 |
| ||
| |||
|
| Math Central | Quandaries & Queries |
|
Question from William, a student: A 3.00 x 10^-3 kg lead bullet travels at a speed of 2.40 x 10^2 m/s and hits a wooden post. If half the heat energy generated remains with the bullet, what is the increase in temperature of the embedded bullet? (specific heat of lead = 1.28 x 10^2 J/kg) |
Hi William.
The bullet's temperature increase is a function of the heat transferred to it as a result of change from kinetic energy to heat. This relationship is Q = cm ΔT where Q is the energy transferred, c is the specific heat of the substance, m is its mass and ΔT is the change in temperature.
So you need to know Q. You are told this is half the heat energy generated, and since all the kinetic energy is transferred into heat (I presume we are ignoring sound and the energy required to penetrate the wood), Q = (1/2)K, where K is the kinetic energy of the bullet.
The kinetic energy of the bullet is simply K = (1/2)mv2, which I'm sure is a formula you've worked with before.
Put it all together and you have an expression for ΔT, the temperature change, in terms of the measurements you have been given.
Hope this helps,
Stephen La Rocque.
| ||
| ||
| ||
 |
 |
|
Math Central is supported by the University of Regina and The Pacific Institute for the Mathematical Sciences.