# What is the final temp of the lead bullet?

A 5-g lead bullet traveling in 20 C air at 300m/s strikes a flat steel plate and stops. What is the final temp of the lead bullet? (Assume bullet retains all heat.) The melting point of lead is 327 C. The specific heat of lead is 0.128 J/g C. The heat of fusion of lead is 24.5 J/g.
a. 227 C b. 260 C c. 293 C d. 327 C

I thought I would be able to solve this by taking the KE + c*m*delta T=0, but it's not working. I'm also not sure how to figure out if the bullet melts or not. I tried that above along with adding the latent heat of fusion for lead*mass, but that's not working either.

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LeonhardEuler
Gold Member
Remember that 1J=1kg m^2/s^2, so you have to convert the bullet's mass to kg in order to be consistant in your units. Your method is correct.

Thanks for the help, but that's evidently not the problem. The answer is d. 327 C, and I keep getting different answers for some reason. Here's my work so far.

0.5*0.005 kg*(300 m/s)^2 + 5 g*0.128 J/gC*(Tf - 20 C) = 0

Any suggestions?

LeonhardEuler
Gold Member
Yeah, I see what the problem is. Once the bullet reaches 327 C, any additional heat goes into the heat of fusion, rather than raising the temperature further until all the lead has melted. What you should do is calculate how much heat it takes to raise the temperature to the melting point. This is (327 - 20)*5*.128=196.48J. This is less than the 225J of KE the bullet had, so it is heated further. Heating it beyond 327 C would require an additional 5*24.5=122.5J, but there are only 225-196.48=28.52J left, so the bullet is only partly melted and remains at 327 C.

Great explanation, thank you very much