According to Stefan's law heat energy
absorbed is given by
$\frac{d Q}{d t}=\varepsilon \sigma A\left(T^4-T_0^4\right)$
where, $T=$ temperature of body
and $T_0=$ temperature of surroundings.
If $m$ be the mass of body and $c$ be the specific heat, then heat loss at temperature $T$ is
$\frac{d Q}{d t}=-m c \frac{d T}{d t}$
From Eqs. (i) and (ii), we get
$\begin{aligned}-m c \frac{d T}{d t} & =\varepsilon \sigma A\left(T^4-T_0^4\right) \\ \Rightarrow \quad-\frac{d T}{d t} & =\frac{\varepsilon \sigma A}{m c}\left(T^4-T_0^4\right)\end{aligned}$
Here, $T=T_0+\Delta T$
$\begin{aligned} \Rightarrow \quad-\frac{d T}{d t} & =\frac{\varepsilon \sigma A}{m c}\left[\left(T_0+\Delta T\right)^4-T_0^4\right] \\ & =\frac{\varepsilon \sigma A}{m c} T_0^4\left[\left(1+\frac{\Delta T}{T_0}\right)^4-1\right] \\ & =\frac{\varepsilon \sigma A}{m c} T_0^4\left[1+\frac{4 \Delta T}{T_0}-1\right] \quad\left(\because \Delta T \ll T_0\right) \\ & =\frac{\varepsilon \sigma A 4 T_0^3}{m c} \Delta T \\ \Rightarrow \quad-\frac{d T}{d t} & =k \Delta T\end{aligned}$
where, $\quad k=\frac{\varepsilon \sigma A 4 T_0^3}{m c}$
$\therefore \quad \frac{d T}{d t}=-k\left(T-T_0\right)$
This is Newton's law of cooling.
Hence, Newton's law of cooling is a special case of Stefan's law.