Date: Sun, 2 Oct 2005 12:48:55 -0400 Reply-To: Sharing resources for high school physics Sender: Sharing resources for high school physics From: David Stern Subject: Re: E=mc^2 To: PHYSHARE@LISTS.PSU.EDU Dear Chris, dear physics teachers all over The equation E = mc^2 is certainly valid in all sorts of energy transformations, but I hope your students have a clear understanding of what it means. When energy is released, mass (i.e. inertia) does change, but that is best viewed as a side effect, rather than a cause. Let me explain. Suppose we have a cocked spring, releasing its potential energy by (say) running a clock. As the spring winds down, its energy is expended to turn cogs and wheel, ultimately changing into heat through friction. One may argue, that the mass of the relaxed spring is smaller than that of the one tightly wound up, by some infinitesimal amount derivable from E = mc^2. That is correct, but incidental: any calculation deriving E better look at the elastic state of the spring before and after. The change in m is different, and in most calculations, it's purely an interesting side effect. Now for a less trivial example: nuclear energy derived from nuclear fission. Too many people view this as a conversion of energy to mass, and if we compare the mass of the unfissioned uranium nucleus to that of the fission fragments, a difference does exist. (This is discussed with the binding energy of nuclei, on http://www.phy6.org/stargaze/Sun7enrg.htm) However, the fission process is better understood if the uranium nucleus is viewed as similar to the cocked spring in the previous example. It contain 92 protons, each carrying the same positive charge, each repelled by the other 91. A lot of electrostatic energy could be released, if only the protons could move apart! Usually they cannot do that, because the strong nuclear force holds them together. That is a short-range force, dominant over the dimensions of the uranium nucleus, but not much further out. Furthermore, its dominance is by a rather slim margin--note that stable nuclei of 93 or more nuclei do not exist (even uranium is unstable, though on the scale of billions of years)--and a small tap, a small extra energy when a thermal neutron attaches itself to this nucleus, is enough to overcome it. Once the nucleus fissions, a great amount of of energy is released--but that is really electrostatic energy, due to two clumps of protons repelling each other. Yes, the mass of the fission fragments is slightly reduced--but students should understand that the energy itself comes from familiar electrostatics, not from some arcane application of E = mc^2. For more: http://www.phy6.org/stargaze/Snuclear.htm Sincerely David P. Stern Greenbelt, Maryland http://www.phy6.org/stargaze/Sstern.htm --------------------------------------------- Christopher Klug wrote: > There was a recent posting on Chemed-L regarding an > article in the New York Times by Brian Greene. > http://www.nytimes.com/2005/09/30/opinion/30greene.html?pagewanted=print > > Someone brought up their impression that in chemical > reactions, there was no conversion of mass to energy, > as opposed to what is stated in the article. I > supplied the following response, which I thought would > also be a good exercise for teachers to have their > classes do. > >>From the article: > > "This isn't true. When you drive your car, E = mc˛ is > at work. As the engine burns gasoline to produce > energy in the form of motion, it does so by converting > some of the gasoline's mass into energy, in accord > with Einstein's formula. When you use your MP3 player, > E = mc˛ is at work. As the player drains the battery > to produce energy in the form of sound waves, it does > so by converting some of the battery's mass into > energy, as dictated by Einstein's formula. As you read > this text, E = mc˛ is at work. The processes in the > eye and brain, underlying perception and thought, rely > on chemical reactions that interchange mass and > energy, once again in accord with Einstein's formula." > > However, my perspective had always been that the > energy contained in the bonds would actually > contribute to the mass of molecules, but imperceptibly > so with all but the most precise of instruments. > > Lets take methane reacting completely with excess > oxygen as an example. > > CH4 + 2O2 --> CO2 + 2H2O > > The C-H bond has an energy of 413 KJ/mole. > 413*4=1652KJ to break up 1 mole methane > > The O=O bond has an energy of 498 KJ/mole. > 498*2=996KJ to break up 2 moles O2 > > The C=O bond has an energy of 805 KJ/mole. > 805*2=1610KJ to break up 1 mole CO2 > > The O-H bond has an energy of 366 KJ/mole. > 366*4=1464KJ to break up 2 moles H2O > > Thus, 1652+996= 2648KJ on the reactants side is > required to break bonds, and when bonds reform on the > products side, 3074KJ are released. Because of this, > overall 426KJ are released per mole of reaction. The > products are lower in energy by 426KJ would be another > way of thinking about this. > > Applying E=mc^2, 426,000J=m*(3*10^8)^2, and > m=4.73*10^-12 kg. Or, 4.73 pg. 4.73 picograms of > mass are converted to energy when 1 mole of methane > reacts with 2 moles of oxygen to form 1 mole of CO2 > and 2 moles of H2O. > > I can't measure picograms in my high school lab. I > can't even tell you by feel whether two things differ > in mass by a few picograms. Thus, these mass changes > in chemical reactions are imperceptible in most > day-to-day activities, and the "conservation of mass" > still generally holds true...afterall, we still have 1 > carbon atom, 4 hydrogen atoms, and 4 oxygen atoms, > don't we? > > Data ref: > http://www.science.uwaterloo.ca/~cchieh/cact/c120/bondel.html > http://www.chem.lsu.edu/htdocs/people/lgbutler/1202_Fall96/WebInit/Ch9_Bonds.html > > Christopher Klug > Chemistry (Regular-Honors-AP)/Cross Country > Havre de Grace High School > 700 Congress Ave. > Havre de Grace, MD 21078 > 410.939.6600 > TA-NCSS2005-BNL > > > > __________________________________ > Yahoo! Mail - PC Magazine Editors' Choice 2005 > http://mail.yahoo.com > > ----- > * Physhare FAQ/Homepage: > * Physhare Archives and Subscription Management: > ----- * Physhare FAQ/Homepage: * Physhare Archives and Subscription Management: