$A$ liquid cools down from $70^{\circ}C$ to $60^{\circ}C$ in $5$ minutes. The time taken to cool it from $60^{\circ}C$ to $50^{\circ}C$ will be

The temperature of hot water falls from $60^oC$ to $50^oC$ in $10 \, min$. It takes another $10 \, min$ to fall from $50^oC$ to $42^oC$. Find the temperature of the surroundings in $^oC$.

$A$ vessel contains hot water at $100^{\circ}C$. If it cools to $80^{\circ}C$ in time $T_1$ and from $80^{\circ}C$ to $60^{\circ}C$ in time $T_2$,then:

Two bottles $A$ and $B$ have radii $R_{A}$ and $R_{B}$ and heights $h_{A}$ and $h_{B}$ respectively,with $R_{B}=2 R_{A}$ and $h_{B}=2 h_{A}$. These are filled with hot water at $60^{\circ} C$. Consider that heat loss for the bottles takes place only from side surfaces. If the time the water takes to cool down to $50^{\circ} C$ is $t_{A}$ and $t_{B}$ for bottles $A$ and $B$,respectively,then $t_{A}$ and $t_{B}$ are best related as:

$A$ body cools from a temperature $60\,^{\circ}\text{C}$ to $40\,^{\circ}\text{C}$ in $10$ minutes. The room temperature is $20\,^{\circ}\text{C}$. Assume that Newton's law of cooling is applicable. The temperature of the body at the end of the next $10$ minutes will be......... $^{\circ}\text{C}$.

The temperature of a body $\theta$ is slightly more than the temperature of the surrounding $\theta_0$. Its rate of cooling $(R)$ versus the temperature of the body $(\theta)$ is plotted. Its shape would be: