A lead bullet penetrates into a solid object and melts. Assuming...

Loss of kinetic energy of the lead bullet=Heat generated to raise its temperature from $127 ° C$ to $327 ° C$.

$40$ loss of the kinetic energy of the lead bullet=Heat gained by the lead bullet to raise its temperature.

$0 . 4\left(\dfrac{1}{2}m{v}^{2}\right) = m \times \left[{S}_{Pb} \times ∆ T + m \times {L}_{Pb}\right]$

$0 . 4 \times \dfrac{1}{2}m{v}^{2} = m{S}_{Pb}\left({T}_{2} - {t}_{1}\right) + m{L}_{Pb}$

$0 . 2m{v}^{2} = m \times \left[125 \times \left(327 - 127\right) + 2 . 5 \times {10}^{4}\right]$

${v}^{2} = 5 \times \left[125 \times 200 + 2 . 5 \times {10}^{4}\right]$

${v}^{2} = 5 \times \left[25000 + 25000\right]$

$v = 500m{s}^{- 1}$