How Fast Must a Bullet Travel to Melt on Impact?

AI Thread Summary
The discussion revolves around calculating the minimum muzzle velocity required for a 7.2-gram lead bullet to melt upon impact. The melting point of lead is 327°C, and the investigator uses the equations for heat transfer and kinetic energy to solve the problem. A misunderstanding arises regarding the conversion of units, particularly the heat of fusion from kilojoules to joules, which leads to incorrect calculations. Once the correct unit conversions are applied, the participant successfully finds the right answer. The importance of maintaining consistent units throughout calculations is emphasized for accurate results.
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Homework Statement


At a crime scene, the forensic investigator notes that the 7.2- lead bullet that was stopped in a doorframe apparently melted completely on impact. Assuming the bullet was shot at room temperature 30, what does the investigator calculate as the minimum muzzle velocity of the gun?

Homework Equations


Q = mc(change in T)
Q = mL
K = .5mv^2

The Attempt at a Solution


So this is really ticking me off because I was so certain that what I was doing was correct. So the melting point of lead is 327C. Heat of fusion (c) is .025J/g. Specific heat is .130J/g.

Basically I plugged these values into the corresponding equations, and

Q(heat of fusion) + Q(specific heat) = .5mv^2

and tried to solve for V but it won't give me the correct answer :(
 
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Are you sure your value for latent heat of fusion is correct?
 


yup...it's 25 kJ/kg but since I'm using grams since the mass of the bullet is in grams, i converted it to .025 J/g
 


What about the kilojoules to joules conversion though?
 


wait...so 25 kJ/kg doesn't equal .025 J/g?? O_O
 


haha wow...sorry...i am so dumb...sorry. :(
 


No. It equals 0.025 KJ[/color]/kg.
 


so Q(heat of fusion) = (7.2g) (25J/g) = 180

and Q(specific heat) = (7.2g) (.130J/g) (327-30C) = 277.992

277.992 + 180 = 457.992 = .5mv^2 = .5(7.2)(v^2)

the problem is when i solve for v, it turns out too small and the wrong answer.
 


The mass on the right hand side should be converted to kilograms.
 
  • #10


thanks, i got the right answer. but would you care to explain why i need to use kilgrams for the .5mv^2 equation when i used grams in all my other calculations? thanks.
 
  • #11


Well the constants in your other equations were given in terms of grams for convenience because the bullet weight was only a few grams. You gained your answer in Joules which is the SI unit of energy, and thus you must use SI units in the kinetic energy equation. The SI unit of mass being the kilogram. Just try and keep the units in mind as you're working through any problem. As you can see they are very important.
 
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