$A$ sphere having temperature $600 \text{ K}$ is losing heat due to radiation. At this temperature its rate of cooling is $R$. The rate of cooling of this sphere at $400 \text{ K}$ is $\frac{x}{243}R$. The value of $x$ is (Temperature of surrounding is $300 \text{ K}$)

The temperature of a body falls from $50^oC$ to $40^oC$ in $10$ minutes. If the temperature of the surroundings is $20^oC$,then the temperature of the body after another $10$ minutes will be ........ $^oC$.

Discuss the experiment verifying Newton's law of cooling.

Two friends $A$ and $B$ are waiting for another friend for tea. $A$ took the tea in a cup and mixed the cold milk and then waits. $B$ took the tea in the cup and then mixed the cold milk when the friend comes. Then the tea will be hotter in the cup of

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:

$A$ 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 $g$)