Newton's Law of Cooling which states that
.
Here, the initial temperature of the body T0 is 33°C; the temperature of the environment Ta is given to be 21°C; and the temperature of the body after 1 hour (T(1)) is 31°C.
Plugging in these numbers give us
![{\displaystyle \displaystyle 31=21+(33-21)e^{-k}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/bc02b013d2b7b332b5c64d7a255e8f891d23102b)
Using this equation, we can solve for k:
![{\displaystyle {\begin{aligned}e^{-k}&={\frac {31-21}{33-21}}={\frac {5}{6}}\\-k&=\ln(5/6)\\k&=-\ln(5/6)\\\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/b74b4d0844fb76bbc695f7759a977faa6b9459b2)
And so now we have found the equation that gives us the temperature of the body at any time t:
![{\displaystyle \displaystyle T(t)=21+12e^{t\ln(5/6)}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/76e78ba5ac29b5cf82e70f78009deb0f3e2026a4)
and we would like to know the time of death, which is the time at which the body temperature was 37°C. For this, we simply solve
![{\displaystyle \displaystyle T(t)=37}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/4d5fe83c49448cdde2a5f5722d9ec0db6ff8a9b2)
that is
![{\displaystyle {\begin{aligned}21+12e^{t\ln(5/6)}&=37\\12e^{t\ln(5/6)}&=16\\e^{t\ln(5/6)}&=16/12\\t\ln(5/6)&=\ln(4/3)\\t&={\frac {\ln(4/3)}{\ln(5/6)}}\\t&\approx -1.578\end{aligned}}}](https://wiki.ubc.ca/api/rest_v1/media/math/render/svg/f89c2db93985cadd83aebaa1e15adf4d309c8dbb)
Therefore, the police arrived
hours after the murder, that is, just over an hour and a half.