An ice block is projected vertically up with a velocity $20m{s}^{-1}$ The amount of ice that melt when it reaches the ground assuming the  loss of P.E. is converted into heat energy if the mass of ice  block is 4.2 kg initially

Since at the Maximum height 

the velocity of the ice block 

becomes It is because 

the kinetic energy is converted 

into the potential (low of conservation of mechanical energy)

$$\begin{gathered} \Rightarrow \text{ gain in potential energy at max height }\mathrm{\Delta U}=\frac{1}{2}{\mathrm{mv}}^2=\frac{1}{2}\times 4.2\times 20\times 20 \\ =840J \\ Since all the penitential energy is covered into heat energy \left(\Delta Q\right) when block returns to ground \\ \Rightarrow \mathrm{\Delta Q}=\mathrm{\Delta U} \\ \mathrm{now}, \mathrm{If} \text{'}\mathrm{m}\text{'} \mathrm{mass} \mathrm{of} \mathrm{ice} \mathrm{block} \mathrm{is} \mathrm{melted} \mathrm{into} \mathrm{water} \\ \Rightarrow \mathrm{\Delta Q}=\mathrm{mL},\mathrm{L}\to \text{ latent heat of water } \\ \Rightarrow \mathrm{\Delta U}=\mathrm{mL} \\ \mathrm{m}=\frac{840}{3.4\times {10}^5}\mathrm{kg} \\ \Rightarrow \mathrm{m}=2.47\mathrm{g}\simeq 2.5\mathrm{g} \end{gathered}$$