# When fired vertically upwards, the lead bullet reached a height of 1200m. When it fell, hitting the ground

**When fired vertically upwards, the lead bullet reached a height of 1200m. When it fell, hitting the ground, it heated up. Assuming that 50 percent of all impact energy went to heating the bullet, calculate how much its temperature increased: Lead = 0.13 * 10 (3) J / kg.k.**

Given:

h = 1200 meters – the height to which the bullet rose;

g = 10 meters per second squared – gravitational acceleration;

n = 50% = 0.5 – the amount of energy on impact, which went to heat the bullet;

c = 0.13 * 103 J / (kg * K) = 130 J / (kg * K) – specific heat of lead.

It is required to determine dT (degree Kelvin) – at what temperature the bullet heated up upon impact.

Let m be the mass of the bullet.

Then the total mechanical energy of the bullet will be equal to:

W = m * g * h.

The amount of energy that will go to heat the bullet is:

Q = W * n.

Then:

W * n = m * c * dT;

m * g * h * n = m * c * dT;

g * h * n = c * dT.

dT = g * h * n / c = 10 * 1200 * 0.5 / 130 = 6000/130 = 46.2 degrees.

Answer: the bullet will heat up to 46.2 degrees.