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 $= 30^{\circ} C$)

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:

$A$ beaker full of hot water is kept in a room. If it cools from $80^{\circ} C$ to $75^{\circ} C$ in $t_1$ minutes,from $75^{\circ} C$ to $70^{\circ} C$ in $t_2$ minutes,and from $70^{\circ} C$ to $65^{\circ} C$ in $t_3$ minutes,then:

$A$ solid sphere and a hollow sphere of the same material and size are heated to the same temperature and allowed to cool in the same surroundings. If the temperature difference between each sphere and its surroundings is $T$,then

$A$ 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:

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