$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes when kept in air at $30^{\circ}C$. In the next $10$ minutes,its temperature will be:

$A$ body cools from $70^{\circ} C$ to $40^{\circ} C$ in $5$ minutes. Calculate the time it takes to cool from $60^{\circ} C$ to $30^{\circ} C$. The temperature of the surroundings is $20^{\circ} C$. (in $min.$)

$A$ hot body is allowed to cool. The surrounding temperature is constant at $30^{\circ} C$. It takes time $t_{1}$ to cool from $70^{\circ} C$ to $68^{\circ} C$ and time $t_{2}$ to cool from $60^{\circ} C$ to $59.5^{\circ} C$. Then:

Two metal spheres,one solid and one hollow,are heated to $300^{\circ}C$ and allowed to cool in the same environment. The rate of heat loss will be:

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

Hot water cools from $80^{\circ} C$ to $60^{\circ} C$ in $1$ minute. How long will it take to cool from $60^{\circ} C$ to $50^{\circ} C$ (in $s$)? (Room temperature $= 30^{\circ} C$)