Equal masses of two liquids are filled in two similar calorimeters. The rate of cooling will

$A$ 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 $cal/sec$?

If the initial temperatures of a metallic sphere and a disc,of the same mass,radius,and material,are equal,then the ratio of their rate of cooling in the same environment will be:

$A$ 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 $W$)

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 ......... $^{\circ}C$

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)