$A$ gun fires a lead bullet of temperature $300 \ K$ into a wooden block. The bullet having a melting temperature of $600 \ K$ penetrates into the block and melts down. If the total heat required for the process is $625 \ J$,then the mass of the bullet is . . . . . . grams. (Latent heat of fusion of lead $= 2.5 \times 10^4 \ J \ kg^{-1}$ and specific heat capacity of lead $= 125 \ J \ kg^{-1} K^{-1}$)

When $1\, g$ of water at $0^{\circ}C$ and $1 \times 10^5\, N/m^2$ pressure is converted into ice of volume $1.091\, cm^3$,the external work done will be (in $, J$)

$A$ glass flask of volume $200 \, cm^3$ is just filled with mercury at $20^{\circ} C$. The amount of mercury that will overflow when the temperature of the system is raised to $100^{\circ} C$ is ........ $cm^3$ $(\gamma_{\text{glass}} = 1.2 \times 10^{-5} /^{\circ}C, \gamma_{\text{mercury}} = 1.8 \times 10^{-4} /^{\circ}C)$

$A$ liquid at $30^{\circ} C$ is poured very slowly into a calorimeter at $110^{\circ} C$. The boiling point of the liquid is $80^{\circ} C$. It is observed that the first $5 \ gm$ of the liquid evaporates completely. After adding another $80 \ gm$ of the liquid,the equilibrium temperature is found to be $50^{\circ} C$. What is the ratio of the latent heat of the liquid to its specific heat? [Neglect heat exchange with the surroundings]

$M \ kg$ of water at $t^{\circ} C$ is divided into two parts so that one part of mass $m \ kg$ when converted into ice at $0^{\circ} C$ would release enough heat to vaporize the other part. Then $\frac{m}{M}$ is equal to (Specific heat of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$,Latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,Latent heat of steam $= 540 \ cal \ g^{-1}$)

Water is fully filled in a container at $4^{\circ}C$. What happens to the water level when the temperature changes?