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:

Steam at $100^{\circ}C$ is added slowly to $1400 \, \text{g}$ of water at $16^{\circ}C$ until the temperature of the water is raised to $80^{\circ}C$. The mass of steam required to do this is $(L_V = 540 \, \text{cal/g})$ ........... $\text{g}$

$2 \ g$ of steam at $100 \ ^\circ C$ is passed into $40 \ g$ of water at $25 \ ^\circ C$. The final temperature of the mixture becomes $54.3 \ ^\circ C$. Find the latent heat of vaporization of steam in $cal/g$.

$A$ block of ice of mass $120\,g$ at temperature $0^{\circ}C$ is put in $300\,g$ of water at $25^{\circ}C$. The $x\,g$ of ice melts as the temperature of the water reaches $0^{\circ}C$. The value of $x$ is.
[Use: Specific heat capacity of water $= 4200\,J\,kg^{-1}K^{-1}$,Latent heat of ice $= 3.5 \times 10^{5}\,J\,kg^{-1}$]

$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

$A$ 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: