$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5\;s$. How long will it take to cool from $40.0^{\circ}C$ 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. The time taken is ....... $s$.

Two bodies $A$ and $B$ of equal masses,surface area,and emissivity are cooling under Newton's law of cooling from the same initial temperature. Their cooling curves are represented by the graph. If $\theta$ is the instantaneous temperature of the body and $\theta_0$ is the temperature of the surroundings,then the relationship between their specific heats $S_A$ and $S_B$ is:

$A$ certain quantity of water cools from $70^\circ C$ to $60^\circ C$ in the first $5$ minutes and to $54^\circ C$ in the next $5$ minutes. The temperature of the surroundings is ..... $^\circ C$

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1$ minute. The ambient temperature is $30\,^{\circ}C$. How long will it take to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$?

$A$ 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 takes $5 \text{ min}$ to cool from $80^{\circ}C$ to $50^{\circ}C$. How many minutes will it take to cool from $60^{\circ}C$ to $30^{\circ}C$,if the room temperature is $20^{\circ}C$?