An ordinary body cools from $4\theta$ to $3\theta$ in $t$ minutes. The temperature of the body after the next $t$ minutes is (Assume Newton's law of cooling and room temperature as $\theta$).

$A$ pan filled with hot food cools from $94^{\circ} C$ to $86^{\circ} C$ in $2$ minutes when the room temperature is $20^{\circ} C$. How long will it take to cool from $71^{\circ} C$ to $69^{\circ} C$?

$A$ hot body,obeying Newton's law of cooling,is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$. It takes $5\,minutes$ to cool down from $80\,^oC$ to $40\,^oC$. How many minutes will it take to cool down from $62\,^oC$ to $32\,^oC$? (Given $\ln 2 = 0.693, \ln 5 = 1.609$)

$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and assuming Newton's law of cooling to hold good,the temperature of the body at the end of the next $10$ minutes will be ......... $^{\circ}C$.

It takes $10$ minutes to cool a liquid from $61^{\circ} C$ to $59^{\circ} C$. If the room temperature is $30^{\circ} C$,then the time taken in cooling from $51^{\circ} C$ to $49^{\circ} C$ is .......... $min$.

The temperature of an iron block falls from $100 \, ^\circ\text{C}$ to $90 \, ^\circ\text{C}$ in time $t_1$,from $90 \, ^\circ\text{C}$ to $80 \, ^\circ\text{C}$ in time $t_2$,and from $80 \, ^\circ\text{C}$ to $70 \, ^\circ\text{C}$ in time $t_3$. Then: