$A$ cup of tea cools from $80^{\circ}C$ to $60^{\circ}C$ in $1$ minute. The ambient temperature is $30^{\circ}C$. In cooling from $60^{\circ}C$ to $50^{\circ}C$,it will take ....... $\text{sec}$.

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$

$A$ solid copper cube of edges $1\;cm$ is suspended in an evacuated enclosure. Its temperature is found to fall from $100^{\circ}C$ to $99^{\circ}C$ in $100\;s$. Another solid copper cube of edges $2\;cm$,with similar surface nature,is suspended in a similar manner. The time required for this cube to cool from $100^{\circ}C$ to $99^{\circ}C$ will be approximately ...... $s$.

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

$A$ metallic sphere cools from $50^{\circ}C$ to $40^{\circ}C$ in $300 \, s$. If the atmospheric temperature is $20^{\circ}C$,then the sphere's temperature after the next $5$ minutes will be close to $.....^{\circ}C$.

$A$ body cools in a surrounding which is at a constant temperature of $\theta_0$. Assuming that it obeys Newton's law of cooling,its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $A(\theta = \theta_1)$ and $B(\theta = \theta_2)$. These tangents meet the time-axis at angles $\alpha_1$ and $\alpha_2$ as shown in the graph. Then: