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A 3.00x10^-3 kg lead bullet enters a target with a speed of 2.00x10^2 m/s and comes to rest within the target. Calculate the rise in temperature of the bullet assuming that 80.0% of the heat produced is absorbed by the bullet. The Specific heat of lead is 1.30x10^2 J/kg K

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    KE=(1/2)mv^2
    KE=(1/2)(.003)(200^2)
    KE=60J*(1cal/4.184J)
    KE=14.34cal

    Q=(%)(KE)
    Q=(.80)(14.34)
    Q=11.47 cal

    Q=mcT
    T=Q/mc
    T=(11.47)/(.003)(1.3x10^2)
    T=29.41 degree celsius

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