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.
oubaas
Favorite Answer
(a) Find the available kinetic energy KE of the bullet.
KE = m/2*V^2 = 24.3*10^-4/2*4.78^2*10^4 = 278 joule
(b) Find the heat required to melt the bullet.
E = 2.43*(0.128*(327-40)+24.5) = 149 joule
Whome
Please remember to select a Best Answer from among your results.
Jim
Ensure all values are in KMS
KE = ½mv² = ½ * 2.43g * (1kg/1000g) * (478m/s)² = 277.60806 J
Now round to 3 sigfigs of 278J
Probably a .22 bullet or equiv bullet weight about 40 gr (2.6 g)
Rick
that's a REALLY SMALL bullet !!!! I use 230 grain !!!!