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## Homework Statement

A 5 g lead bullet moving at 343 m/s is stopped in a block of wood. All of the kinetic energy goes into heat energy added to the bullet. The initial temperature of the bullet is 24 degrees C. What is the final temperature of the bullet in degrees C? (cPb = 0.128 kJ/kg . K, LfPb = 24.7 kJ/kg, MPPb = 600 K)

## Homework Equations

K = 1/2*m*v^2

Q = mc[tex]\Delta[/tex]T

Answer = 327C

## The Attempt at a Solution

1/2mv[tex]^{2}[/tex]=mc[tex]\Delta[/tex][tex]T[/tex]

1/2v[tex]^{2}[/tex] = c[tex]\Delta[/tex]T

Using this equation [tex]\Delta[/tex]T = 460C

I realise that the mass must come into it with the other provided information but can not work out how.

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