$A$ metal sphere cools at a rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. When the temperature of the sphere is $40^{\circ} C$,its rate of cooling is $0.3^{\circ} C / min$. The temperature of the surrounding $\left(\theta_0\right)$ is (in $^{\circ} C$)

$A$ liquid in a beaker has temperature $\theta(t)$ at time $t$ and $\theta_0$ is the temperature of the surroundings. According to Newton's law of cooling,which of the following is the correct graph between $\log_e(\theta - \theta_0)$ and $t$?

$A$ hot liquid kept in a beaker cools from $80\,^{\circ}C$ to $70\,^{\circ}C$ in $2\, \text{min}$. If the surrounding temperature is $30\,^{\circ}C$,then the time taken for the same liquid to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$ is ........ $\text{sec}$.

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.

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

There are two spherical balls $A$ and $B$ of the same material with the same surface,but the diameter of $A$ is half that of $B$. If $A$ and $B$ are heated to the same temperature and then allowed to cool,then