$2 \ kg$ of ice at $-20^{\circ} C$ is mixed with $5 \ kg$ of water at $20^{\circ} C$. What is the final amount of water in the mixture (in $kg$)? (Given: specific heat of ice $= 0.5 \ cal/g^{\circ} C$,specific heat of water $= 1 \ cal/g^{\circ} C$,latent heat of fusion of ice $= 80 \ cal/g$)

$A$ hammer of mass $200 \ kg$ strikes a steel block of mass $200 \ g$ with a velocity $8 \ ms^{-1}$. If $23 \%$ of the energy is utilized to heat the steel block,the rise in temperature of the block is (specific heat capacity of steel,$s = 460 \ J \ kg^{-1} \ K^{-1}$) (in $K$)

$A$ block of steel of mass $2 \,kg$ slides down a rough inclined plane of inclination $\sin ^{-1}\left(\frac{3}{5}\right)$ at a constant speed. Assuming that the mechanical energy lost due to friction is used to increase the temperature of the block, calculate the rise in temperature of the block as it slides through $80 \,cm$. (Specific heat capacity of steel $= 420 \,J kg^{-1} K^{-1}$ and acceleration due to gravity $= 10 \,ms^{-2}$) (in $^{\circ} C$)

An amount of ice of mass $10^{-3} \ kg$ and temperature $-10^{\circ} C$ is transformed to vapour of temperature $110^{\circ} C$ by applying heat. The total amount of heat required for this conversion is,(Take,specific heat of ice $= 2100 \ J \ kg^{-1} \ K^{-1}$,specific heat of water $= 4180 \ J \ kg^{-1} \ K^{-1}$,specific heat of steam $= 1920 \ J \ kg^{-1} \ K^{-1}$,Latent heat of ice $= 3.35 \times 10^5 \ J \ kg^{-1}$ and Latent heat of steam $= 2.25 \times 10^6 \ J \ kg^{-1}$) (in $J$)

When $54 \ g$ of ice at $-20^{\circ} C$ is mixed with $25 \ g$ of steam at $100^{\circ} C$,then the final mixture at thermal equilibrium contains

If two balls of the same metal weighing $5\, g$ and $10\, g$ strike a target with the same velocity,and the heat energy developed is used solely for raising their temperature,then the temperature rise will be higher: