Hot water cools from $60^oC$ to $50^oC$ in the first $10$ minutes and to $42^oC$ in the next $10$ minutes. The temperature of the surrounding is ......... $^oC$.

Describe the procedure to demonstrate that the rate of loss of heat from a hot body is directly dependent on the temperature difference between the body and its surroundings.

$A$ container contains hot water at $100^{\circ}C$. If in time $T_1$ the temperature falls to $80^{\circ}C$ and in time $T_2$ the temperature falls to $60^{\circ}C$ from $80^{\circ}C$,then:

$A$ mass of $50\,g$ of water in a closed vessel,with surroundings at a constant temperature,takes $2\,minutes$ to cool from $30\,^oC$ to $25\,^oC$. $A$ mass of $100\,g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30\,^oC$ to $25\,^oC$. The specific heat of the liquid is .......... $kcal/(kg \cdot ^oC)$ (The water equivalent of the vessel is $30\,g$).

$A$ metal ball of mass $1 \ kg$ is heated using a $40 \ W$ heater in a room at $30^{\circ} C$. The temperature of the ball becomes steady at $70^{\circ} C$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} C$ is (in $W$)

$A$ metal rod cools at the rate of $4^{\circ}C/min$ when its temperature is $90^{\circ}C$ and at the rate of $1^{\circ}C/min$ when its temperature is $30^{\circ}C$. The temperature of the surrounding is: (in $^{\circ}C$)