$5 \ kg$ of water at $20^{\circ} C$ is added to $10 \ kg$ of water at $60^{\circ} C$. Neglecting the heat capacity of the vessel and other losses,the resultant temperature will be nearly: (in $^{\circ} C$)

The temperatures of equal masses of three different liquids $A, B$ and $C$ are $15^{\circ} C, 24^{\circ} C$ and $30^{\circ} C$ respectively. The resultant temperature when liquids $A$ and $B$ are mixed is $20^{\circ} C$ and when liquids $B$ and $C$ are mixed is $26^{\circ} C$. Then the ratio of specific heat capacities of the liquids $A, B$ and $C$ is

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 $g$)

$A$ $50 \,g$ ice cube at $-10^{\circ} C$ is added to $200 \,g$ of water at $30^{\circ} C$. The final temperature of the mixture is (specific heat of water $= 1 \,cal \,g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$,specific heat of ice $= 0.5 \,cal \,g^{-1} {}^{\circ} C^{-1}$). (in $^{\circ} C$)

The specific heat of alcohol is about half that of water. Suppose you have identical masses of alcohol and water. The alcohol is initially at temperature $T_A$. The water is initially at a different temperature $T_W$. Now the two fluids are mixed in the same container and allowed to come into thermal equilibrium,with no loss of heat to the surroundings. The final temperature of the mixture will be :-

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 ......... $g$.