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Temperature change in an object

  1. Jan 2, 2012 #1
    1. A lead bullet with a mass of 0.03 kg traveling at 198 m/s strikes an armor plate and comes to a stop. If all the energy is converted to heat and absorbed by the bullet, what is the temperature change?

    2. Q=KE=1/2mv^2; Q=CmΔT

    3. Heat capacity (C) of lead = 130J/kg-°C, mass of the bullet = .03kg, velocity of the bullet = 198m/s

    I first wanted to calculate the KE of the bullet (.5)(.03kg)(198m/s)^2 and obtained 588.06 kg m/s/s but the closest example to this problem in my book does not follow that equation. It instead would calculate it as follows:
    Q = KE = 1/2mv^2: 1/2*m*(198m/s)^2; = (19602J/kg) * m [How does it become a J/kg?]
    Then the increase in temperature caused by this much heat is:
    Q = 19602J/kg * m = CmΔT (the masses will cancel out here)
    Q = 19602J/kg = 130J/kg-°C * ΔT
    Q = 19602J/kg / 130J/kg-°C = ΔT
    ΔT = 150.78°C

    I have a couple of questions about this equation (even if correct)
    Why/how did the m/s/s part of the equation change to J/kg?
    Why does the example in the book leave the mass out of the equation when solving for the heat?
  2. jcsd
  3. Jan 2, 2012 #2

    this question is from thermal physics. in the physics education, you are taught dimensional analysis at the very beginning. so by the time you are solving problems from thermal physics, you should be very clear on dimensional analysis.

    when you do (.5)(.03kg)(198m/s)^2 , the units you get is [itex]kg.\frac{m^2}{s^2}[/itex]
    and not [itex]kg.\frac{m}{s^2}[/itex] as you have done. And [itex]kg.\frac{m^2}{s^2}[/itex]
    is the unit of energy (here, kinetic form). In the honor of english physicist Thomas Joule, who made imporatant contributions to the thermal physics, this unit is called Joule. So

    [tex]1\mbox{ Joule }=1 \; kg.\frac{m^2}{s^2}[/tex]

    So if the Q is energy, the unit of Q is J. We can write Q as [itex]Q=\frac{Q}{m}.\;m[/itex] . Now the units of [itex]\frac{Q}{m}[/itex]
    would be J/kg . So we can write Q as product of some number (whose unit is J/kg) and m (the mass, with unit kg).

    The reason the example leaves the mass separate like this is , as you can see, it cancels out . So it just saves the number of
    numerical computations you have to do.......
    Last edited: Jan 2, 2012
  4. Jan 2, 2012 #3
    Thanks for the reply and for pointing out my error on the units. I knew there was something in common with the J and kg m2/s2, I just couldn't find it in my notes.

    Otherwise, I believe the math to be correct? Thanks for the speedy reply. I really love math, but it's so hard to pick up again when you haven't done it in over 20 years.
  5. Jan 2, 2012 #4
    good luck with your education. remember to always present your work. if you ask a question and don't present your work, people will tend to not answer it since this is not a
    tutoring website.
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