$A$ beaker is filled with water at $4\,^{\circ}C$. At one time the temperature is increased by a few degrees above $4\,^{\circ}C$ and at another time it is decreased by a few degrees below $4\,^{\circ}C$. One shall observe that:

$A$ child running a temperature of $101\,^{\circ} F$ is given an antipyrin (i.e. a medicine that lowers fever) which causes an increase in the rate of evaporation of sweat from his body. If the fever is brought down to $98\,^{\circ} F$ in $20$ minutes,what is the average rate of extra evaporation caused by the drug (in $g/min$)? Assume the evaporation mechanism to be the only way by which heat is lost. The mass of the child is $30\; kg$. The specific heat of the human body is approximately the same as that of water,and the latent heat of evaporation of water at that temperature is about $580\; cal \;g^{-1}$.

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

$1 \ gm$ of ice at $0^o C$ is mixed with $1 \ gm$ of steam at $100^o C$. What will be the final temperature $(^o C)$?

The temperature of Bhakhra dam water at the ground level with respect to the temperature at a high level should be:

The densities of wood and benzene at $0^{\circ} C$ are $880 \ kg \ m^{-3}$ and $900 \ kg \ m^{-3}$,respectively. The coefficients of volume expansion are $1.2 \times 10^{-3} \ ^{\circ}C^{-1}$ for wood and $1.5 \times 10^{-3} \ ^{\circ}C^{-1}$ for benzene. The temperature at which a piece of wood just sinks in benzene is (in $^{\circ} C$)