Question

A 4.2 g lead bullet moving at 275 m/s strikes a steel plate and stops. If all the kinetic energy is converted to thermal energy and none leaves the bullet, what is its temperature change?

Answer 1

𝑚 = 4.2 𝑔 = 0.00420 𝑘𝑔

𝑣𝑖 = 275 𝑚⁄𝑠

𝑣𝑓 = 0

𝑐 = 130 𝐽⁄𝑘𝑔 ∙ ℃

∆𝑇 =?

All of the kinetic energy of the bullet is converted into thermal energy, ⸫

∆𝐾𝐸 = mc∆T

∆T = $\frac{∆KE}{mc}$

= - $\frac{(1/2)m(v_f^2-v_i^2)}{mc}$ = - $\frac{(1/2)(v_f^2-v_i^2)}{c}$

= - $\frac{(1/2)(0-275m/s^2)}{130J/Kg. °C}$

∆T = 290°C