Question

If a lead bullet goes fast enough, it can melt completely when it comes to a halt suddenly, and all it's kinetic energy is converted into heat via friction. Find the minimum speed of a lead bullet (initial temp. = 30.0 C) needed for this to happen.

So I related that Q = KE therefore Q = (0.5)mv^2

So I related that to the equation Q = cm(delta)T

which gave me (0.5)mv^2 = cm(delta)T

I algebraically get rid of m

resulting in (0.5)v^2 = c (Tf - To)

The melting point for lead is 327.3 which is Tf and the specific heat for lead is 128

So I solve for V = sqrt(2(128)(327.3-30)) which gives me 276m/s

The correct answer is 350m/s what am I doing wrong?

Answer 1

Besides the heat energy needed to raise the temperature of the lead to the 327.4 C melting poing, you need to provide the heat of fusion of lead, in order to melt it.
The heat of fusion of lead is 27 kJ/kg. See if that raises the required velocity enough to give the "correct" answer.

Answer 2

mv^2/2 = cmΔT +λm,
v = sqrt[2•(c•ΔT +λ)] =
= sqrt[2•(1300•(327.5-30) +25000)] =
= 907 m/s.