Question

Solve the differential equation
Suppose the temperature of a body when discovered is 85° F. Two hours later, the temperature is 74°F and the room temperature is 68°F. Find the time when the body was discovered after death (assume the body temperature to be 98.6°F at the time of death.)

Answer 1

Answer on Question #76037 – Math – Differential Equations

Question

Solve the differential equation.

Suppose the temperature of a body when discovered is $85{}^{\circ}$ F. Two hours later, the temperature is $74{}^{\circ}$ F and the room temperature is $68{}^{\circ}$ F. Find the time when the body was discovered after death (assume the body temperature to be $98.6{}^{\circ}$ F at the time of death.)

Solution

Using Newton's Law of Cooling tells us:

$$\frac{dT}{dt} = k(68 - T), \quad T_0 = 98.6, \quad T(t_1) = T_1 = 85, \quad T_2 = T(t_1 + 2) = 74.$$ $$\frac{dT}{dt} = k(68 - T), \quad \frac{dT}{dt} = -k(T - 68), \quad \frac{dT}{T - 68} = -kdt.$$

Now we can integrate both sides of equations, variables are separated.

$$\int \frac{dT}{T - 68} = \int -kdt$$ $$\ln(T - 68) = -kt + C.$$ $$T - 68 = C e^{-kt}, \quad T = 68 + C e^{-kt}.$$

We can use initial condition to find the value of $C$:

$$T(0) = 68 + C = 98.6, \quad C = 98.6 - 68 = 30.6.$$

The solution of this differential equation is $T = 68 + 30.6e^{-kt}$.

Now we need to find the value of $k$.

$$85 = 68 + 30.6e^{-kt_1}, \quad 30.6e^{-kt_1} = 17, \quad e^{-kt_1} = \frac{17}{30.6}$$ $$\begin{array}{l} 74 = 68 + 30.6e^{-k(t_1 + 2)}, \quad 30.6e^{-kt_1 - 2k} = 6, \quad 30.6e^{-kt_1}e^{-2k} = 6.17e^{-2k} = 6, \quad -2k = \ln \frac{6}{17}, \\ k = \frac{1}{2} \ln \frac{17}{6}. \end{array}$$

Now we can find $t_1$:

$$e^{-kt_1} = \frac{17}{30.6}, \quad -kt_1 = \ln \frac{17}{30.6}, \quad t_1 = \frac{1}{k} \ln \frac{30.6}{17} = 2 \frac{\ln \frac{30.6}{17}}{\ln \frac{17}{6}} = 1.13 \text{ (hours)}.$$

Answer: body was found in 1.13 hours after death.

Answer provided by https://www.AssignmentExpert.com