$A$ bowl filled with very hot soup cools from $98^{\circ}C$ to $86^{\circ}C$ in $2$ minutes when the room temperature is $22^{\circ}C$. How many minutes will it take to cool from $75^{\circ}C$ to $69^{\circ}C$?

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 .......... $min$.

$A$ container with $1 \ kg$ of water is kept in sunlight, which causes the water to get warmer than the surroundings. The average energy per unit time per unit area received from sunlight is $700 \ W \ m^{-2}$ and it is absorbed by the water over an effective area of $0.05 \ m^2$. Assuming that the heat loss from the water to the surroundings is governed by Newton's law of cooling, the difference (in $^{\circ}C$) in the temperature of water and the surroundings after a long time will be. . . . . . . . (Ignore the effect of the container, and take the constant for Newton's law of cooling $k = 0.001 \ s^{-1}$, Heat capacity of water $s = 4200 \ J \ kg^{-1} \ K^{-1}$)

Describe the procedure to demonstrate that the rate of loss of heat from a hot body is directly dependent on the temperature difference between the body and its surroundings.

$A$ 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 $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

$A$ 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 $.....^{\circ}C$.