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

Question

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 · Accepted 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 · Answer

Please remember to select a Best Answer from among your results.

Jim · Answer

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 · Answer

that's a REALLY SMALL bullet !!!!   I use 230 grain !!!!

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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.?

4 Answers