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 ${27}^{\circ }C$ and its melting point is ${327}^{\circ }C$.

Latent heat of fusion of lead $=2.5\times {10}^{4}Jk{g}^{-1}$ and

Specific heat capacity of lead $=125Jk{g}^{-1}{K}^{-1}$.

• A. $500m{s}^{-1}$
• B. $5000m{s}^{-1}$
• C. $100m{s}^{-1}$
• D. $1000m{s}^{-1}$

Answer 1

The correct option is A

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

Given:
Initial temperature of the bullet = ${27}^{\circ }C$
Final temperature of the bullet = ${327}^{\circ }C$
Latent fusion of fusion of lead, $L_f=2.5\times {10}^{4}Jk{g}^{-1}$
Specific heat capacity of lead, $S=125Jk{g}^{-1}{K}^{-1}$
Let the mass of the bullet = $m$ and
Q be the total heat required, $Q_1$ be the heat required to raise the temperature of bullet from ${27}^{\circ }C$ to ${327}^{\circ }C$ and $Q_2$ be the heat required to melt the bullet. Then:
$Q=Q_1+Q_2$
$Q=mS\Delta T+m\times L_f$
$Q=m[125\times (327-27)+2.5\times {10}^4]$
$Q=6.25\times {10}^4\times m$

If the initial speed by $v$, the kinetic energy is $\frac{1}{2}m{v}^2$ and only $50$% of kinetic energy is used in heating the bullet.

$\frac{1}{2}\left(\frac{1}{2}m{v}^2\right)=m\times 6.25\times {10}^4$
$v=500m{s}^{-1}$