The temperature of $100 \,g$ of water is to be raised from $24^{\circ} C$ to $90^{\circ} C$ by adding steam at $100^{\circ} C$ to it. The mass of the steam required in this process is (latent heat of steam is $540 \,cal \,g^{-1}$). (in $\,g$)

The temperature of a copper piece of mass $50 \ g$ is raised by $10 \ ^\circ C$. If the same amount of heat is given to $10 \ g$ of water,the rise in its temperature is = ...... $^\circ C$ (Specific heat of copper $= 420 \ J/kg \cdot ^\circ C$,Specific heat of water $= 4200 \ J/kg \cdot ^\circ C$).

The temperatures of equal masses of three different liquids $x$,$y$,and $z$ are $10^{\circ}C$,$20^{\circ}C$,and $30^{\circ}C$ respectively. The temperature of the mixture when $x$ is mixed with $y$ is $16^{\circ}C$,and the temperature when $y$ is mixed with $z$ is $26^{\circ}C$. The temperature of the mixture when $x$ and $z$ are mixed will be ...... $^{\circ}C$.

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 $^{\circ} C$)

When $0.15 \; kg$ of ice at $0^{\circ} C$ is mixed with $0.30 \; kg$ of water at $50^{\circ} C$ in a container,the resulting temperature is $6.7^{\circ} C$. Calculate the latent heat of fusion of ice. (Given: $s_{\text{water}} = 4186 \; J \; kg^{-1} \; K^{-1}$)

Steam of mass $60 \ g$ at a temperature $100^{\circ} C$ is mixed with water of mass $360 \ g$ at a temperature $40^{\circ} C$. The ratio of the masses of steam and water in equilibrium is (Latent heat of steam is $540 \ cal \ g^{-1}$ and specific heat capacity of water is $1 \ cal \ g^{-1} {}^{\circ} C^{-1}$)