A body cools from $${50^ \circ }C$$  to $${49.9^ \circ }C$$  in $$5\,s.$$  How long will it take to cool from $${40^ \circ }C$$  to $${39.9^ \circ }C$$  ? (Assume the temperature of surroundings to be $${30.0^ \circ }C$$  and Newton’s law of cooling to be valid)

Solution :
According to Newton's law of cooling, the rate of loss of heat of a body is directly proportional to the difference in temperatures of the body and the surroundings, provided the difference in temperature is small, not more than $${30^ \circ }C.$$
∴ Average rate of fall of temperature $$ \propto $$ average temperature excess
i.e., $$\frac{{dT}}{{dt}} \propto \left( {{T_t} - {T_s}} \right)$$
$$ \Rightarrow \frac{{dT}}{{dt}} = K\left( {{T_t} - {T_s}} \right)$$
According to question for 1^st and 2^nd case,
$$\eqalign{ & \frac{{50.1 - 49.9}}{5} = k\left[ {\frac{{50.1 + 49.9}}{2} - 30} \right]\,......\left( {\text{i}} \right) \cr & \frac{{40.1 - 39.9}}{{t'}} = k\left[ {\frac{{40.1 + 39.9}}{2} - 30} \right]\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing Eq. (i) by Eq. (ii), we get
$$\frac{2}{5} \times \frac{{t'}}{2} = \frac{{20}}{{10}} \Rightarrow t' = 10\,s$$