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 $30^{\circ}C$.

$A$ liquid cools from $50^oC$ to $45^oC$ in $5$ minutes and from $45^oC$ to $41.5^oC$ in the next $5$ minutes. The temperature of the surrounding is ...... $^oC$.

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:

The circuit below is used to heat water kept in a bucket. Assuming heat loss only by Newton's law of cooling, the variation in the temperature of the water in the bucket as a function of time is depicted by

$A$ metal sphere cools at a rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. When the temperature of the sphere is $40^{\circ} C$,its rate of cooling is $0.3^{\circ} C / min$. The temperature of the surrounding $\left(\theta_0\right)$ is (in $^{\circ} C$)

The temperature of a body in air falls from $40^{\circ} \text{C}$ to $24^{\circ} \text{C}$ in $4 \text{ minutes}$. The temperature of the air is $16^{\circ} \text{C}$. The temperature of the body in the next $4 \text{ minutes}$ will be $:$