$A$ black body calorimeter filled with hot water cools from $60^{\circ}C$ to $50^{\circ}C$ in $4 \text{ min}$ and from $40^{\circ}C$ to $30^{\circ}C$ in $8 \text{ min}$. The approximate temperature of the surrounding is ........ $^{\circ}C$.

Assuming Newton's law of cooling to be valid,a body at temperature $50^{\circ} C$ in a surrounding of temperature $20^{\circ} C$ achieves a steady state with the help of a $100 \ W$ heater. If the same body has a temperature of $35^{\circ} C$ in the same surrounding,then the power of the heater required to maintain a steady state is ........ $W$.

For a small temperature difference between the body and the surroundings,the relation between the rate of loss of heat $R$ and the temperature of the body $\theta$ is depicted by:

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

$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1$ minute. The ambient temperature is $30\,^{\circ}C$. How long will it take to cool from $60\,^{\circ}C$ to $50\,^{\circ}C$?

$A$ body takes $5 \text{ min}$ to cool from $80^{\circ}C$ to $50^{\circ}C$. How many minutes will it take to cool from $60^{\circ}C$ to $30^{\circ}C$,if the room temperature is $20^{\circ}C$?