Assume the total chemical energy stored in the propellant in ammunition is 6500J

Question

Assume the total chemical energy stored in the propellant in ammunition is 6500J; the bullet weighs 6.35g and the barrel is made of steel with a heat capacity of .45j/(gC). Kinetic energy is expressed as 1/2mv2 and one J=1(kgm2)/s2.

1. If the conversion of chemical energy to kinetic energy is 30% efficient, estimate the velocity of the bullet in meters/second.

2. If the barrel weighs 18.2g and 10% of the chemical energy is lost to heating the barrel, what will be the final temperature of the barrel? assume that it starts at 25 C

Explanation / Answer

1) assuming 30 % efficiency kinetic energy = ( 30 /100) x 6500 J = 1950 J

KE = 1950 J = 1950 kgm2/s2

E = 1/2 m x v^2            where m = mass of bullet = 6.35 g = 0.00635 kg

1950 kgm2/s2 = 1/2 x 0.00635 kg x v^2

v = 783.7 m/s is velocity of bullet

2) 10 % energy is lost which means energy lost = ( 10 /100) x 6500 J = 650 J

this heat is absorbed by barrel

heat absorbed = specific heat of barrel x temperature rise x mass of barrel

650 J = 0.45 J/gC x ( T-25) x 18.2 g

T = final temp of barrel = 104..4 C

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