According to Newton's law of cooling, the rate of loss of heat, $\left(-\frac{d}{d t}\right)$ of the body is directly proportional to the difference of temperature $\Delta \mathrm{T}=\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$ of the body and the surroundings.

The law holds good only for small difference of temperature.

The loss of heat by radiation depends upon the nature of the surface of the body and the area of the exposed surface.

Hence, we can write $-\frac{d}{d t}=k\left(\mathrm{~T}_{2}-\mathrm{T}_{1}\right) \ldots$ $(1)$

where $k$ is a positive constant depending upon the area and nature of the surface of the body. Suppose a body of mass $m$ and specific heat capacity $s$ is at temperature $\mathrm{T}_{2} .$ Let $\mathrm{T}_{1}$ be the temperature of the surroundings.

If the temperature falls by a small amount $d \mathrm{~T}_{2}$ in time $d t$, then the amount of heat lost is, $d \mathrm{Q}=m s \mathrm{dT}$

$\therefore$ Rate of loss of heat is given by

$\frac{d \mathrm{Q}}{d t}=m s \frac{d \mathrm{~T}}{d t}$

$\ldots$ $(2)$

From Equ.$(1)$ and $(2)$ we have,

$-m s \frac{d \mathrm{~T}}{d t}=\mathrm{k}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)$

$\therefore \frac{d \mathrm{~T}}{\mathrm{~T}_{2}-\mathrm{T}_{1}}=\frac{-\mathrm{k}}{m s} \cdot d t$ $…(3)$

$\therefore \frac{d \mathrm{~T}}{\mathrm{~T}_{2}-\mathrm{T}_{1}}=-\mathrm{K} d t$$\text { where } \mathrm{K}=\frac{k}{m s}$

On integrating,

$\log _{\mathrm{e}}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right)=-\mathrm{K} t+\mathrm{C}$

or $\mathrm{T}_{2}=\mathrm{T}_{1}+\mathrm{e}^{-k t}$; where $\mathrm{C}^{\prime}=\mathrm{e}^{\mathrm{C}}$

Equation enables to calculate the time of cooling of a body through a particular range of temperature.

For small temperature differences, the rate of cooling, due to conduction, convection, and radiation combined, is proportional to the difference in temperature.