If a body cools from a temperature of $62^{\circ} C$ to $50^{\circ} C$ in $10 \, \text{minutes}$ and to $42^{\circ} C$ in the next $10 \, \text{minutes}$, then the temperature of the surroundings is (in $^{\circ} C$)

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

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

Two bodies $A$ and $B$ of equal masses,surface area,and emissivity are cooling under Newton's law of cooling from the same initial temperature. Their cooling curves are represented by the graph. If $\theta$ is the instantaneous temperature of the body and $\theta_0$ is the temperature of the surroundings,then the relationship between their specific heats $S_A$ and $S_B$ is:

If a piece of metal is heated to temperature $\theta$ and then allowed to cool in a room which is at temperature $\theta_0$,the graph between the temperature $T$ of the metal and time $t$ will be closest to:

$A$ 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 $\,s$)