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 $g$)

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 $20\,g$ of water at $10\,^{\circ}C$. When water acquires a temperature of $80\,^{\circ}C$,the mass of water present will be ........ $g$. [Take specific heat of water $= 1\,cal\,g^{-1}\,^{\circ}C^{-1}$ and latent heat of steam $= 540\,cal\,g^{-1}$]

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 $^{\circ} C$)

$A$ copper ball of mass $100 \ gm$ is at a temperature $T$. It is dropped into a copper calorimeter of mass $100 \ gm$,filled with $170 \ gm$ of water at room temperature. Subsequently,the temperature of the system is found to be $75 ^\circ C$. $T$ is given by......$^\circ C$ (Given: room temperature $= 30 ^\circ C$,specific heat of copper $= 0.1 \ cal/gm ^\circ C$)

$A$ small quantity of water of mass $m$ at temperature $\theta^{\circ} C$ is mixed with a large mass $M$ of ice which is at its melting point. If $s$ is the specific heat capacity of water and $L$ is the latent heat of fusion of ice,then the mass of ice melted is: