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 $^{\circ} C$)

$300 \text{ g}$ of water at $25^{\circ}C$ is added to $100 \text{ g}$ of ice at $0^{\circ}C$. The final temperature of the mixture is (in $^{\circ}C$)

$5 \ kg$ 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 $^{\circ} C$)

We have half a bucket $(6\,L)$ of water at $20\,^oC$. If we want water at $40\,^oC$,how much steam at $100\,^oC$ should be added to it?

Ice in a freezer is at $-7^{\circ} C$. $100 \, g$ of this ice is mixed with $200 \, g$ of water at $15^{\circ} C$. Take the freezing temperature of water to be $0^{\circ} C$,the specific heat of ice equal to $2.2 \, J/g^{\circ} C$,specific heat of water equal to $4.2 \, J/g^{\circ} C$,and the latent heat of ice equal to $335 \, J/g$. Assuming no loss of heat to the environment,the mass of ice in the final mixture is closest to .......... $g$.

Three copper blocks of masses $M_1, M_2$ and $M_3$ $kg$ respectively are brought into thermal contact until they reach equilibrium. Before contact,they were at temperatures $T_1, T_2, T_3$ $(T_1 > T_3)$. Assuming there is no heat loss to the surroundings,the equilibrium temperature $T$ is ($s$ is the specific heat of copper).