$A$ body cools in $7$ minutes from $60^{\circ}C$ to $40^{\circ}C$. What time (in minutes) does it take to cool from $40^{\circ}C$ to $28^{\circ}C$ if the surrounding temperature is $10^{\circ}C$? Assume Newton's Law of cooling holds.

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120^{\circ}C$,then

$A$ bowl filled with very hot water cools from $98^\circ C$ to $86^\circ C$ in $2$ minutes when the room temperature is $22^\circ C$. How long will it take to cool from $75^\circ C$ to $69^\circ C$?

$A$ solid cube and a solid sphere of the same material have equal surface area. Both are at the same temperature $120\ ^oC$,then:

Assuming Newton's law of cooling to be valid,a body at temperature $50^{\circ} C$ in a surrounding of temperature $20^{\circ} C$ achieves a steady state with the help of a $100 \ W$ heater. If the same body has a temperature of $35^{\circ} C$ in the same surrounding,then the power of the heater required to maintain a steady state is ........ $W$.

$A$ 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 $\,s$)