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

$A$ kettle with $2\, L$ of water at $27\, ^oC$ is heated by a coil heater with a power of $1\, kW$. The heat is lost to the atmosphere at a constant rate of $160\, J/s$ when its lid is open. In how much time will the water be heated to $77\, ^oC$ (specific heat of water $= 4.2\, kJ/(kg \cdot K)$) with the lid open?

We have half a bucket $(6 \text{ litre})$ of water at $20^{\circ}C$. If we want water at $40^{\circ}C$,how much steam at $100^{\circ}C$ should be added to it?

The water equivalent of a calorimeter is $10 \ g$ and it contains $50 \ g$ of water at $15^{\circ} C$. Some amount of ice,initially at $-10^{\circ} C$,is dropped in it and half of the ice melts till equilibrium is reached. What was the initial amount of ice that was dropped (given specific heat of ice $= 0.5 \ cal \ g^{-1} {}^{\circ} C^{-1}$,specific heat of water $= 1.0 \ cal \ g^{-1} {}^{\circ} C^{-1}$ and latent heat of melting of ice $= 80 \ cal \ g^{-1}$) (in $g$)?

$1\,\,kg$ of ice at $-10^{\circ}C$ is mixed with $4.4\,\,kg$ of water at $30^{\circ}C$. The final temperature of the mixture is $........^{\circ}C$ (specific heat of ice is $2100\,\,J/kg/K$ and latent heat of fusion of ice is $336000\,\,J/kg$)

The specific heat of water is $4200 \, J \, kg^{-1} \, K^{-1}$ and the latent heat of ice is $3.4 \times 10^{5} \, J \, kg^{-1}$. $100 \, g$ of ice at $0^{\circ} C$ is placed in $200 \, g$ of water at $25^{\circ} C$. The amount of ice that will melt as the temperature of the water reaches $0^{\circ} C$ is close to (in grams):