Steam at $100\,^{\circ}C$ is passed into $22\,g$ of water at $20\,^{\circ}C$. The mass of water that will be present when the water acquires a temperature of $90\,^{\circ}C$ (Latent heat of steam is $540\,cal/g$) is ......... $g$.

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

When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

$2 \; g$ of steam condenses when passed through $40 \; g$ of water initially at $25 \; ^\circ C$. The condensation of steam raises the temperature of water to $54.3 \; ^\circ C$. What is the latent heat of steam in $cal/g$?

$M$ 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:

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