Temperature change in an object

[kieslingrc]

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?

 

[issacnewton]

hi

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 $kg.\frac{m^2}{s^2}$
and not $kg.\frac{m}{s^2}$ as you have done. And $kg.\frac{m^2}{s^2}$
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

$$1\mbox{ Joule }=1 \; kg.\frac{m^2}{s^2}$$

So if the Q is energy, the unit of Q is J. We can write Q as $Q=\frac{Q}{m}.\;m$ . Now the units of $\frac{Q}{m}$
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...

 

[kieslingrc]

hi

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 $kg.\frac{m^2}{s^2}$
and not $kg.\frac{m}{s^2}$ as you have done. And $kg.\frac{m^2}{s^2}$
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

$$1\mbox{ Joule }=1 \; kg.\frac{m^2}{s^2}$$

So if the Q is energy, the unit of Q is J. We can write Q as $Q=\frac{Q}{m}.\;m$ . Now the units of $\frac{Q}{m}$
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...

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.

 

[issacnewton]

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.

 

[asteriatic]

I would like to clarify and provide an explanation for the questions raised in the response.

Firstly, the m/s/s unit in the calculation is not changed to J/kg. The unit of J/kg is used to represent the heat capacity of the material, which is the amount of heat required to raise the temperature of 1 kg of the material by 1 degree Celsius. In this case, the heat capacity of lead is 130J/kg-°C, which means that 130 joules of heat is required to raise the temperature of 1 kg of lead by 1 degree Celsius.

Secondly, the example in the book may have left out the mass in the equation because it is assumed to be constant and known. It is common practice in physics and chemistry to cancel out units that are common on both sides of an equation, which may be the reason why the mass was not explicitly included in the calculation.

Lastly, it is important to note that the equations used in this problem are simplified models and may not accurately represent the actual scenario. In reality, some of the energy from the bullet may be lost to other forms, such as sound or deformation of the armor plate, and the heat may not be evenly distributed throughout the bullet, resulting in a slightly different temperature change. These factors should be taken into account when interpreting the results of this calculation.