$A$ 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 $^{\circ} C$)

$A$ body takes $5 \ min$ to cool from $90^{\circ}C$ to $60^{\circ}C$. If the temperature of the surroundings is $20^{\circ}C$,how much time (in $min$) will it take to cool from $60^{\circ}C$ to $30^{\circ}C$?

$A$ solid copper cube of edges $1\;cm$ is suspended in an evacuated enclosure. Its temperature is found to fall from $100^{\circ}C$ to $99^{\circ}C$ in $100\;s$. Another solid copper cube of edges $2\;cm$,with similar surface nature,is suspended in a similar manner. The time required for this cube to cool from $100^{\circ}C$ to $99^{\circ}C$ will be approximately ...... $s$.

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\,minutes$ and to $42\,^{\circ}C$ in the next $10\,minutes$. The temperature of the surroundings is ......... $^{\circ}C$

If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_0$,the graph between the temperature $T$ of the metal and time $t$ will be closest to:

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