Question

A crime scene investigator finds a body at 8 P.M., in a room that is kept at a constant temperature of 70◦F. The temperature of the body is 88.8◦F at that time. Thirty minutes later the temperature of the body is 87.5◦F.

(a) Set up a differential equation describing $\frac{dT}{dt}$, and solve it to get a formula for the temperature of the body t minutes after the time of death. Your answer will involve a proportionality constant k. (b) Use the information in the problem to determine k.

(c) Assuming that the body had a normal temperature of 98.6◦F at the time of death, estimate the time of death of the victim.

Short answer

(a) $\mathrm{The}\mathrm{final}\mathrm{equation}\mathrm{is}=70+18.8e^{-kt}$

(b) $k=-0.1433$

(c)t=-2.93

Step-by-step solution

Step 1. Given 

A crime scene investigator finds a body at 8 P.M., in a room that is kept at a constant temperature of 70◦F. The temperature of the body is 88.8◦F at that time. Thirty minutes later the temperature of the body is 87.5◦F.

Part(a) Step 2. 

Note that the problem resembles Newton's Law of Cooling and Heating models. Recall that the differential equation modelling Newton's Law of Cooling and Heating for ambient temperature A is given by

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

In the above equation k is the constant of proportionality. Take the time 8 P.M. as the initial time of the problem, that is 8 P.M. corresponds to t = 0 Also, measure the time in hours; so, the temperature of the body measured after 30 minutes means after half an hour. Hence, as per given information in the problem T(0) = 88.8 degrees * F , A=70°F and temperature of body after half an hour = 87.5° F. Further, note that the rate of change of temperature is negative (since the temperature of body is reducing). Use this information in equation (1) and write down the differential equation representing the problem as

$$\begin{gathered} \frac{dT}{dt}=-k(70-T) \\ =k(T-70) \end{gathered}$$

Now, proceed to solve the differential equation.

Observe that the differential equation does not contain the independent variable at all, so solve it by using the method of anti-differentiating. Transfer the term containing T to the left hand side and integrate both the sides

$$\begin{gathered} \int \frac{dT}{T-70}=k\int dt \\ \ln \left|T-70\right|=kt+C \\ T-70=e^{kt+C} \\ \mathrm{T}=70+{\mathrm{Ae}}^{\mathrm{kt}} \\ \mathrm{Use}\mathrm{the}\mathrm{initial}\mathrm{conditions},\mathrm{namely}T=88.8fort=0andsolvetoevaluatethecons\tan tA \\ 88.8=70+A \\ A=18.8 \\ \mathrm{The}\mathrm{final}\mathrm{equation}\mathrm{is}=70+18.8e^{-kt} \end{gathered}$$

Part(b)  Step 3. Calculating the proportionality constant

$$\begin{gathered} TakeT=87.5,t=\frac{1}{2}inequation\left(2\right)ansolvefork \\ 87.5=70+18.8e^{\frac{1}{2}k} \\ \mathrm{solving}\mathrm{this}k=-0.1433 \end{gathered}$$

Part (c) Step 4. Calculation

Substitute the value of k obtained in part (b) in equation (2) to find the solution of the model

$87.5=70+18.8e^{\frac{1}{2}k}$

when T(t)= 98.6

t=-2.93

The negative sign of t is justified as t = 0 has been taken to correspond to the time 8 P.M. and the death had occurred prior to this time. Thus, the death had occurred 2.93 hours (or 2 hours and 56 minutes) before 8 P.M., that is, just after 5 P.M.