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?

$100 \text{ g}$ of ice at $0^{\circ}C$ is mixed with $100 \text{ g}$ of water at $100^{\circ}C$. The final temperature of the mixture is. [Take,$L_f = 3.36 \times 10^5 \text{ J kg}^{-1}$ and $S_w = 4.2 \times 10^3 \text{ J kg}^{-1} \text{ K}^{-1}$] (in $^{\circ}C$)

$1 \,kg$ of ice at $-20^{\circ} C$ is mixed with $2 \,kg$ of water at $90^{\circ} C$. Assuming that there is no loss of energy to the environment,the final temperature of the mixture is ............ $^{\circ} C$. (Assume,latent heat of ice $= 334.4 \,kJ/kg$,specific heat of water and ice are $4.18 \,kJ \,kg^{-1} K^{-1}$ and $2.09 \,kJ \,kg^{-1} K^{-1}$,respectively.)

$A$ $10 \,kW$ drilling machine is used to drill a bore in an aluminium block of mass $25 \,kg$. If the machine is on for $3 \,minutes$ and $50 \%$ of the heat liberated is absorbed by the block, what is the rise in temperature of the block (in $^{\circ} C$)? (Specific heat of aluminium is $900 \,J \,kg^{-1} \,K^{-1}$)

If $540 \, g$ of ice at $0^{\circ}C$ is mixed with $540 \, g$ of water at $80^{\circ}C$,what will be the final temperature of the mixture in $^{\circ}C$?

$10 \ kg$ of ice at $-10^{\circ}C$ is added to $100 \ kg$ of water to lower its temperature from $25^{\circ}C.$ Consider no heat exchange to surroundings. The decrement to the temperature of water is . . . . . . $^{\circ}C.$ (specific heat of ice $= 2100 \ J/kg.^{\circ}C$, specific heat of water $= 4200 \ J/kg.^{\circ}C$, latent heat of fusion of ice $= 3.36 \times 10^{5} \ J/kg$)