How Much Ice Melts When Shot with a Lead Bullet?

[Bennigan88]

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
$$\dfrac{1}{2} m v^2 + m_{bullet} c_{bullet} \left|\Delta \right| T = L_f \Delta m$$

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

Then calculate the amount of ice that melts using:
$$L \Delta m = -m_{bullet} c_{bullet} \left( T_f - T_i \right)$$
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?

 

[haruspex]

Changing the variable names to be consistent, you have:
$$\dfrac{1}{2} m_b v^2 = m_bc_b \left( T_h-T_{30}\right)$$
$$L \Delta m = -m_b c_b \left( T_f - T_h \right)$$
Subtracting one from the other gives
$$\dfrac{1}{2} m_b v^2 - L \Delta m = m_bc_b \left( T_f-T_{30}\right)$$
Rearranging
$$\dfrac{1}{2} m_b v^2 - m_bc_b \left( T_f-T_{30}\right) = L \Delta m$$
For the bullet, $$\Delta T = T_f-T_{30} < 0$$, so $$- m_bc_b \left( T_f-T_{30}\right) = - m_bc_b \Delta T = +m_bc_b |\Delta T|$$
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.