Question

A 5.80-g lead bullet traveling at 600 m/s is stopped by a large tree. If half the kinetic energy of the bullet is transformed into internal energy and remains with the bullet while the other half is transmitted to the tree, what is the increase in temperature of the bullet?

Answer in degrees celcius

PLEASE EXPLAIN ALL STEPS.

Answer 1

The kinetic energy is
K = 1/2 m v² = 0.5 x 5.80 x 10^-3 x 600² =1044 J

Half this energy is
E = K/2 = 1044 / 2 = 522 J

The increase in temperature comes from thermal equation
E = m c ΔT
where m is the mass and c the specific heat of lead = 128 J/kg°K

Then
ΔT = E/mc = 522 / (5.80 x 10^-3 x 128) = 703 °C

However, such a temperature increase is physically impossible, because lead melts at 327°C, and even in a polar climate an increase of 703°C would bring the bullet well beyond the melting point.
What we can state is that part of the energy will be spent to melt the bullet, and this can be computed as
Ef = λf m
where λf is the fusion heat of lead = 25000 J/kg. Thus:
Ef = 25000 x 5.80 x 10^-3 = 145 J

The available energy for temperature increase is therefore
E - Ef = 522- 145 = 377 J
which is still too much to be accounted for by temperature increase of solid lead only.
This means that some energy will be spent in increasing temperature of molten lead.
But since solid and molten lead have different specific heats, and since we don't know the initial temperature of the bullet, it's impossible to state how much of this energy will be spent on solid or on molten lead respectively: it's therefore impossible to carry out the calculation.