Steam at $100^{\circ} C$ is passed into $114 \ g$ of water at $30^{\circ} C$. The mass of water present in the mixture when the temperature of the mixture becomes $70^{\circ} C$ is (Latent heat of steam $= 540 \ cal \ g^{-1}$; specific heat capacity of water $= 1 \ cal \ g^{-1} {}^{\circ} C^{-1}$) (in $g$)

$540 \ g$ of ice at $0 \ ^\circ C$ is mixed with $540 \ g$ of water at $80 \ ^\circ C$. The final temperature of the mixture is

$2\,kg$ of metal at $100\,^{\circ}C$ is cooled by $1\,kg$ of water at $0\,^{\circ}C$. If the specific heat capacity of the metal is $\frac{1}{2}$ of the specific heat capacity of water,the final temperature of the mixture would be:

Steam at $100^{\circ} C$ is passed into $1 \, kg$ of water contained in a calorimeter at $9^{\circ} C$ until the temperature of the water and calorimeter increases to $90^{\circ} C$. The mass of the steam condensed is nearly (water equivalent of calorimeter $= 0.1 \, kg$, specific heat of water $= 1 \, cal \cdot g^{-1} \cdot {}^{\circ} C^{-1}$, and latent heat of vaporisation $= 540 \, cal \cdot g^{-1}$). (in $g$)

Steam 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 $90^{\circ}C$ is ........ $g$. (Latent heat of steam is $540 \, cal/g$)

An electric kettle (rated at $2.5\, kW$) is used to heat $3\, kg$ of water from $15\, ^oC$ to its boiling point. It takes $9.5$ minutes. The amount of heat lost is: