1. A lead bullet with a mass of 0.03 kg traveling at 198 m/s strikes an armor plate and comes to a stop. If all the energy is converted to heat and absorbed by the bullet, what is the temperature change? 2. Q=KE=1/2mv^2; Q=CmΔT 3. Heat capacity (C) of lead = 130J/kg-°C, mass of the bullet = .03kg, velocity of the bullet = 198m/s I first wanted to calculate the KE of the bullet (.5)(.03kg)(198m/s)^2 and obtained 588.06 kg m/s/s but the closest example to this problem in my book does not follow that equation. It instead would calculate it as follows: Q = KE = 1/2mv^2: 1/2*m*(198m/s)^2; = (19602J/kg) * m [How does it become a J/kg?] Then the increase in temperature caused by this much heat is: Q = 19602J/kg * m = CmΔT (the masses will cancel out here) Q = 19602J/kg = 130J/kg-°C * ΔT Q = 19602J/kg / 130J/kg-°C = ΔT ΔT = 150.78°C I have a couple of questions about this equation (even if correct) Why/how did the m/s/s part of the equation change to J/kg? Why does the example in the book leave the mass out of the equation when solving for the heat?