$A$ body cools from a temperature $60\,^{\circ}\text{C}$ to $40\,^{\circ}\text{C}$ in $10$ minutes. The room temperature is $20\,^{\circ}\text{C}$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be......... $^{\circ}\text{C}$.

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120\ ^oC$,then:

Two liquids of equal volume are cooled under identical conditions from $60^{\circ}C$ to $50^{\circ}C$ in $324 \; sec$ and $810 \; sec$ respectively. If the ratio of their specific heats is $3:4$,find the ratio of their densities. (The water equivalent of the calorimeter is negligible.)

$A$ body cools from $80^{\circ} C$ to $50^{\circ} C$ in $5 \text{ min}$. In the next time of $t \text{ min}$, the body continues to cool from $50^{\circ} C$ to $30^{\circ} C$. The total time taken by the body to cool from $80^{\circ} C$ to $30^{\circ} C$ is
[The temperature of the surroundings is $20^{\circ} C$.] (in $\text{ min}$)

The temperature of a body $\theta$ is slightly more than the temperature of the surrounding $\theta_0$. Its rate of cooling $(R)$ versus the temperature of the body $(\theta)$ is plotted. Its shape would be:

An object is cooled from $75^{\circ} C$ to $65^{\circ} C$ in $2 \text{ min}$. The time it takes to cool from $55^{\circ} C$ to $45^{\circ} C$ is [The temperature of the surrounding is $30^{\circ} C$] (in $\text{ min}$)