$A$ piece of ice falls from a height $h$ so that it melts completely. Only one-quarter of the heat produced is absorbed by the ice and all energy of ice gets converted into heat during its fall. The value of $h$ is (Latent heat of ice is $L = 3.4 \times 10^{5} \text{ J/kg}$ and $g = 10 \text{ N/kg}$) (in $\text{ km}$)

If $60 \%$ of the kinetic energy of water falling from a $210 \ m$ high waterfall is converted into heat,the rise in temperature of the water at the bottom of the fall is nearly (specific heat of water $= 4.2 \times 10^3 \ J \ kg^{-1} \ K^{-1}$ and $g = 10 \ m/s^2$):

$A$ 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 $^{\circ} C$)

$1 \ gm$ of ice at $0^o C$ is mixed with $1 \ gm$ of steam at $100^o C$. What will be the final temperature $(^o C)$?

$A$ steel ball of mass $200 \,g$ falls freely from a height of $20 \,m$ and bounces to a height of $10.8 \,m$ from the ground. If the energy lost in this process is absorbed by the ball, the rise in its temperature is ($g = 10 \,ms^{-2}$, specific heat capacity of steel is $460 \,Jkg^{-1} K^{-1}$). (in $^{\circ} C$)

When $1\, g$ of water at $0^{\circ}C$ and $1 \times 10^5\, N/m^2$ pressure is converted into ice of volume $1.091\, cm^3$,the external work done will be (in $, J$)