A 5.90-g lead bullet traveling at 410 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?

rise in temperature = 333.53°C

Explanation:

mass of lead bullet, m = 5.9 g

initial velocity, u = 410 m/s

specific heat of lead = 0.126 J/gm - K = 126 J/kg K

Initial kinetic energy of the bullet

K = 1/2 mu^2 = 0.5 x 5.9 x 10^-3 x 410 x 410 = 495.895 J

Half of the kinetic energy is used to raise the temperature of lead bullet

K / 2 = mass of bullet x specific heat of lead x rise in temperature

247.95 = 5.9 x 10^-3 x 126 x rise in temperature

rise in temperature = 333.53°C

Thus, the rise in temperature of lead bullet is 333.53°C.