Skip to main content

Ordinary Differential Equations

Section 1.3 Newton’s law of heating and cooling

The rate of change in an object’s temperature is proportional to the difference in its temperature and the temperature of its surrounding medium. Let \(T(t)\) be the temperature of an object at time \(t\) and \(T_m\) be the (constant) temperature of the surrounding environment. Then,
\begin{equation} \dfrac{\mathrm{d}T}{\mathrm{d}t} = k(T-T_m),\tag{1.3.1} \end{equation}
where \(k\) is a constant of proportionality called the heat transfer coefficient. We can check that for any constant \(T_0\text{,}\) all functions of the form \(T(t)=T_m+(T_0-T_m)e^{-kt}\) are solutions to (1.3.1).

Example 1.3.1. Newton’s law of heating and cooling model setup.

Suppose that an object initially having a temperature of 25°F is placed in a large temperature controlled room of 89°F and one hour later the object has a temperature of 37°F. Write a differential equation for the temperature \(T(t)\) of the object at time \(t\) (in hours) assuming \(T(t)\) satisfies Newton’s law of heating and cooling.