$A$ body cools down from $52.5^{\circ} C$ to $47.5^{\circ} C$ in $5 \ min$ and from $47.5^{\circ} C$ to $42.5^{\circ} C$ in $7.5 \ min$. Then,the temperature of the surroundings is (in $^{\circ} C$)

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

An object is cooled from $75^{\circ}C$ to $65^{\circ}C$ in $2$ minutes in a room at $30^{\circ}C$. The time taken to cool another object from $55^{\circ}C$ to $45^{\circ}C$ in the same room in minutes is

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

$A$ body cools from $80\,^{\circ} C$ to $50\,^{\circ} C$ in $5$ minutes. Calculate the time (in $min$) it takes to cool from $60\,^{\circ} C$ to $30\,^{\circ} C$. The temperature of the surroundings is $20\,^{\circ} C$.

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