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

A solid cube and a solid sphere of identical material and equal masses are heated to the same temperature and left to cool in the same surroundings. Then,

Cooling rate of a sphere of $600\,K$ at external environment $(200\,K)$ is $R$ . When the temperature of sphere is reduced to $400\,K$ then cooling rate of the sphere becomes

An object is cooled from $75°C$ to $65°C$ in $2$ minutes in a room at $30°C$ . The time taken to cool another object from $55°C$ to $45°C$ in the same room in minutes is

In $5\, minutes,$ a body cools from $75^{\circ} {C}$ to $65^{\circ} {C}$ at room temperature of $25^{\circ} {C}$. The temperature of body at the end of next $5\, minutes$ is $......\,{ }^{\circ} {C} .$

The top of a lake gets frozen at a place where the surrounding air is at a temperature of $-20\,^oC$. Then