$A$ cup of tea cools from $80\,^{\circ}C$ to $60\,^{\circ}C$ in $1\,min$. The ambient temperature is $30\,^{\circ}C$. In cooling from $60\,^{\circ}C$ to $50\,^{\circ}C$,it will take ....... $sec$.

$A$ body cools from $50.0^{\circ}C$ to $49.9^{\circ}C$ in $5 \, s$. How long will it take to cool from $40.0^{\circ}C$ to $39.9^{\circ}C$ (in $, s$)? The temperature of the surroundings is $30^{\circ}C$. Assume Newton's Law of Cooling applies.

An object cools from $100^{\circ} C$ to $40^{\circ} C$ in $10$ minutes, when the surrounding temperature is $10^{\circ} C$. Then the time taken by the object to cool from $70^{\circ} C$ to $20^{\circ} C$ is
$ [\text{Take } \ln 2=0.7, \ln 3=1.1, \ln 6=1.8 ]$ (in $min$)

An object kept in a large room having air temperature of $25^{\circ}C$ takes $12 \text{ minutes}$ to cool from $80^{\circ}C$ to $70^{\circ}C$. The time taken to cool for the same object from $70^{\circ}C$ to $60^{\circ}C$ would be nearly.....$min$

$A$ cup of coffee cools from $90^{\circ} C$ to $80^{\circ} C$ in $t$ minutes when the room temperature is $20^{\circ} C$. The time taken by the similar cup of coffee to cool from $80^{\circ} C$ to $60^{\circ} C$ at the same room temperature is $:$

$A$ mass of $50\,g$ of water in a closed vessel,with surroundings at a constant temperature,takes $2\,minutes$ to cool from $30\,^oC$ to $25\,^oC$. $A$ mass of $100\,g$ of another liquid in an identical vessel with identical surroundings takes the same time to cool from $30\,^oC$ to $25\,^oC$. The specific heat of the liquid is .......... $kcal/(kg \cdot ^oC)$ (The water equivalent of the vessel is $30\,g$).