It takes $1 \, min$ for hot water to cool from $80^{\circ}C$ to $60^{\circ}C$. Find the time taken (in $sec$) to cool from $60^{\circ}C$ to $50^{\circ}C$. The temperature of the surroundings is $30^{\circ}C$.

$A$ body cools from a temperature $3\theta$ to $2\theta$ in $10 \text{ minutes}$. The room temperature is $\theta$. The temperature of the body at the end of the next $10 \text{ minutes}$ is '$x$'. Assuming that Newton's law of cooling is applicable, the value of '$x$' will be

$A$ body takes $5 \text{ min}$ to cool from $80^{\circ}C$ to $50^{\circ}C$. How many minutes will it take to cool from $60^{\circ}C$ to $30^{\circ}C$,if the room temperature is $20^{\circ}C$?

$A$ 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:

If a liquid cools from $95^{\circ}C$ to $90^{\circ}C$ in $30 \, s$ and from $55^{\circ}C$ to $50^{\circ}C$ in $70 \, s$,then the temperature of the surroundings is ...... $^{\circ}C$. (in $.5$)

In a room where the temperature is $30^{\circ}C$,a body cools from $61^{\circ}C$ to $59^{\circ}C$ in $4$ minutes. The time (in min) taken by the body to cool from $51^{\circ}C$ to $49^{\circ}C$ will be ....... $\text{min}$.