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

$A$ 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?

$10 \, g$ of ice at $0 \, ^oC$ is mixed with $m \, g$ of water at $50 \, ^oC$. What is the minimum value of $m$ (in $g$) so that the ice melts completely? (Given: $L_f = 80 \, cal/g$ and $S_W = 1 \, cal/g \cdot ^oC$)

$A$ $50 \,g$ ice cube at $-10^{\circ} C$ is added to $200 \,g$ of water at $30^{\circ} C$. The final temperature of the mixture is (specific heat of water $= 1 \,cal \,g^{-1} {}^{\circ} C^{-1}$,latent heat of fusion of ice $= 80 \,cal \,g^{-1}$,specific heat of ice $= 0.5 \,cal \,g^{-1} {}^{\circ} C^{-1}$). (in $^{\circ} C$)

When $100\,g$ of a liquid $A$ at $100\,^oC$ is added to $50\,g$ of a liquid $B$ at temperature $75\,^oC$,the temperature of the mixture becomes $90\,^oC$. The temperature of the mixture,if $100\,g$ of liquid $A$ at $100\,^oC$ is added to $50\,g$ of liquid $B$ at $50\,^oC$,will be ........$^oC$

$A$ piece of copper of mass $250 \ g$ at $500^{\circ} C$ is put inside a calorimeter of water equivalent $50 \ g$ containing $200 \ g$ of water at $20^{\circ} C$. At thermal equilibrium,the temperature of the mixture is $60^{\circ} C$. The specific heat of copper (in $J / kg \cdot ^{\circ} C$) is approximately: [Specific heat of water $= 4200 \ J / kg \cdot ^{\circ} C$]