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How Much Ice Melts When Shot with a Lead Bullet?

  1. Nov 16, 2012 #1
    I am trying to find the amount of ice that melts when a lead bullet traveling at 240 m/s at 30 °C. This textbook claims I should use the formula
    [tex] \dfrac{1}{2} m v^2 + m_{bullet} c_{bullet} \left|\Delta \right| T = L_f \Delta m[/tex]

    What I don't understand is why I shouldn't first calculate how hot the bullet gets after embedding into the ice using
    [tex] \dfrac{1}{2} m v^2 = mc \left( T_f-30^{\circ}C \right)[/tex]

    Then calculate the amount of ice that melts using:
    [tex] L \Delta m = -m_{bullet} c_{bullet} \left( T_f - T_i \right) [/tex]
    Where Tf is 0°C and Ti is the Tf after the embedding in the ice.

    The data is that the mass of the bullet is 3.00g, so (1/2)mv^2 = mc(Tf-Ti) gives a Tf of 481 deg Celsius, which would cause a change in mass (melting of the ice into water) of .56 g but this is incorrect. Can anyone explain why this method is wrong?
     
  2. jcsd
  3. Nov 16, 2012 #2

    haruspex

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    Changing the variable names to be consistent, you have:
    [tex] \dfrac{1}{2} m_b v^2 = m_bc_b \left( T_h-T_{30}\right)[/tex]
    [tex] L \Delta m = -m_b c_b \left( T_f - T_h \right)[/tex]
    Subtracting one from the other gives
    [tex] \dfrac{1}{2} m_b v^2 - L \Delta m = m_bc_b \left( T_f-T_{30}\right)[/tex]
    Rearranging
    [tex] \dfrac{1}{2} m_b v^2 - m_bc_b \left( T_f-T_{30}\right) = L \Delta m[/tex]
    For the bullet, [tex]\Delta T = T_f-T_{30} < 0[/tex], so [tex]- m_bc_b \left( T_f-T_{30}\right) = - m_bc_b \Delta T = +m_bc_b |\Delta T|[/tex]
    That gives the equation in the book, so you ought to get the same result either way. If you did not then you must have made a numerical error.
     
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