A lead bullet moving with a speed of 400 ~m/s strikes a rigid wall and comes to a complete halt. Assuming that 50 % of the kinetic energy lost is absorbed by the bullet as heat, calculate the rise in temperature of the bullet. (Given: specific heat capacity of lead = 125 ~J/(kg K) )

Step-by-step solution

Let the mass of the bullet be m . The initial kinetic energy of the bullet is K = 1 2 mv^2 . The heat absorbed by the bullet is 50 % of this kinetic energy: Q = 0.5 ( 1 2 mv^2 ) = 0.25 m v^2 The heat absorbed is also given by Q = ms T , where s is the specific heat capacity and T is the rise in temperature. Equating the two expressions for Q : 0.25 m v^2 = ms T The mass m cancels out: T = 0.25 v^2 s Substitute the given values ( v = 400 ~m/s and s = 125 ~J/(kg K) ): T = 0.25 (400)^2 125 T = 0.25 160000 125 T = 4000