As the temperature differences are small, we can use Newton's law of cooling.

$\frac{d\theta }{dt}=-kt(\theta -{\theta }_0)$

$\frac{d\theta }{\theta -{\theta }_0}=-kdt$

or,                                               ...(i)

Where k is a constant, $\theta$ is the temperature of the body at time t and ${\theta }_0={16}^{\circ }C$ is the temperature of the surrounding. We have,

${\int }_{{40}^{\circ }C}^{{36}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt\left(5min\right)$

$ln\frac{{36}^{\circ }C-{16}^{\circ }C}{{40}^{\circ }C-{16}^{\circ }C}=-kt\left(5min\right)$

or,

                                     $k=-\frac{ln\left(\frac{5}{6}\right)}{5min}$

or, .

If t be the time required for the temperature to fall from 36${}^{\circ }$C to 32${}^{\circ }$C then by (i),

${\int }_{{36}^{\circ }C}^{{32}^{\circ }C}\frac{d\theta }{\theta -{\theta }_0}=-kt$

or,        $ln\frac{{32}^{\circ }C-{16}^{\circ }C}{{36}^{\circ }C-{16}^{\circ }C}=-\frac{ln\left(\frac{5}{6}\right)}{5min}$

or,          t=$\frac{ln\left(\frac{4}{5}\right)}{\left(ln\right(\frac{5}{6})}\times 5min$

= 6.1 min.

Alternative method

The mean temperature of the body as it cools from 40${}^{\circ }$C to 36${}^{\circ }$C is  $\frac{{40}^{\circ }C+{36}^{\circ }C}{2}={38}^{\circ }C$ The rate of decrease of temperature is

  $\frac{{40}^{\circ }C+{36}^{\circ }C}{5min}={0.80}^{\circ }C/min$ .

Newton's law of cooling is

$\frac{d\theta }{dt}=-k(\theta -{\theta }_0)$

or, -0.8${}^{\circ }$C/min = - k(38${}^{\circ }$C - 16${}^{\circ }$C)=-k(22${}^{\circ }$C)

or, $k=\frac{0.8}{22}mi{n}^{-1}$.

Let the time taken for the temperature to become 32${}^{\circ }$C be t.

During this period,.

The mean temperature is

Now,