A 2.43-g lead bullet traveling at 478 m/s strikes a target, converting its kinetic energy into thermal energy. ?

A 2.43-g lead bullet traveling at 478 m/s strikes a target, converting its kinetic energy into thermal energy. Its initial temperature is 40.0°C. The specific heat is 128 J/(kg · °C), latent heat of fusion is 24.5 kJ/kg, and the melting point of lead is 327°C.

(a) Find the available kinetic energy of the bullet.


(b) Find the heat required to melt the bullet.

NCS2020-08-08T03:32:09Z

Favorite Answer

(a) KE = ½mv² = ½ * 0.00243kg * (478m/s)² = 277.6 J

(b) Q = m*(c*ΔT + L)
Q = 0.00243kg * (128J/kg·ºC * (327-40.0)ºC + 24500J/kg)
Q = 148.8 J

bonus:
so the bullet melts in its entirety

Hope this helps!