$A$ $2 \, kg$ 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 = $80 \, kcal/kg$)

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$, specific heat of water $= 4200 \ J/kg.^{\circ}C$, latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)

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?

If an electric heater is rated at $1000\,W$,then the time required to heat one litre of water from $20\,^{\circ}C$ to $60\,^{\circ}C$ is

$A$ calorimeter has a heat capacity of $100 \ J/K$ and is at $30^{\circ}C$. $100 \ g$ of water at $40^{\circ}C$ (specific heat capacity $4200 \ J/kg \cdot K$) is poured into the calorimeter. What is the final temperature of the mixture in the calorimeter in $^{\circ}C$?

$5\, g$ of ice at $0^\circ C$ is dropped in a beaker containing $20\, g$ of water at $40^\circ C$. The final temperature will be........ $^\circ C$.