A liquid in a beaker has temperature $\theta (t)$ at time $t$ and $\theta_0$ is temperature of surroundings, then according to Newton's law of cooling the correct graph between loge $log_e(\theta – \theta_0) $ and $t$ is

A body cools in a surrounding which is at a constant temperature of ${\theta _0}$. Assume 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 $P(\theta = {\theta _1})$ and $Q(\theta = {\theta _2})$. These tangents meet the time axis at angles of ${\varphi _2}$and ${\varphi _1}$, as shown

For a system with newtons law of cooling applicable the initial rate of cooling is $R^0\  C/sec$ find the time when temperature diff. $\Delta T_0 =$ initial temperature difference, is reduced to half.

A pan filled with hot food cools from $94\,^{\circ} C$ to $86\,^{\circ} C$ in $2$ minutes when the room temperature is at $20\,^{\circ} C$. How long (in $sec$) will it take to cool from $71\,^{\circ} C$ to $69\,^{\circ} C ?$

It takes $10$ minutes to cool a liquid from $61^oC$ to $59^oC$. If room temperature is $30^oC$ then time taken in cooling from $51^oC$ to $49^oC$ is ……. $\min$