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Wouldn't also some mass be converted into energy? Probably not much though
by spaetzleesser 6y ago
Wouldn't also some mass be converted into energy? Probably not much though
- Toutouxc 6y agoHow? What do you mean by that?
- brippalcharrid 6y agoTrace amounts of radioisotopes found in the candle will be subject to decay while the candle is burned (much as they would be while the candle isn't burning), so there would presumably be some negligible amount of alpha, beta and/or gamma radiation.
- snissn 6y agovery very very little
- nostromo 6y agoNone, actually. It’s a chemical reaction which means all mass is preserved.
- dheera 6y agoNo, actually, very very little, not zero. The candle (summed with its gas emissions) loses mass equal to the relativistic mass of the photons emitted. For something like a candle it's an undetectably small amount but it's not zero.
- nostromo 6y agoGood catch, thanks.
- spaetzleesser 6y agoIs heat basically a bunch of photons?
- Cerium 6y agoFor normal non-nuclear candles the energy obtained through breaking chemical bonds and creating outputs with less energy contained in the connections between atoms. The matter is not destroyed.
- PeterisP 6y agoWell, sure, but energy=mass; just as an atom's mass is not equal to the sum of the protons, neutrons and electrons masses but depends also on their binding energy, in the same manner, a compound with highly energetic chemical bounds has a (very, very slightly) different inertia and gravity than the separate components. The difference of this effect between atomic and chemical bounds [spelling? English isn't my main language] is quantitative, not qualitative; chemical bounds have much less energy, so the effect is much smaller, so for most purposes it can be considered as insignificantly small, but it's still there and not zero. Any system that loses energy (e.g. releases an energetic photon - no matter if it's because of a nuclear or chemical reaction) also loses mass, though E=mc^2 means that losing reasonable amounts of energy mean losing very, very small amounts of mass.
- smithza 6y agoenergy != mass; e=mc^2 declares there is a proportional relationship between energy and mass while the speed of light, c, squared is the constant of proportionality.
- carlob 6y agoYou can set c to 1 if you want, then energy == mass. As a matter of fact this is pretty common in particle physics where all masses are given in electronvolts. https://en.wikipedia.org/wiki/Geometrized_unit_system https://en.wikipedia.org/wiki/Geometrized_unit_system
- PeterisP 6y ago"there is a proportional relationship between energy and mass" -> exactly. the mass of the system is strictly proportional to its total energy, and changes in the energy of a system will have a corresponding proportional effect on its mass. If you split the system into two components and release X energy, then the two components will weigh X/c^2 less than the whole system did; if splitting the system consumed X energy, then the separate components will weigh X/c^2 more than the original thing.
- 0_____0 6y agothink about the energy in a candle released via burning vs the energy of that same mass viewed through E=mc^2. that's the ratio
- graerg 6y agoNo this is a chemical reaction, not a nuclear reaction.
- metric10 6y agoYes. Back of envelope, took physics in college many years ago[1] analysis: E=mc^2, or energy = mass * (speed of light)^2. According to Wikipedia a candle can produce 77 watts of energy "combined." I guess that means 77 joules (1 watt = 1 J/s). So we have: 77 = m * (299792458)^2 Solving for m via Wolfram Alpha: m = 11 / 12839369696240252 Which is in grams. That's a _very_ small amount, but it's not zero. edit: [1] If I'm being honest, I got E=mc^2 from watching the Twilight Zone as a kid, not college physics.
- Ovah 6y agoKind of pedantic but I've never understood why it's E=mc^2 and not E=Δmc^2. In Einsteins original paper he derives the equation with a delta m: a change of mass corresponds to some amount of energy. To me that is different than to say that a whole mass corresponds to a some amount of energy. I've never seen a justification why the delta can be omitted and why the equation still would hold true.
- colinmhayes 6y agoThe delta is implied. If you're being rigorous yea you'd include it. But you never see it used rigorously.
- Ovah 6y agoPhysics is usually pretty rigorous. In the context of science for a general audience, or printing pretty T-shirts, I get why it's omitted. I just find it funny that when it comes to E=mc^2 physics suddenly lack some rigour even in the college physics classes I've taken. It's probably OK to skip it I'm just curious as to why. I don't think I've ever seen the delta mentioned except in Einstein's original paper.
- snissn 6y agousually the momentum is really low so it's fine, you could read this wikipedia article about it, share questions if that doesn't clear things up https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relation https://en.wikipedia.org/wiki/Energy%E2%80%93momentum_relati...
- magicroot75 6y agoIsn't that energy what we call "heat?"