A hot body, obeying Newton’s law of cooling is cooling down from its peak value ${80}^{\circ }C$ to an  ambient temperature of ${30}^{\circ }C$ .It takes 5 minutes in cooling down from ${80}^{\circ }C$ to  ${40}^{\circ }C$. How much time will it take to cool down from  ${62}^{\circ }C$ to  ${32}^{\circ }C$
in minutes? (Given In2=0.693, In 5=1.609) 
 

From Newton’s law of cooling,
$t=\frac{1}{k}{\log }_e\left(\frac{{\theta }_2-{\theta }_0}{{\theta }_1-{\theta }_0}\right)$ 
From question and above equation,
$5=\frac{1}{k}{\log }_e\frac{(40-30)}{(80-30)}$ 
And, t= $\frac{1}{k}{\log }_e\frac{(32-30)}{(62-30)}$
Dividing equation (2) by (1),
$\frac{t}{5}=\frac{\frac{1}{k}{\log }_e\frac{(32-30)}{(62-30)}}{\frac{1}{k}{\log }_e\frac{(40-30)}{(80-30)}}$ 
On solving we get, time taken to cool down from  ${62}^{\circ }C$ to  ${32}^{\circ }C$, t=8.6 minutes