If the initial temperatures of a metallic sphere and a disc,of the same mass,radius,and material,are equal,then the ratio of their rate of cooling in the same environment will be:

$A$ solid sphere and a hollow sphere of the same material and same radius are heated to the same temperature. They are placed in an environment of the same temperature. If the temperature difference between both and the surroundings is $T$,then .......

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:

$A$ 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 $s = 4200 \ J \ kg^{-1} \ K^{-1}$)

$A$ body cools in $7$ minutes from $60^{\circ}C$ to $40^{\circ}C$. What time (in minutes) does it take to cool from $40^{\circ}C$ to $28^{\circ}C$ if the surrounding temperature is $10^{\circ}C$? Assume Newton's Law of cooling holds.

If a liquid takes $30 \; s$ in cooling from $80^{\circ} C$ to $70^{\circ} C$ and $70 \; s$ in cooling from $60^{\circ} C$ to $50^{\circ} C$,then find the room temperature. (in $^{\circ} C$)