$A$ $2 \, kg$ 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 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 = $4200 \text{ J/kg} \cdot \text{K}$)

The time taken for a calorimeter containing $75 \ g$ of water at $62^{\circ} C$ to cool to $58^{\circ} C$ is $9 \ minutes$. When the calorimeter contains $105 \ g$ of water,it takes $12 \ minutes$ to cool from $62^{\circ} C$ to $58^{\circ} C$. The water equivalent of the calorimeter is $.........$ (in $g$)

$A$ 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 $= 4200 \ J / kg \cdot ^{\circ} C$]

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 $1 \ cal \ g^{-1} {}^{\circ} C^{-1}$)

$10\,g$ of ice at $-20\,^{\circ}C$ is dropped into a calorimeter containing $10\,g$ of water at $10\,^{\circ}C$. The specific heat of water is twice that of ice. When equilibrium is reached,the calorimeter will contain