=40e^{1947 t}\] and solve for \(t\)

Answer 2

sorry typo on the last line, should be
\[\large 58.6=40e^{-.1947 t}\]

Answer 3

you know how to solve this?

Answer 4

no i dont actually, could you help me out?

Answer 5

sure

Answer 6

mhhm

Answer 7

step one, divide both sides by \(40\) and get
\[\frac{58.9}{40}=e^{-.1947t}\]

Answer 8

ok i see

Answer 9

quick question when you add or subtract an inequality do you flip the sign?

Answer 10

damn another typo, i meant
\[\frac{58.6}{40}=e^{-1947t}\] or
\[1.465=e^{-1947t}\]

Answer 11

I got 1.4725 when I divided by 40, but what do I do with the e^-.1947t ?

Answer 12

no you only flip the inequality sign when you multiply or divide. @Karina1

Answer 13

yeah i made a typo there, it should be \(58.6\div 40=1.465\)

Answer 14

oh okay, and what do I do for the next step?

Answer 15

is the multiplication (does the) have a impact on the flip, like i know you flip it, bury I'm confused on that part

Answer 16

so you have
\[e^{-.1947t}=1.465\] in equivalent logarithmic form you have
\[-.1947t=\ln(1.465)\]

Answer 17

then divide to get
\[t=-\frac{\ln(1.465)}{.1947}\]

Answer 18

thanks @ali1029

Answer 19

i get \(-1.96\) so about 2 hours before 10

Answer 20

I divided that and got -1.96, is that right?

Answer 21

so it would be 8 am?

Answer 22

yes that is what i get too

Answer 23

thank you so much! :)