Question

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)

Answer 1

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.

Answer 2

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

Answer 3

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