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Specific heat physics help

  1. Sep 10, 2008 #1
    1. The problem statement, all variables and given/known data
    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?

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

    3. 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 :(
     
  2. jcsd
  3. Sep 10, 2008 #2

    Kurdt

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    Re: Thermodynamics...:(

    Are you sure your value for latent heat of fusion is correct?
     
  4. Sep 10, 2008 #3
    Re: Thermodynamics...:(

    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
     
  5. Sep 10, 2008 #4

    Kurdt

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    Re: Thermodynamics...:(

    What about the kilojoules to joules conversion though?
     
  6. Sep 10, 2008 #5
    Re: Thermodynamics...:(

    wait...so 25 kJ/kg doesn't equal .025 J/g?? O_O
     
  7. Sep 10, 2008 #6
    Re: Thermodynamics...:(

    haha wow...sorry...i am so dumb...sorry. :(
     
  8. Sep 10, 2008 #7

    Kurdt

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    Re: Thermodynamics...:(

    No. It equals 0.025 KJ/kg.
     
  9. Sep 10, 2008 #8
    Re: Thermodynamics...:(

    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.
     
  10. Sep 10, 2008 #9

    Kurdt

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    Re: Thermodynamics...:(

    The mass on the right hand side should be converted to kilograms.
     
  11. Sep 10, 2008 #10
    Re: Thermodynamics...:(

    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.
     
  12. Sep 10, 2008 #11

    Kurdt

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    Re: Thermodynamics...:(

    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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