Subsection 10.3.7 Newton’s Law of Cooling
The rate of loss of heat by a body is directly proportional to the temperature difference between the body and its surroundings, provided that the temperature difference is small.
Let \(\theta\) and \(\theta_{o}\) be the temperature of a body and its surroundings, respectively. Let \(\frac{\,dQ}{\,dt}\) be the rate of loss of heat, then from Newton’s law of cooling -
\begin{equation*}
\frac{\,dQ}{\,dt} \propto \left(\theta-\theta_{o}\right)
\end{equation*}
\begin{equation}
\text{or,}\quad \frac{\,dQ}{\,dt} =\kappa \left(\theta-\theta_{o}\right) \tag{10.3.8}
\end{equation}
where \(\kappa\) is a constant. If \(m\) be the mass of a body and \(c\) be its specific heat capacity, then
\begin{equation*}
Q=mc\theta
\end{equation*}
\begin{equation}
\therefore\quad \frac{\,dQ}{\,dt} = mc \frac{\,d\theta}{\,dt} \tag{10.3.9}
\end{equation}
\begin{equation*}
mc \frac{\,d\theta}{\,dt} =\kappa \left(\theta-\theta_{o}\right)
\end{equation*}
\begin{equation*}
\text{or,}\quad \frac{\,d\theta}{\,dt} = \frac{\kappa}{mc}\left(\theta-\theta_{o}\right)
\end{equation*}
\begin{equation*}
\therefore\quad \frac{\,d\theta}{\,dt} \propto \left(\theta-\theta_{o}\right)
\end{equation*}
This is known as Newton’s law of cooling. Here \(\kappa,\, m\) and \(c\) are constant.
