17. Applications of Differential Equations

b. Newtonian Heating and Cooling

In addition to discovering the laws of inertia and gravity, Sir Isaac Newton also discovered the law which describes the heating or cooling of an object when its temperature is different from that of its surroundings.

The rate of change of the temperature of an object is proportional to the difference between the temperature of the object \(T\) and the temperature of its surroundings \(T_A\), called the ambient temperature. Thus the temperature satisfies the differential equation \[ \dfrac{dT}{dt}=-k(T-T_A) \qquad \text{where} \quad k>0 \] The minus sign says that if the temperature is greater than the ambient temperature, \(T \gt T_A\), then the temperature decreases, while if the temperature is less than the ambient temperature, \(T \lt T_A\), then the temperature increases.

Find the general solution of: \[ \dfrac{dT}{dt}=-k(T-T_A) \qquad \text{where} \qquad k>0\] and then find the specific solution satisfying the initial condition: \[ T(0)=T_o \]

Since \(T_A\) is a constant, notice that \(\dfrac{dT_A}{dt}=0\). So the equation can be written as \[ \dfrac{d}{dt}(T-T_A)=-k(T-T_A) \] But this is an exponential decay equation for the variable \((T-T_A)\). So, the solution is \[ T-T_A=Ae^{-kt} \quad \text{or} \quad T=T_A+Ae^{-kt} \] If the initial temperature is \(T(0)=T_o\), then \(T_o=T_A+Ae^0\) and so \(A=T_o-T_A\). The solution becomes \[ T=T_A+(T_o-T_A)e^{-kt} \]

Notice that the general solution has \(3\) constants: \(T_A\), \(T_o\) and \(k\). So in addition to the ambient temperature, \(T_A\), and the initial temperature, \(T_o\), we need the temperature at one later time to find \(k\). If we don't know \(T_o\), then we need the temperature at two later times to find \(k\) and \(T_o\).

A medical examiner arrived at the scene of a murder and determined that the corpse had a body temperature of \(30^{\circ}\text{C}\). One hour later, its temperature was \(27^{\circ}\text{C}\). The temperature in the room was \(12^{\circ}\text{C}\). Assuming that the temperature in the room has been constant, how long was the victim dead before the body was discovered? Normal body temperature is \(37^{\circ}\text{C}\).

Let \(T(t)\) be the body temperature in \(^{\circ}\text{C}\) at time \(t\) measured in hours. We know \(T(0)=30\) and \(T(1)=27\). Further the ambient temperature is \(T_A=12\). We want to know the time when the temperature was \(37\).

By Newton's law the temperature satisfies the differential equation: \[ \dfrac{dT}{dt}=-k(T-12) \] where we do not yet know \(k\). We solve it anyway. This equation is both separable and linear. We choose to separate and integrate: \[ \int \dfrac{dT}{T-12}=-\int k\,dt \qquad \Longrightarrow \qquad \ln|T-12|=-kt+C \] We solve for \(T\): \[ |T-12|=e^C e^{-kt} \qquad \Longrightarrow \qquad T=12+Ae^{-kt} \] where \(A=\pm e^C\). The initial condition \(T(0)=30\) implies that \(A=18\). The secondary condition \(T(1)=27\) implies \(27=12+18e^{-k}\), so that: \[ e^{-k}=\dfrac{15}{18}=\dfrac{5}{6} \qquad \text{and} \qquad k=-\ln\dfrac{5}{6}=\ln\dfrac{6}{5} \] Thus the solution is: \[ T=12+18e^{-t\ln(6/5)} \] We want to know the time when the temperature was \(37\). So we solve the equation: \[ 37=12+18e^{-t\ln(6/5)} \] as follows: \[\begin{aligned} e^{-t\ln(6/5)}=\dfrac{25}{18} &\qquad \Longrightarrow \qquad -t\ln\dfrac{6}{5}=\ln\dfrac{25}{18} \\[5pt] &\qquad \Longrightarrow \qquad t=-\dfrac{\ln(25/18)}{\ln(6/5)}\approx -1.8018 \end{aligned}\] So the murder occurred only \(1.8\) hours before the medical examiner arrived.

The equation \(\dfrac{dT}{dt}=-k(T-T_A)\) can also be solved by rewriting it in standard linear form as \(\dfrac{dT}{dt}+kT=kT_A\), in which case the integrating factor is \(I=e^{kt}\) and the equation becomes \(\dfrac{d}{dt}(Te^{kt})=kT_Ae^{kt}\). Integrating, we get: \[ Te^{kt}=T_Ae^{kt}+C \] So the solution is: \[ T=T_A+Ce^{-kt} \]

The equation \(\dfrac{dT}{dt}=-k(T-T_A)\) can also be solved by remembering that the derivative of a constant is zero. So \(\dfrac{d(-T_A)}{dt}=0\). Adding this to the left side, the differential equation becomes: \[ \dfrac{d(T-T_A)}{dt}=-k(T-T_A) \] This is just the exponential decay equation for \(T-T_A\). So the solution is: \[ T-T_A=Ae^{-kt} \]

This is not a realistic problem because cooking a real turkey depends on the size of the turkey, how quickly the heat can penetrate to the center of the turkey and whether the center of the turkey is still frozen.