Question

A lead bullet penetrates into a solid object and melts. Assuming that $50$% of its kinetic energy was used to heat it, calculate the initial speed of the bullet. The initial temperature of the bullet is ${27}^{\circ }\text{C}$ and its melting point is ${327}^{\circ }\text{C}$.
Latent heat of fusion of lead $=2.5\times {10}^4\text{J}{kg}^{-1}$ and specific heat capacity of lead $=125\text{J /kg K}$.

• A. $100\text{m/s}$
• B. $200\text{m/s}$
• C. $400\text{m/s}$
• D. $500\text{m/s}$

Answer 1

The correct option is D $500\text{m/s}$

Let the mass of the bullet is $m$.

Heat required to take the bullet from ${27}^{\circ }\text{C}$ to ${327}^{\circ }\text{C}$:
$Q_1=m{s}_{lead}\Delta T=m\times 125\times 300=m\times (3.75\times {10}^4)\text{J}$

Heat required to melt the bullet:
$Q_2=m{l}_{lead}=m\times (2.5\times {10}^4)\text{J}$

As per the question, If initial speed be $v$, the kinetic energy is $\frac{1}{2}m{v}^2$ and hence, the heat developed is $\frac{1}{2}K.E$.

Thus,
$\frac{1}{4}m{v}^2=Q_1+Q_2=m(3.75+2.5)\times {10}^4$
or, $v=500\text{m/s}$