When $0.4 \ kg$ of water at $30^{\circ} C$ is mixed with $0.15 \ kg$ of water at $25^{\circ} C$ in a calorimeter,the final temperature is found to be $27^{\circ} C$. Find the water equivalent of the calorimeter. (in $kg$)

$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

If equal masses of $10$ liquids of specific heats $s, 2s, 3s, \ldots, 10s$ at temperatures $10^{\circ} C, 20^{\circ} C, 30^{\circ} C, \ldots, 100^{\circ} C$ respectively are mixed,the resultant temperature of the mixture is . . . . . . . (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 $g$)

$4 \text{ g}$ of steam at $100^{\circ} C$ is added to $20 \text{ g}$ of water at $46^{\circ} C$ in a container of negligible mass. Assuming no heat is lost to the surroundings, the mass of water in the container at thermal equilibrium is. (Latent heat of vaporisation $= 540 \text{ cal/g}$, Specific heat of water $= 1 \text{ cal/g}^{\circ} C$):- (in $\text{ g}$)

How many grams of a liquid with a specific heat of $0.2 \, cal/g^{\circ}C$ at a temperature of $40^{\circ}C$ must be mixed with $100 \, g$ of a liquid with a specific heat of $0.5 \, cal/g^{\circ}C$ at a temperature of $20^{\circ}C$,so that the final temperature of the mixture becomes $32^{\circ}C$?