Posted by **Eric** on Friday, July 19, 2013 at 7:51pm.

A 3.00 g lead bullet traveling at 640 m/s strikes a target, converting its kinetic energy into thermal energy. It's initial temperature is 40.0°C.

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

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

- physics -
**bobpursley**, Friday, July 19, 2013 at 10:18pm
KE=1/2 massbullet* velocity^2

heat required to melt=heat required to change temp to melting poing+heat to change lead to liquid

= mass*specifHeatLead*(Tm-40)+mass*HeatfusionLead

look up the specifice heat of lead, the temperature of melting lead, and finally the heat of fusion for lead.

- physics -
**Graham**, Friday, July 19, 2013 at 10:29pm
bullet mass: M = 3.00g

initial velocity: V0 = 640m/s

initial temperature: T0 = 313.15K

final velocity: V1 = 0m/s

melting point of Lead: T2 = 600.61K

specific heat of lead: C = 0.128J/g/K

latent heat of melting: L = 22.4J/g

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

∆E1 = (1/2)(M)(V1^2-V0^2)

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

∆E2 = M C (T2-T0) + M L

- physics -
**mh**, Sunday, November 29, 2015 at 9:33pm
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