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

Newton's Law of Cooling which states that

$T(t)=T_a+(T_0-T_a)e^{-kt}$.

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

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

Using this equation, we can solve for k:

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

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

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

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

$T(t)=37$

that is

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

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