$A$ metal sphere cools at the rate of $1.5^{\circ} C / min$ when its temperature is $80^{\circ} C$. At what rate will it cool when its temperature falls to $50^{\circ} C$? [Temperature of surrounding is $30^{\circ} C$]

$A$ solid sphere and a hollow sphere of the same material and same radius are heated to the same temperature. They are placed in an environment of the same temperature. If the temperature difference between both and the surroundings is $T$,then .......

$A$ metal ball of mass $1 \ kg$ is heated using a $40 \ W$ heater in a room at $30^{\circ} C$. The temperature of the ball becomes steady at $70^{\circ} C$. Assuming Newton's law of cooling, the rate of loss of heat to the surrounding when the ball is at $40^{\circ} C$ is (in $W$)

The temperature of a body $\theta$ is slightly more than the temperature of the surrounding $\theta_0$. Its rate of cooling $(R)$ versus the temperature of the body $(\theta)$ is plotted. Its shape would be:

Hot water cools from $60\,^{\circ}C$ to $50\,^{\circ}C$ in the first $10\, min$ and to $42\,^{\circ}C$ in the next $10\, min$. The temperature of the surrounding is ........ $^{\circ}C$.

$A$ heating element of mass $100 \,g$ and having specific heat of $1 \,J/(g^{\circ}C)$ is exposed to surrounding air at $27^{\circ}C$. The element attains a steady state temperature of $127^{\circ}C$, while absorbing $100 \,W$ of electric power. If the power is switched off, then the approximate time taken by the element to cool down to $126^{\circ}C$ will be (neglect radiation). (in $\,s$)