Newtons Law of Cooling in Forensic Science Question goes: Law enforcement would like to know the time at which a person died. The investigator arrived on the scene at 8:15pm, which we will call $t$ hours after death. At 8:15 (i.e $t$ hours after death), the temp of the body was found to be $27.4°C$ (Degrees). One hour later, $t+ 1$ hours after death, the body was found to be $26.1°C$. Known constants are $T_s=21°C$, $T_o=36.8°C$.
At what time did the victim die?
MY WORKING
Formula: $T(t)=T_s+(T_o-T_s)e^{-kt}$
1. $T(t)=T_s+(T_o-T_s)e^{-kt} \quad \rightarrow \quad  
27.4=21+15.8e^{-kt}$
2.$T(t)=T_s+(T_o-T_s)e^{-kt} \quad \rightarrow \quad 26.1=21+15.8e^{-kt}$


*

*$27.4=21+15.8e^{-kt}\rightarrow  
6.4=15.8e^{-kt}\rightarrow  
\ln(6.4/15.8)=-kt\rightarrow  
-0.903=-kt$

*$26.1=21+15.8e^{-kt} \rightarrow  
5.1=15.8e^{-k(t+1)} \rightarrow 
ln(5.1/15.8)=-k(t+1)  \rightarrow $
$-1.131=-k(t+1)$
This is as far as I have got and I believe I should be doing simultaneous equations but am totally unsure if thats correct. Any help would be appreciated.

Your work is correct but it is better to write the two starting equations as:
1. $T(t)=T_s+(T_o-T_s)e^{-kt} $
2. $T(t+1)=T_s+(T_o-T_s)e^{-k(t+1)} $
You have found the system
$$
\begin{cases}
0.903=kt\\
1.131=kt+k
\end{cases}
$$
so, substituting $kt$ in the second equation you find $k=0.228$ and , from the first, $t=3,93$ that is the time , in hours, after death.
