The temperature of $100 \,g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required in this process is (latent heat of steam is $540 \,cal \,g^{-1}$). (in $\,g$)

$A$ calorimeter contains $0.5 \,kg$ of water at $30^{\circ} C$. When $0.3 \,kg$ of water at $60^{\circ} C$ is added to it, the resulting temperature is found to be $40^{\circ} C$. The water equivalent of the calorimeter is (in $\,kg$)

$10 \, kg$ of water is to be heated from $20^{\circ}C$ to $80^{\circ}C$ in $1 \, hour$. Steam at $150^{\circ}C$ is passed through a copper coil immersed in the water. The steam condenses and returns to the boiler at $90^{\circ}C$. How much steam is required per hour? (Specific heat of water $= 1 \, cal/g^{\circ}C$,Latent heat of vaporization $= 540 \, cal/g$)

$50\, g$ of ice at $0\,^{\circ}C$ is dropped into a calorimeter containing $100\, g$ of water at $30\,^{\circ}C$. If the thermal capacity of the calorimeter is zero,then the amount of ice left in the mixture at equilibrium is ........ $g$.

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

$M$ grams of steam at $100^{\circ} C$ is mixed with $200 \; g$ of ice at its melting point in a thermally insulated container. If it produces liquid water at $40^{\circ} C$ [heat of vaporization of water is $540 \; cal/g$ and heat of fusion of ice is $80 \; cal/g$],the value of $M$ is: