A 3.00 gram lead bullet travels at a speed of 240 m/s and hits a wooden post. If half the heat energy generated remain with the bullet, what is the increase in temperature of the embedded bullet? (clead = 128 J/kg C)

Answer 1

KE= 1/2 mv^2 = 1/2 (3x10^-3)(240)^2 = 86.4J
Q=mct
t= Q/mc = 86.4J/ (3x10^-3)(128J) = 225celsius
Divide by half, 112.5, round it off to 113 degrees celsius.