# A coroner arrives at 1 AM, immediately takes the temperature of the body, and finds it to be 95...

## Question:

A coroner arrives at 1 AM, immediately takes the temperature of the body, and finds it to be {eq}95^{\circ} {/eq}. She waits for 2 hrs, takes the temperature again, and finds it to be {eq}85^{\circ} {/eq}.

She also notes that the temperature of the room is {eq}70^{\circ} {/eq}.

When was the murder committed?

## Newton's Law of Cooling:

Newton's Law of Cooling suggests that the rate of cooling (or heating) of an object is proportional to the temperature difference of its own body and of its surrounding environment, expressed as:

$$\frac{dT}{dt} = -k (T - T_s) $$

where {eq}T {/eq} is the temperature of the object, {eq}T_s {/eq} is the temperature of its surroundings, and {eq}k {/eq} is a cooling (or heating) constant specific to the body.

Manipulating and integrating the differential equation above, the temperature {eq}T {/eq} of the object at any time {eq}t {/eq} when its initial temperature is {eq}T_0 {/eq} is given by the equation:

$$T = T_s + (T_0 - T_s) e^{-kt} $$

## Answer and Explanation:

At 1 AM, the coroner finds the temperature of the body to be {eq}95^\circ F {/eq}. After {eq}t = 2 \,hrs {/eq}, she again takes the temperature of the body and reads an {eq}85^\circ F {/eq} temperature. She also notes that the temperature of the room is {eq}70^\circ F {/eq}.

First, we will lay out some assumptions:

- the body died in the mentioned room;
- the temperature of the said room did not change from the time of death until the temperature reading at 3 AM, {eq}2 \,hrs {/eq} after the coroner's arrival;

- the body's composition remained unchanged such that its cooling constant {eq}k {/eq} is still indeed, constant;

- the body has been dead for {eq}t = t_0 \,hrs {/eq} when the coroner arrived at 1 AM; and,

- that the temperature of the body at the time of death is equal to the normal average body temperature of {eq}T_0 = 98.6^\circ F {/eq}.

Based on the given and all the assumptions, solve for the cooling constant {eq}k {/eq} of the body using the temperature readings at 1 AM and 3 AM:

{eq}\begin{align*} &t = 2 \,hrs & \text{[From 1 AM to 3 AM]} \\ &\Rightarrow T = T_s + (T_0 - T_s) e^{-kt} & \text{[Substitute all known values]} \\ &\Rightarrow 85 = 70 + (95 - 70) e^{-2k} \\ &\Rightarrow 15 = 25 e^{-2k} \\ &\Rightarrow e^{-2k} = \frac{3}{5} \\ &\Rightarrow -2k = \ln \left( \frac{3}{5} \right) \\ &\Rightarrow k = - \frac{1}{2} \ln \left( \frac{3}{5} \right) \\ &\Rightarrow k \approx 0.25541 & \text{[Cooling constant of the body]} \end{align*} {/eq}

Next, determine the time elapsed since the time of death up to the time of arrival of the coroner at 1 AM:

{eq}\begin{align*} &t = 2 \,hrs & \text{[From time of death to 1 AM]} \\ &\Rightarrow T = T_s + (T_0 - T_s) e^{-kt} & \text{[Substitute all known values]} \\ &\Rightarrow 95 = 70 + (98.6 - 70) e^{-0.25541 t_0} \\ &\Rightarrow 25 = 28.6 e^{-0.25541 t_0} \\ &\Rightarrow e^{-0.25541 t_0} = \frac{125}{143} \\ &\Rightarrow -0.25541 t_0 = \ln \left( \frac{125}{143} \right) \\ &\Rightarrow t_0 = - \frac{1}{0.25541} \ln \left( \frac{125}{143} \right) \\ &\Rightarrow \boxed{ t_0 \approx 0.53 \,hrs = 32 \,mins } \end{align*} {/eq}

Since the coroner arrived at 1 AM, {eq}t = t_0 = 32 \,mins {/eq} after death, the **murder must have been committed at approximately {eq}\boxed{ 12:28 \,AM } {/eq}.**

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from College Chemistry: Help and Review

Chapter 6 / Lesson 5