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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.
1 Answer
- ?Lv 78 months agoFavorite 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!
