$A$ liquid of mass $2m$ and specific heat $C$ is heated to a temperature $4T$. Another liquid of mass $m$ and specific heat $2C$ is heated to a temperature $T$. If these two liquids are mixed,the resulting temperature of the mixture is:

One kilogram of ice at $0\,^{\circ}C$ is mixed with one kilogram of water at $80\,^{\circ}C$. The temperature of the mixture is ........ $^{\circ}C$ (specific heat of water $= 4200\,J\,kg^{-1}\,K^{-1}$ and latent heat of ice $= 336\,kJ\,kg^{-1}$).

Steam at $100^{\circ}C$ is added slowly to $1400 \, \text{g}$ of water at $16^{\circ}C$ until the temperature of the water is raised to $80^{\circ}C$. The mass of steam required to do this is $(L_V = 540 \, \text{cal/g})$ ........... $\text{g}$

Steam at $100^{\circ}C$ is passed into $1.1 \, kg$ of water contained in a calorimeter of water equivalent $0.02 \, kg$ at $15^{\circ}C$ until the temperature of the calorimeter and its contents rises to $80^{\circ}C$. The mass of the steam condensed in $kg$ is:

$5 \ kg$ 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 heat required to raise the temperature of $50 \, g$ of copper by $10^\circ C$ is given to $10 \, g$ of water. What is the rise in temperature of the water in $^\circ C$? (Specific heat of copper $= 420 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$,Specific heat of water $= 4200 \, J \cdot kg^{-1} \cdot ^\circ C^{-1}$)