A coroner arrives at a murder scene at 9 p.m. She finds the temperature of the body to be $86.2^{\circ} \mathrm{F}$. She waits one hour, takes the temperature again, and finds it to be $81.3^{\circ} \mathrm{F}$. She notes that the room temperature is $74^{\circ} \mathrm{F}$. Assuming that the temperature of the body was $98.6^{\circ} \mathrm{F}$ when the murder occured, when was the murder committed?
The murder was commited at approximately approximately
(Do not round until the final answer. Then round to the nearest hour.)

Step 1 ：Given that the temperature of the body at 9 p.m. is \(86.2^\circ F\), and at 10 p.m. it is \(81.3^\circ F\). The ambient temperature is \(74^\circ F\) and the initial temperature of the body when the murder occured was \(98.6^\circ F\).

Step 2 ：We can use Newton's law of cooling to solve this problem. The formula for Newton's law of cooling is \[T(t) = T_a + (T_0 - T_a) e^{-kt}\] where \(T(t)\) is the temperature of the object at time \(t\), \(T_a\) is the ambient temperature, \(T_0\) is the initial temperature of the object, and \(k\) is the cooling constant.

Step 3 ：First, we can use the two temperature readings to solve for \(k\). We know that \(T_1 = 86.2\), \(T_2 = 81.3\), \(T_a = 74\), \(T_0 = 98.6\), and the time difference \(\Delta t = 1\) hour. Substituting these values into the formula, we get \(k = 0.513561603584866\).

Step 4 ：Next, we can use the value of \(k\) to solve for the time of the murder. Substituting \(k\) and the other known values into the formula, we get \(t = 1.365582018405732\) hours.

Step 5 ：Since the coroner arrived at the scene at 9 p.m., the murder must have occurred \(t\) hours before that time. Therefore, the murder time is \(9 - t = 7.634417981594268\) p.m.

Step 6 ：Rounding to the nearest hour, we find that the murder was committed at approximately \(\boxed{8}\) p.m.