A moving lead bullet just melts when stopped by an obstacle. Assuming that  25 % of the heat is absorbed by the obstacle, then the velocity of the bullet if its initial temperature is ${27}^{\circ }\mathrm{C}$  (melting point of lead $={327}^{\circ }\mathrm{C}$  , specific heat of lead $=30\mathrm{cal}/\mathrm{kg}{/}^0\mathrm{C}$ , latent heat of fusion of lead $=6000\mathrm{cal}/\mathrm{kg},\mathrm{J}=4.2\text{ joules }/\mathrm{cal}.)$ (approximately)

Let the initial K.E. of lead bullet be $1/2{\mathrm{mv}}^2$ J where m is in kg and v in  ${\mathrm{ms}}^{-1}$
 25% of heat is absorbed in the obstacle.
 75% of K.E. is used up in melting the bullet
Mechanical energy available $=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$4$}\right.\cdot \raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.{\mathrm{mv}}^2=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$8$}\right.{\mathrm{mv}}^2\mathrm{J}$
Total heat=heat for raising temperature from 27°C to 327°C  without melting Latent heat taken up by the bullet.
$=\left[\mathrm{ms}\left({\mathrm{t}}_2-{\mathrm{t}}_1\right)+\mathrm{mL}\right]\times 4.2\mathrm{J}$
$=[\mathrm{m}\times 30\times (327-27)+6000\mathrm{m}]\times 4.2$
 $=63.000\mathrm{m}$
$\frac{3}{8}{\mathrm{mv}}^2=63,000\mathrm{m}$
Hence $\mathrm{v}=410\mathrm{m}/\mathrm{s}$