$A$ 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 $50\,^{\circ}C$?

Two circular discs $A$ and $B$ with equal radii are blackened. They are heated to the same temperature and are cooled under identical conditions. What inference do you draw from their cooling curves shown in the figure?

It takes $1 \, min$ for hot water to cool from $80^{\circ}C$ to $60^{\circ}C$. Find the time taken (in $sec$) to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the surroundings is $30^{\circ}C$.

$A$ vessel contains hot water at $100^{\circ}C$. If it cools to $80^{\circ}C$ in time $T_1$ and from $80^{\circ}C$ to $60^{\circ}C$ in time $T_2$,then:

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

Hot water kept in a beaker placed in a room cools from $70^{\circ}C$ to $60^{\circ}C$ in $4$ minutes. The time taken by it to cool from $69^{\circ}C$ to $59^{\circ}C$ will be