$100 \text{ g}$ 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 $^{\circ}C$)

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:

$A$ 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?

$A$ $2 \, kg$ block of ice at $-20^{\circ}C$ is added to $5 \, kg$ of water at $20^{\circ}C$. What will be the total mass of water in $kg$? (Specific heat of water = $1 \, kcal/kg/^{\circ}C$,specific heat of ice = $0.5 \, kcal/kg/^{\circ}C$,latent heat of fusion of ice = $80 \, kcal/kg$)

$A$ piece of copper of mass $250 \ g$ at $500^{\circ} C$ is put inside a calorimeter of water equivalent $50 \ g$ containing $200 \ g$ of water at $20^{\circ} C$. At thermal equilibrium,the temperature of the mixture is $60^{\circ} C$. The specific heat of copper (in $J / kg \cdot ^{\circ} C$) is approximately: [Specific heat of water $= 4200 \ J / kg \cdot ^{\circ} C$]

An aluminium piece of mass $50 \,g$ initially at $300^{\circ} C$ is dipped quickly and taken out of $1 \,kg$ of water,initially at $30^{\circ} C$. If the temperature of the aluminium piece immediately after being taken out of the water is found to be $160^{\circ} C$,the temperature of the water is ............ $^{\circ} C$. The specific heat capacities of aluminium and water are $900 \,J \,kg^{-1} K^{-1}$ and $4200 \,J \,kg^{-1} K^{-1}$,respectively.