Question

A lead bullet initially at 32°C just melts upon striking a target. Assuming that all of the initial kinetic energy of the bullet goes into the internal energy of the bullet to raise its temperature and melt it, calculate the speed of the bullet upon impact.

Answer 1

We can use the definition of kinetic energy to express the
speed of the bullet upon impact in terms of its kinetic energy. The heat absorbed
by the bullet is the sum of the heat required to warm the bullet from 305 K to its
melting temperature of 600 K and the heat required to melt it. We can use the first
law of thermodynamics to relate the impact speed of the bullet to the change in its
internal energy.
Using the first law of
thermodynamics, relate the change in
the internal energy of the bullet to
the work done on it by the target:
ΔEint = Qin +Won
or, because Qin = 0,
ΔEint = Won = ΔK = -(Kf - Ki )
Substitute for ΔEint, Kf, and Ki to
obtain:
mc_PbΔT_Pb + mL_f,_Pb= -(0 - mv²/2) = mv²/2
Solving for v yields: v = (2*(mc_PbΔT_Pb + L_f,_Pb))^1/2
v=[2{(128)*(600-305)+24700}]^1/2
v=353.44 m/s