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 $g$)

$15 \text{ g}$ of ice at $0^{\circ}\text{C}$ is added to a vessel containing water at $40^{\circ}\text{C}$. The mass of water and water equivalent of the vessel is $60 \text{ g}$. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be $[L_{\text{ice}} = 80 \text{ cal/g}, S_{\text{water}} = 1 \text{ cal/g}]$ (in $^{\circ}\text{C}$)

One kilogram of ice at $0^{\circ}C$ is mixed with one kilogram of water at $80^{\circ}C$. The final temperature of the mixture is ........ $^{\circ}C$. (Take: specific heat of water $= 4200 \ J \ kg^{-1} \ K^{-1}$,latent heat of ice $= 336 \ kJ \ kg^{-1}$)

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:

$A$ water heater of power $2000\,W$ is used to heat water. The specific heat capacity of water is $4200\,J\,kg^{-1}\,K^{-1}$. The efficiency of the heater is $70\%$. The time required to heat $2\,kg$ of water from $10^{\circ}C$ to $60^{\circ}C$ is $..........s$. (Assume that the specific heat capacity of water remains constant over the temperature range).

We have half a bucket $(6\,L)$ of water at $20\,^oC$. If we want water at $40\,^oC$,how much steam at $100\,^oC$ should be added to it?