Question

A lead bullet penetrates into a solid object and melts. Assuming that $50\%$ of its kinetic energy was used to heat it, the initial speed (in $m/s$) of the bullet is
(the initial temperature of the bullet is ${27}^{\circ }C$ and its melting point is ${327}^{\circ }C$). Latent heat of fusion of lead $=2.5\times {10}^{4}J/kg$ and specific heat capacity of lead $=125J/kgK$

Answer 1

We know that: $Q=ms\Delta \theta ,Q=mL$

Given: ${\theta }_1={27}^{\circ }C,{\theta }_2={327}^{\circ }C,L=2.5\times {10}^{4}J/kg,s=125J/kgK$

Half of kinetic energy got converted into rising temperature and then change state of body.

$\frac{1}{2}m{v}^2\times \frac{1}{2}=ms\Delta \theta +mL$

$\frac{v^2}{4}=125(327-27)+2.5\times {10}^4$

$\frac{v^2}{4}=125\times 300+2.5\times {10}^4$

$v=500m/s$

Final answer: $500m/s$