Science:Math Exam Resources/Courses/MATH100/December 2011/Question 02 (a)/Solution 1

Newton's Law of Cooling which states that

$\displaystyle T(t)=T_{a}+(T_{0}-T_{a})e^{-kt}$.

Here, the initial temperature of the body T₀ is 33°C; the temperature of the environment T_a 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

$\displaystyle 31=21+(33-21)e^{-k}$

Using this equation, we can solve for k:

${\begin{aligned}e^{-k}&={\frac {31-21}{33-21}}={\frac {5}{6}}\\-k&=\ln(5/6)\\k&=-\ln(5/6)\\\end{aligned}}$

And so now we have found the equation that gives us the temperature of the body at any time t:

$\displaystyle T(t)=21+12e^{t\ln(5/6)}$

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

$\displaystyle T(t)=37$

that is

${\begin{aligned}21+12e^{t\ln(5/6)}&=37\\12e^{t\ln(5/6)}&=16\\e^{t\ln(5/6)}&=16/12\\t\ln(5/6)&=\ln(4/3)\\t&={\frac {\ln(4/3)}{\ln(5/6)}}\\t&\approx -1.578\end{aligned}}$

Therefore, the police arrived $\ln(4/3)/\ln(5/6)$ hours after the murder, that is, just over an hour and a half.