(II) At a crime scene, the forensic investigator notes that the 6.2-g lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature (20°C), what does the investigator calculate as the minimum muzzle velocity of the gun?

When the phase of a material changes from solid to liquid or from liquid to gas, the material absorbs some amount of heat energy. The amount of heat required to change the phase of a material of unit mass at constant temperature is termed latent heat.

The amount of heat required to change the phase of 1.0 kg of a substance from solid to liquid is termed latent heat of fusion \(\left( {{L_{\rm{F}}}} \right)\). Latent heat of fusion of lead is \({L_{\rm{F}}} = 0.25 \times {105}\;{\rm{J/kg}}\).

Mass of the lead bullet is \(m = 6.2\;{\rm{g}} = 6.2 \times {10{ - 3}}\;{\rm{kg}}\).

The temperature of the room is \({T_1} = 20\circ {\rm{C}}\).

The melting point of lead is \({T_2} = 327\circ \;{\rm{C}}\).

Specific heat of lead is \(c = 130\;{\rm{J/kg}} \cdot {\rm{\circ C}}\).

When the bullet is stopped by the doorframe, then the whole of the kinetic energy gets converted into heat energy. This heat energy is partially used by the bullet to increase its temperature from room temperature to its melting point and partially used in melting, that is, in changing its phase from solid to liquid.

The lead bullet takes heat \(\left( { = mc\left( {{T_1} - {T_2}} \right)} \right)\) to increase its temperature from \({T_1}\)to \({T_2}\) and then melts into the liquid lead at temperature \({T_2}\) by taking heat \(\left( { = m{L_{\rm{F}}}} \right)\).

The total heat energy taken is

\(\begin{array}{c}Q = mc\left( {{T_1} - {T_2}} \right) + m{L_{\rm{F}}}\\ = m\left[ {c\left( {{T_1} - {T_2}} \right) + {L_{\rm{F}}}} \right]\end{array}\).

Substitute the values into the above expression.

\(\begin{array}{c}Q = 6.2 \times {10{ - 3}}\;{\rm{kg}}\left[ {\left( {130\;{\rm{J/kg}} \cdot {\rm{\circ C}}} \right)\left( {327 - 20} \right)\circ {\rm{C}} + \left( {0.25 \times {{10}5}\;{\rm{J/kg}}} \right)} \right]\\ = 402.442\;{\rm{J}}\end{array}\)

Let v be the minimum muzzle velocity of the gun.

When the bullet is shot from the gun, then it moves with some kinetic energy. The kinetic energy of the bullet is

\(KE = \frac{1}{2}m{v2}\).

According to the law of conservation of energy, the total energy remains conserved in an isolated system.

So, the kinetic energy of the bullet must have converted into heat energy.

\(\begin{array}{c}KE = Q\\\frac{1}{2}m{v2} = Q\\v = \sqrt {\frac{{2Q}}{m}} \end{array}\)

Substitute the values into the above expression.

\(\begin{array}{c}v = \sqrt {\frac{{2 \times 402.442\;{\rm{J}}}}{{6.2 \times {{10}{ - 3}}\;{\rm{kg}}}}} \\ = \sqrt {129,820} \;{\rm{m/s}}\\ \approx 360\;{\rm{m/s}}\end{array}\)

Thus, the minimum muzzle velocity of the gun is \(360\;{\rm{m/s}}\).