$A$ body takes $10$ minutes to cool down from $62^{\circ}C$ to $50^{\circ}C$. If the temperature of the surrounding is $26^{\circ}C$,then in the next $10$ minutes,the temperature of the body will be ......... $^{\circ}C$.

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. After the next $6$ minutes,its temperature will be (Temperature of the surroundings is $10^{\circ} C$). (in $^{\circ} C$)

$A$ 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:

$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)

What is Newton's law of cooling used to determine in a laboratory?

$A$ metal sphere cools at the rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. At what rate will it cool when its temperature falls to $50^{\circ} C$? [Temperature of surrounding is $30^{\circ} C$]