Steam at $100^{\circ}C$ is passed into $1.1 \, kg$ of water contained in a calorimeter of water equivalent $0.02 \, kg$ at $15^{\circ}C$ until the temperature of the calorimeter and its contents rises to $80^{\circ}C$. The mass of the steam condensed in $kg$ is:

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

The temperature of water of mass $100 \ g$ is raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam to it. The mass of the steam added is (Latent heat of steam $= 540 \ cal \ g^{-1}$ and specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

$10\; g$ of ice cubes at $0\;^{\circ} C$ are released in a tumbler (water equivalent $55\; g$) at $40\;^{\circ} C$. Assuming that negligible heat is taken from the surroundings,the temperature (in $^{\circ} C$) of water in the tumbler becomes nearly $(L_f = 80\; cal/g)$.

The time required to raise the temperature of $3 \text{ litre}$ of water from $0^{\circ} C$ to $80^{\circ} C$ by a heater operated under $200 \text{ V}$ having resistance of $50 \Omega$ is
[specific heat capacity of water is $4200 \text{ J kg}^{-1} \text{ K}^{-1}$] [density of water $= 1000 \text{ kg/m}^3$] (in $\text{ min}$)

Water of volume $2\, L$ in a closed container is heated with a coil of $1\, kW$. While water is heated,the container loses energy at a rate of $160\, J/s$. In how much time will the temperature of water rise from $27\, ^\circ C$ to $77\, ^\circ C$? (Specific heat of water is $4.2\, kJ/kg\cdot K$ and that of the container is negligible)