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Another Bullet problem

  1. Jul 3, 2017 #1
    1. The problem statement, all variables and given/known data
    A lead bullet flies at a speed of 450 m / s and, striking a wall perpendicularly, melts
    completely. What was the bullet temperature before the blow, if for melting it consumes only half of its mechanical energy? Latent melting heat and the specific and melting temperature of the lead are λt
    = 25 kJ / kg, c = 130 J / (kg · K) Tmelting= 600 K.

    2. Relevant equations
    ½ x ½mv² = mcΔT + mλ

    3. The attempt at a solution
    The bullet (mass m) is first be heated to its melting temperature. Call the temperature-increase ΔT. This requires energy = mcΔT.

    The bullet is then melted which requires energy = mλ

    The bullet's initial kinetic (mechanical) energy was ½mv². Half of this heats and melts the bullet so:
    ½ x ½mv² = mcΔT + mλ

    v²/4 = cΔT + λ

    λ = 25000 J/kg
    c = 130 J/(kg·K)

    450²/4 = 130ΔT + 25000
    ΔT = (67500 - 25000)/130
    . . . = 327K

    Therefore the temperature has risen 327K to reach melting point (600K) so the initial temperature was 600 - 327 = 273K (i.e. 0°C).



    The problem is that the answer is supossed to be aproximately 400 K but i can't get it.
    I feel like i'm missing something but don't know what.

    Any help will be appreciated.
     
  2. jcsd
  3. Jul 3, 2017 #2
    I think the problem is just the math in the end.
    Everything is right up to:
    [tex]\frac{450^2}{4}\text{=130$\Delta $T+25000}[/tex]
    but [itex]\frac{450^2}{4}=50625[/itex], not [itex]67500[/itex] as you put in your solution.
    So then:
    [tex]\text{$\Delta $T=}\frac{(50625-25000)}{130}\text{=197.115}[/tex]
    So [itex]T=600-197.115=402.885K[/itex]
     
  4. Jul 3, 2017 #3

    gneill

    User Avatar

    Staff: Mentor

    Looks like it may be a calculator problem. Try that calculation again.
     
  5. Jul 3, 2017 #4
    I shoulb be way more attentive. Sorry for bothering and thank you for help.
     
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