- #1
phhasek
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A 20g lead bullet at 30°C and moving at 350m/s embeds itself in a wodden block.
If 70% of the initial kinetic energy becomes internal energy of the bullet, what is its final temperature?
My attempt:
Ek = (mV^2)/2 = 1225J
70% of this energy = 857.5J
So the available energy to heat the bullet is 857.5J Right?
dQ = m c dt
c - lead = 130J/kg K (specific heat)
dt = (Tf-30°C)
m = 0.02Kg
dQ = 857.5J
Tf = final temperture
So:
857.5 = 0.02 * 130 * (Tf-30)
Tf = 359.8°C
The correct answer should be 327°C
What am I missing?
If 70% of the initial kinetic energy becomes internal energy of the bullet, what is its final temperature?
My attempt:
Ek = (mV^2)/2 = 1225J
70% of this energy = 857.5J
So the available energy to heat the bullet is 857.5J Right?
dQ = m c dt
c - lead = 130J/kg K (specific heat)
dt = (Tf-30°C)
m = 0.02Kg
dQ = 857.5J
Tf = final temperture
So:
857.5 = 0.02 * 130 * (Tf-30)
Tf = 359.8°C
The correct answer should be 327°C
What am I missing?