Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Heat and Electron question

  1. Mar 4, 2005 #1
    Here's the question,

    A beam of electrons, travelling at [tex]9x10^6ms^{-1}[/tex] In every second, [tex]1x10^{14}[/tex] electrons hit a metal target and all their energy is converted into heat. How long would it take to heat up the metal from 18 degrees celsius to 80 degrees celsius if the heat capacity of the metal is [tex] 0.15JK^{-1}[/tex]. The mass of the electron is [tex]9x10^{-31} kg[/tex].

    How i solved this problem was to find the momentum of each individual electron using the formula [tex]p=mv[/tex]. Then i found the energy of each electron using the relation [tex]E^2=(mc^2)^2+(pc)^2[/tex]. Then afterwards i found the heat energy in joules required to heat up the metal to 80 degrees celsius.

    The thing is, i got 10.2 for the energy required to heat the metal object up. And the energy per second gained is [tex]8.08886[/tex]. Therefore the answer i got is 1.26 seconds. The problem is, the answer is [tex]1.44x10^5 s[/tex].

    Where did I go wrong? I've got a feeling the momentum-energy relation in SR is a little outta place but I can't think of any other relation which relates momentum and energy. Thanks alot.
  2. jcsd
  3. Mar 4, 2005 #2
    the energy you were using is the kinetic energy + the rest mass energy....... do you see what is your problem?
  4. Mar 4, 2005 #3


    User Avatar
    Science Advisor

    {Kinetic Energy Each Electron} = (1/2)mv2 = (1/2){9.0e(-31) kg}{9.0e(6) m/s}2 =
    = (3.645e(-17) Joules)
    {Power Dissipated In Metal Target} = {Kinetic Energy Ea Electron}*{# Electrons/Sec Absorbed} =
    = {3.645e(-17)}*{1.0e(14)} = (3.645e(-3) Watts)
    {Rate of Temperature Change} = {Power Dissipated In Metal Target}/{0.15 J/oK} =
    = (3.645e(-3) Watts)/{0.15 J/oK} = (0.0243 oK/sec)

    {Time To Heat Metal 18 to 80 oC} = (80 - 18 oC)/(0.0243 oK/sec) = (2551 sec)

    Last edited: Mar 4, 2005
  5. Mar 4, 2005 #4
    ohhhhh yes ! that solved the problem, thanks alot....
  6. Mar 4, 2005 #5
    How does that solved your problem? as Xanthym pointed out, your textbook answer is wrong... so after you know how to do this problem, your answer still can't match the textbook answer....
  7. Mar 4, 2005 #6
    The method i used is different from Xanthym's one as i did not use the formula [tex] KE=\frac{1}{2}mv^2[/tex]. What i did was first to find momentum, then relate it with energy using the relativistic momentum-energy relation.

    The kinetic energy per electron I got is [tex]6.0479974_E -19[/tex] and the kinetic energy hitting the target in one second is [tex]6.0479974_E -5[/tex]. To find the energy required to heat the metal object, i got [tex]9.3 J[/tex].

    So, [tex] \frac{9.3}{60479974_E -5}[/tex], i get answer [tex]143,519.0944[/tex] which is approximately equal to [tex] 1.44_E 5[/tex], which is the textbook answer.
  8. Mar 4, 2005 #7


    User Avatar
    Science Advisor
    Homework Helper

    The approximation using KE=1/2mv^{2} is excellent.No need to apply relativistic formulas..

Share this great discussion with others via Reddit, Google+, Twitter, or Facebook