physics

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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 -

    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 -

    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 -

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