$A$ calorimeter contains $10 \ g$ of water at $20^{\circ}C$. The temperature falls to $15^{\circ}C$ in $10 \ min$. When the calorimeter contains $20 \ g$ of water at $20^{\circ}C$,it takes $15 \ min$ for the temperature to become $15^{\circ}C$. The water equivalent of the calorimeter is: (in $g$)

$A$ body cools from $60^{\circ} C$ to $40^{\circ} C$ in $6$ minutes. After the next $6$ minutes,its temperature will be (Temperature of the surroundings is $10^{\circ} C$). (in $^{\circ} C$)

$A$ body cools from $80^{\circ} C$ to $60^{\circ} C$ in $5$ minutes. The temperature of the surrounding is $20^{\circ} C$. The time it takes to cool from $60^{\circ} C$ to $40^{\circ} C$ is........... $s$

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5 \, s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

Assuming Newton's law of cooling to be valid,a body at temperature $50^{\circ} C$ in a surrounding of temperature $20^{\circ} C$ achieves a steady state with the help of a $100 \ W$ heater. If the same body has a temperature of $35^{\circ} C$ in the same surrounding,then the power of the heater required to maintain a steady state is ........ $W$.

In an experiment to verify Newton's law of cooling,a graph is plotted between the temperature difference $(\Delta T)$ of the water and surroundings and time as shown in the figure. The initial temperature of water is taken as $80^{\circ}C$. The value of $t_{2}$ as mentioned in the graph will be...........