Hot water kept in a beaker placed in a room cools from $70^{\circ}C$ to $60^{\circ}C$ in $4$ minutes. The time taken by it to cool from $69^{\circ}C$ to $59^{\circ}C$ will be

$A$ container is filled with a liquid that cools from $100^{\circ}C$ to $70^{\circ}C$ in $5 \text{ min}$, when kept at a room temperature of $30^{\circ}C$. The time that it must have taken to cool down to $80^{\circ}C$ from its initial temperature is approximately: (in $\text{ min}$)

In Newton's experiment of cooling,the water equivalent of two similar calorimeters is $10 \ g$ each. They are filled with $350 \ g$ of water and $300 \ g$ of a liquid (equal volumes) separately. The time taken by water and liquid to cool from $70^{\circ}C$ to $60^{\circ}C$ is $3 \ min$ and $95 \ s$ respectively. The specific heat of the liquid will be ...... $Cal/g^{\circ}C$.

$A$ body takes $10 \, \text{minutes}$ to cool from $60 \, ^\circ\text{C}$ to $50 \, ^\circ\text{C}$. The temperature of the surroundings is constant at $25 \, ^\circ\text{C}$. Then, the temperature of the body after the next $10 \, \text{minutes}$ will be approximately ....... $^\circ\text{C}$

$A$ container contains hot water at $100^{\circ}C$. If in time $T_1$ the temperature falls to $80^{\circ}C$ and in time $T_2$ the temperature falls to $60^{\circ}C$ from $80^{\circ}C$,then:

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: