$A$ vessel contains $110 \, g$ of water. The heat capacity of the vessel is equivalent to $10 \, g$ of water. The initial temperature of water in the vessel is $10^{\circ}C$. If $220 \, g$ of hot water at $70^{\circ}C$ is poured into the vessel,the final temperature,neglecting radiation loss,will be........ $^{\circ}C$

$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

$m$ g of water at $30^{\circ} C$ is mixed with $5$ g of ice at $-20^{\circ} C$. If the final equilibrium temperature of the mixture is $6^{\circ} C$,find the value of $m$. (Given: specific heat capacity of ice $= 0.5 \text{ cal g}^{-1} {}^{\circ} C^{-1}$,specific heat capacity of water $= 1 \text{ cal g}^{-1} {}^{\circ} C^{-1}$,and latent heat of fusion of ice $= 80 \text{ cal g}^{-1}$) (in $g$)

$A$ 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:

$300 \text{ g}$ of water at $25^{\circ}C$ is added to $100 \text{ g}$ of ice at $0^{\circ}C$. The final temperature of the mixture is (in $^{\circ}C$)

Water is kept in a thermally insulated container at a temperature of $-10^{\circ}C$. If a small piece of ice is dropped into it, what will be the ratio of the mass of ice formed from the water to the initial mass of water?