$A$ vessel contains $110\,g$ of water. The heat capacity of the vessel is equivalent to $10\,g$ of water. The initial temperature of the water in the vessel is $10\,^{\circ}C$. If $220\,g$ of hot water at $70\,^{\circ}C$ is poured into the vessel,the final temperature,neglecting radiation loss,will be nearly equal to ........ $^{\circ}C$.

$15 \text{ g}$ of ice at $0^{\circ}\text{C}$ is added to a vessel containing water at $40^{\circ}\text{C}$. The mass of water and water equivalent of the vessel is $60 \text{ g}$. Assuming that negligible heat is taken from the surroundings, the final temperature of the mixture will be $[L_{\text{ice}} = 80 \text{ cal/g}, S_{\text{water}} = 1 \text{ cal/g}]$ (in $^{\circ}\text{C}$)

When $M_1$ gram of ice at $-10\,^{\circ}C$ (specific heat $= 0.5\, cal\, g^{-1}\,^{\circ}C^{-1}$) is added to $M_2$ gram of water at $50\,^{\circ}C$,finally no ice is left and the water is at $0\,^{\circ}C$. The value of latent heat of ice,in $cal\, g^{-1}$ is

When $50 \ g$ of water at $10^{\circ} C$ is mixed with $50 \ g$ of water at $100^{\circ} C$,the resultant temperature is: (in $^{\circ} C$)

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 for this purpose is ........... $g$.

$A$ block of ice at $-20\,^oC$ having a mass of $2\,kg$ is added to $3\,kg$ of water at $15\,^oC$. Neglecting heat losses and the heat capacity of the container,what is the final state of the system?