$A$ 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 $= 3.5 \times 10^{5}\,J\,kg^{-1}$]

Steam is passed into $22 \, g$ of water at $20^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90^{\circ}C$ is ........ $g$. (Latent heat of steam is $540 \, cal/g$)

$5 \ g$ of ice at $-30^{\circ} C$ and $20 \ g$ of water at $35^{\circ} C$ are mixed together in a calorimeter. The final temperature of the mixture is (Neglect heat capacity of the calorimeter,specific heat capacity of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion of ice $= 80 \ cal \ g^{-1}$,and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$). (in $^{\circ} C$)

$A$ piece of metal of $850 \text{ K}$ is dropped into $1 \text{ kg}$ of water at $300 \text{ K}$. If the equilibrium temperature of the mixture is $350 \text{ K}$, then the heat capacity of the metal expressed in $\text{J/K}$ is (Specific heat of water = $4200 \text{ J/kg} \cdot \text{K}$)

If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

Steam of mass $60 \ g$ at a temperature $100^{\circ} C$ is mixed with water of mass $360 \ g$ at a temperature $40^{\circ} C$. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is $540 \ cal \ g^{-1}$ and specific heat capacity of water is $1 \ cal \ g^{-1} {}^{\circ} C^{-1}$)