ok second attempt:
55 grain 193 lead core bullet weighs: 3.6 grams = .0036 Kg
bullet speed leaving the chamber is: 938m/s
initial bullet temperature is: 267.22C
assuming the bullet is all lead (ignoring copper because it's a small amount of the weight) our specific heat is: 0.127 J/gm K
so we'll go about this by calculating the maximum temperature the bullet could achieve, aka hitting a solid object upon leaving the barrel:
Kinetic Energy = .5mv^2 = .5*0.0036*938^2 = 879,844 Joules
if:
Q = m * c * ΔT
where Q is the heat energy put into or taken out of the substance, m is the mass of the substance, c is the specific heat capacity, and ΔT is the temperature differential.
then: ΔT = Q / (m * c)
Change in Temperature = 879,844J / (0.127*.0036) = 1,924,418,197.73 Degrees Celsius is our maximum temperature of our bullet on impact?....but wait there's more!
add our initial temperature of 267.22 to the mix (now seems almost laughable)
we get:
1,924,418,464.95 Degrees Celsius!
In conclusion:
Thank god all of the energy of a bullet doesn't turn into heat when it impacts, because the lead would be instantly vaporized and you'd get lead poisoning in about a single bullet.
DOUBTERS TAKE NOTE: ok yes, this temperature is not realistic. However the calculations are sound i double checked, the difference is that the bullet often fragments and warps. If the bullet did not warp at all and stopped instantly then this would be the final temperature
it surprised even me
-Kirk