Pure water supercooled to $-15^{\circ} C$ is contained in a thermally insulated flask. $A$ small amount of ice is thrown into the flask. The fraction of water frozen into ice is:

$200 \,g$ of ice at $-20^{\circ} C$ is mixed with $500 \,g$ of water at $20^{\circ} C$ in an insulating vessel. The final mass of water in the vessel is ........... $g$ (specific heat of ice $= 0.5 \,cal \,g^{-1} {^{\circ}C}^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$).

$200 \, g$ of a solid ball at $20 \, ^oC$ is dropped into an equal amount of water at $80 \, ^oC$. The resulting temperature is $60 \, ^oC$. This means that the specific heat of the solid is

$A$ piece of ice (heat capacity $= 2100 \ J \ kg^{-1} \ ^{\circ}C^{-1}$ and latent heat $= 3.36 \times 10^5 \ J \ kg^{-1}$) of mass $m$ grams is at $-5^{\circ}C$ at atmospheric pressure. It is given $420 \ J$ of heat so that the ice starts melting. Finally,when the ice-water mixture is in equilibrium,it is found that $1 \ g$ of ice has melted. Assuming there is no other heat exchange in the process,the value of $m$ is:

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be:

$A$ metal block of mass $3.3 \ kg$ is heated to a temperature of $400^{\circ} C$ and then placed on a large ice block. The specific heat of the metal is $0.4 \ J \ g^{-1} \ K^{-1}$ and the latent heat of fusion of water is $330 \ J \ g^{-1}$. The maximum amount of ice that can melt is: (in $kg$)