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?

$50 \, g$ of ice at $0^\circ C$ is placed in an insulated vessel. $50 \, g$ of water at $100^\circ C$ is mixed into it. Neglecting heat loss,what is the final temperature of the mixture?

$150 \,g$ of ice is mixed with $100 \,g$ of water at temperature $80^{\circ} C$. The latent heat of ice is $80 \,cal/g$ and the specific heat of water is $1 \,cal/g^{\circ} C$. Assuming no heat loss to the environment,the amount of ice which does not melt is ........... $g$.

$A$ calorimeter of water equivalent $20 \, g$ contains $180 \, g$ of water at $25^{\circ} C$. '$m$' grams of steam at $100^{\circ} C$ is mixed in it until the temperature of the mixture is $31^{\circ} C$. The value of '$m$' is close to (Latent heat of water $= 540 \, \text{cal} \, g^{-1}$,specific heat of water $= 1 \, \text{cal} \, g^{-1} {}^{\circ} C^{-1}$)

$A$ thermos flask contains $250 \ g$ of coffee at $90^{\circ} C$. To this $20 \ g$ of milk at $5^{\circ} C$ is added. After equilibrium is established,the temperature of the liquid is (Assume no heat loss to the thermos bottle. Take specific heat of coffee and milk as $1.00 \ cal/g^{\circ} C$) (in $^{\circ} C$)

$A$ copper block of mass $5.0 \, kg$ is heated to a temperature of $500^{\circ} C$ and is placed on a large ice block. What is the maximum amount of ice (in $kg$) that can melt? [Specific heat of copper: $0.39 \, J g^{-1} {}^{\circ} C^{-1}$ and latent heat of fusion of water: $335 \, J g^{-1}$]