Two hot bodies $A$ and $B$ have temperatures $100^{\circ}C$ and $80^{\circ}C$ respectively. The surrounding temperature is $40^{\circ}C$. The ratio of their rates of cooling $R_1 : R_2$ at $t = 0$ is:

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

$A$ 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 $39.9^{\circ}C$? Assume the temperature of surroundings to be $30.0^{\circ}C$ and Newton's law of cooling to be valid. The time taken is ....... $s$.

$A$ 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 $g$)

The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ 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 ......... $^{\circ}C$.