$A$ body cools from $60^{\circ}C$ to $50^{\circ}C$ in $10$ minutes. If the room temperature is $25^{\circ}C$ and Newton's law of cooling is applicable,what will be the temperature of the body after another $10$ minutes in $^{\circ}C$?

According to Newton's law of cooling,the rate of cooling of a body is proportional to $(\Delta \theta )^n$,where $\Delta \theta$ is the difference of the temperature of the body and the surroundings,and $n$ is equal to

Two identical beakers $A$ and $B$ contain equal volumes of two different liquids at $60\,^oC$ each and are left to cool down. The liquid in $A$ has a density of $8 \times 10^2\, kg/m^3$ and a specific heat of $2000\, Jkg^{-1}K^{-1}$,while the liquid in $B$ has a density of $10^3\, kg/m^3$ and a specific heat of $4000\, Jkg^{-1}K^{-1}$. Which of the following best describes their temperature versus time graph schematically? (Assume the emissivity of both the beakers to be the same.)

$A$ body takes $5$ minutes for cooling from $50^oC$ to $40^oC$. Its temperature comes down to $33.33^oC$ in the next $5$ minutes. The temperature of the surroundings is ....... $^oC$.

An object is cooled from $75^{\circ} C$ to $65^{\circ} C$ in $2 \text{ min}$. The time it takes to cool from $55^{\circ} C$ to $45^{\circ} C$ is [The temperature of the surrounding is $30^{\circ} C$] (in $\text{ min}$)

$A$ can is taken out from a refrigerator at $0^{\circ}C$. The atmospheric temperature is $25^{\circ}C$. If $t_1$ is the time taken to heat from $0^{\circ}C$ to $5^{\circ}C$ and $t_2$ is the time taken to heat from $10^{\circ}C$ to $15^{\circ}C$,then: