A body was found at 6 a.m. outdoors on a day when the temperature was 50oF. The medical examiner found the temperature of the body to be 66oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.
http://i.imgur.com/G3nSe8u.png

Midnight (12 a.m.)

5 a.m.

2 a.m.

3 a.m.

Question

A body was found at 6 a.m. outdoors on a day when the temperature was 50oF. The medical examiner found the temperature of the body to be 66oF. What was the approximate time of death? Use Newton's law of cooling, with k = 0.1947.
http://i.imgur.com/G3nSe8u.png

       Midnight (12 a.m.)

       5 a.m.

       2 a.m.

       3 a.m.

anonymous · Answer

tt = 50 + (16)e(^-0.1947 x change in time) => tt -50 /16 = e(^-0.1947 x change in time) => change in time = ln (tt - 50/16) / -0.1947 (k) hmm without a change in time, or a temp of body at time of death I don't see how to do this, very sorry :C I did find a useful tool that may help though http://www.endmemo.com/physics/coollaw.php

perl · Answer

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