$A$ body cools from a temperature of $60^{\circ} C$ to $50^{\circ} C$ in $10 \text{ minutes}$ and $50^{\circ} C$ to $40^{\circ} C$ in $15 \text{ minutes}$. The time taken in minutes for the body to cool from $40^{\circ} C$ to $30^{\circ} C$ is

$A$ copper cube of side $a$ is heated and allowed to cool in a vacuum. It takes time $t$ to cool from temperature $\theta_1$ to $\theta_2$. Now,another copper cube of side $2a$ is allowed to cool in the same environment. How much time will it take to cool from $\theta_1$ to $\theta_2$?

$A$ hot body,obeying Newton's law of cooling,is cooling down from its peak value $80\,^oC$ to an ambient temperature of $30\,^oC$. It takes $5\,minutes$ to cool down from $80\,^oC$ to $40\,^oC$. How many minutes will it take to cool down from $62\,^oC$ to $32\,^oC$? (Given $\ln 2 = 0.693, \ln 5 = 1.609$)

$A$ body takes $5$ minutes for cooling from $50^oC$ to $40^oC$. Its temperature comes down to $33.33^oC$ in the next $5$ minutes. The temperature of the surroundings is ....... $^oC$.

$A$ body cools in a surrounding which is at a constant temperature of $\theta_0$. Assuming that it obeys Newton's law of cooling,its temperature $\theta$ is plotted against time $t$. Tangents are drawn to the curve at the points $A(\theta = \theta_1)$ and $B(\theta = \theta_2)$. These tangents meet the time-axis at angles $\alpha_1$ and $\alpha_2$ as shown in the graph. Then:

$A$ 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 $W$)