$A$ liquid is filled in a vessel which is kept in a room with a temperature of $20^{\circ}C$. When the temperature of the liquid is $80^{\circ}C$,it loses heat at the rate of $60 \; cal/sec$. What will be the rate of loss of heat when the temperature of the liquid is $40^{\circ}C$ in $cal/sec$?

Two spheres of the same metal have radii $r$ and $2r$. They are heated to the same temperature and placed in the same environment. The ratio of their rates of cooling will be:

$A$ certain quantity of water cools from $70\,^{\circ}C$ to $60\,^{\circ}C$ in the first $10\, minutes$ and to $54\,^{\circ}C$ in the next $10\, minutes$. The temperature of the surrounding is ......... $^{\circ}C$.

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

$A$ metal sphere cools at the rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. At what rate will it cool when its temperature falls to $50^{\circ} C$? [Temperature of surrounding is $30^{\circ} C$]

Two bodies $A$ and $B$ of equal masses,surface area,and emissivity are cooling under Newton's law of cooling from the same initial temperature. Their cooling curves are represented by the graph. If $\theta$ is the instantaneous temperature of the body and $\theta_0$ is the temperature of the surroundings,then the relationship between their specific heats $S_A$ and $S_B$ is: