Matt Springer is a graduate student of physics at Texas A&M university. He is also an occasional writer and tinkerer, and he is probably too curious for his own good.
when you order from Amazon, and a fraction of the purchase price will be sent to me at zero cost to you. Much obliged, and thanks for your patronage.What's the temperature inside a microwave oven?
I've seen some thermodynamics textbooks start off with a preliminary definition of temperature that amounts to "The temperature is what a thermometer says it is", since temperature is really a concept that fundamentally is derived from energy and entropy. So the books like to discuss those at length before talking about the real definition of temperature despite the fact that temperature is what people are accustomed to seeing.
So let's put a thermometer in the microwave. If it's a mercury thermometer it will probably rocket up to off-scale high before breaking. An electronic thermometer will simply die instantly. Strictly from the thermometer definition, either a microwave is tremendously hot or hasn't got a temperature at all. That's answer number one, and it's not a very good one. But if you could insulate the thermometer from the radiation and let it reach equilibrium with the air in the oven, it would be pretty much at room temperature. We know this because if you let a microwave run for a little while and then open it up, the air inside isn't hot. And since air can't exactly cool instantly, it can't have been hot while the oven was running.
That's answer number two. If you define temperature as the average kinetic energy of the air molecules in the microwave, it's at or near room temperature. This is why it's very difficult to bake things in a microwave. They'll get hot, but they don't brown from external hot air as in a conventional oven. And since "average kinetic energy of the molecules" is the actual physics definition of temperature, this is the answer.
But food placed in an operating microwave will get hot quickly, and that doesn't happen to objects just sitting in room temperature air. The resolution to this is the fact that food in a microwave oven isn't getting its heat by conducting heat from the kinetic energy of the molecules in its environment. Instead, it absorbs energy directly from the microwave radiation. Air doesn't, or at least not particularly well. Thus the temperature of the oven environment is largely immaterial to the cooking process.
Now if you want to you can extend this a little bit, as writers occasionally do in answering questions about the temperature of space. You'll often hear about the 2.7K cosmic background radiation. This is talking about the microwave radiation that saturates the whole universe, and they say it has a temperature of 2.7 Kelvin because a blackbody heated up to 2.7 Kelvin would radiate the same wavelengths. All other effects aside, a blackbody left floating in space far from any other light source or hot objects will reach that same temperature because that's the temperature at which it's radiating energy away at the same pace it's absorbing energy from the microwave background.
In space near the earth there's this huge star not terribly far away, and so the energy emitted by the sun will make a blackbody considerably hotter. As a rough estimate, you can use the Stefan-Boltzmann law, plug in the solar constant, and solve for T. Doing this, I get that the blackbody "temperature" of space near the earth is about 123 degrees Celsius. Which is pretty hot. Things like the Space Shuttle don't get that hot because they're not blackbodies: they reflect much of the incoming radiation away.
Physical Science
Comments
Temperature is not a meaningful measure within a microwave oven cavity because the radiation within it does not have a thermal spectrum. It is strongly peaked in a fairly narrow band. Because this radiation field is fairly strongly coupled with certain materials, there is an energy transfer between the field and those materials, but the thermodynamic concept of temperature cannot be usefully applied.
As for the space shuttle, it is sitting in a blackbody radiation field, it will warm up. It doesn't matter how much you reflect away, some will get through, and you must equilibrate to the temperature of the surroundings. There are three things that keep the space shuttle from heating up to above boiling. First, it has a fairly high reflectivity, so it takes a long time to come to thermal equilibrium. Second, because the shuttle is limited to fairly low inclination orbits and low altitudes, it necessarily spends about half of each 90 minute orbit in shadow, and finally, it has a big radiator in the cargo bay that it uses to reject heat to the surroundings.
The shuttle radiators represent 111 square metres of heat rejection area, heat is carried there by two water loops and two freon loops. Late in ascent, before the shuttle bay doors can be opened and the radiators exposed to space, flash evaporators are used that simply boil water off into space, cooling the shuttle by the heat of vaporisation. The same thing happens when the doors are closed before re-entry. However, if the bay doors cannot be opened fairly promptly on arrival in orbit, the mission has to be scrubbed because the shuttle can't reject heat quickly enough.
It's also worth noting that the shuttle gets heat not only from impinging solar radiation, but also from the fact that there are exothermic chemical reactions going on aboard, most notably the human crew and the fuel cells.
Posted by: Winter Toad | August 15, 2008 11:22 AM
Also, spacecraft thermal engineers do wacky things like defining the properties alpha and e as "absorptance" and "emittance" of a surface, respectively. "Huh??" I go, knowing from freshman physics that the radiative and absorptive properties of a surface are equal.
I figured it out before they told me: alpha is defined at the wavelengths that will be absorbed and heat up the spacecraft (i.e. visible light from the sun), and e is defined for longer infrared wavelengths, where most of the radiated energy from the spacecraft will be carried away by radiative cooling. Okay, that makes sense then. Still, I consider thermal engineers a bit wacky anyway (in the best sense of the term).
Posted by: Johnny Vector | August 15, 2008 12:34 PM
I always just thought the "how hot inside a microwave" question was simply ill posed. Places don't have temperatures. Things do. Thus:
- how hot is the air in a microwave?
- how hot is the food/dish/walls/thermometer in a microwave?
And these all have obvious answers.
Posted by: kevin | August 15, 2008 12:35 PM
You miscomputed the blackbody temperature of space near earth.
The solar constant at earth is 1,366 W/m^2.
A sphere, which is the default shape we consider, has a cross-section of pi*r^2, and a surface area of 4 * pi * r^2. It absorbs based on cross-section, radiates based on area. Thus, it will retain constant temperature if it radiates away 342W/m^2. Thus, we solve:
T^4 = 342 / 5.67e-8 = 6.03e9
T = 278K = 5C
Posted by: Anthony | August 15, 2008 12:43 PM
Consider a Pyrex measuring cup filled with tap water placed in a microwave oven. As the magnetron irradiates the cavity, the water will warm quickly. At the point where the water boils, the borosilicate glass will be warmed by conduction of heat from the water, while the air inside will be slightly warmed by the evaporation of heated water. You've got three objects in there, each at a different temperature.
You can measure these by opening the door and putting your hand inside to feel the air temperature and the glass temperature. By seeing the water boiling, you know the temperature is 100 C.
So there isn't a single temperature, not in this example.
(Yeah, you guessed it, this is my method for preparing green tea.)
Posted by: 6EQUJ5 | August 15, 2008 12:43 PM
Anthony, hence "as a rough estimate". ;)
Actually, my motivation was more to examine the equilibrium on the side absorbing the radiation. Heat isn't going to propagate through the structure instantly, and so as a matter of measuring the temperature of something directly exposed to the solar light, my approximation will be better. For instance, the daylight side of the moon is generally around 107 C. But your point as regards small, good conductors of heat is well taken.
Kevin, precisely so. The question is ill-posed. There's a few ways that the question can be better posed, and I focused on three of them. You and 6EQUJ5 have posed additional valid ways. Thanks!
Posted by: Matt Springer | August 15, 2008 1:04 PM
Earth's atmosphere is not a blackbody absorber. It has a perversely large absorbance for vacuum UV. An active sun's marginally cooler sunspots are surrounded by signficantly warmer faculae, then Stefan’s law T^4 emission plus Wien’s displacement law. Then add very hot mass ejections as "controlled nuclear fusion" is empirically gainsaid by positive feedback magnetohydrodynamics. The past 20 years enjoyed roaring huge local energy deposition.
We are now enjoying a dead quiet sun in what should Officially be the most active sunspot maximum in recorded history. Uncle Al goes on record that by 2015 (phase lag - the hottest part of the day is ~1500 hrs not 1200 hrs, that of the year ~August not the June solstice) catastrophic Global Cooling will be ripping global harvests to shreds. Drink your biofuels and emit as much CO2 as you can.
How much Carbon Tax on Everything will it cost to unbury deep saline aquifer CO2 depots?
Posted by: Uncle Al | August 15, 2008 1:26 PM
I didn't think microwaves heated by excitation, I always thought they created a field that caused the water molecules in food to vibrate...
Posted by: Alex | August 15, 2008 5:24 PM
That's vibrational and rotational excitation. The microwave oven is run off-resonance for water so it penetrates about 3/4" rather than charring the surface. With suitable lossy inductors surroudned by foamy or fibrous insulation you can have a small kiln for doing brazing and glazes. SiC is interesting for becoming more lossy with increasing temp, limitng runaways.
Microwave for condensed phase chemistry can be leveraged. By putting polar reactants in a microwave-transparent solvent you can thermally slam the reaction then use the solvent as a thermal sink to save the product.
Posted by: Uncle Al | August 15, 2008 8:06 PM
Alex: there is nothing special about the frequency of microwave radiation produced in a microwave oven as it might relate to any natural frequencies of the water molecule. The frequency of operation of a consumer microwave oven is 2.45 GHz, and was chosen simply because it lies in an unregulated part of the electromagnetic spectrum allocated for non-communication uses. Commercial microwave ovens operate at another frequency, typically 0.915 GHz, penetrating deeper into objects, and this is also not near any particular water resonance. Microwave energy will heat anything that has mobile charge carriers and non-zero resistance, of which water is only one example.
In my opinion, one of the most interesting graphs in physics can be found on page 291 of Jackson's "Classical Electrodynamics", the second edition. This graph shows the absorption coefficient of pure water to light of frequencies from about 100 MHz up to ultraviolet, and again around 1E+17 Hz and upward. At 2.45 GHz, the absorption coefficient is a smoothly increasing line, meaning that if you were to reduce the frequency somewhat, the microwave energy would penetrate farther, and if you increased it somewhat, it would penetrate less. There are no resonances or other unusual features in that region of the graph. The remarkable (to me) feature of this graph is that it shows with amazing clarity why the frequencies that humans can see are what they are, as the absorption coefficient of water crashes a remarkable 7 orders of magnitude over less than a factor 10 in frequency, stays low over the visible range, then climbs 9 orders of magnitude over about another factor of 2 in frequency.
Posted by: Winter Toad | August 16, 2008 9:56 AM
What's amazing to me is that in this article and through the comments no one has considered how microwaves actually "cook" the occupants of the cavity. Microwaves use high frequency switching (not EM waves) to change the magnetic field of the cavity. It's not like it's generating enough EM radiation (i.e. enough kWatts of power) to actually cook something by radiation alone-- rather, the high frequency switching of the magnetic field causes dipoles of the contents of the cavity to align and realign with the magnetic field, which does actually generate heat. In other words, the more water you have in a food (H20 has a dipole moment), the hotter it becomes, relative to a drier food, over the same cooking time. This makes sense. As an experiment (one I've inadvertently repeated every thanksgiving since my youth), take a piece of dried turkey and a piece of moist turkey, and cook them for the same amount of time in the oven. You will most definitely find that the dry piece is lukewarm and the moist piece is scalding hot. You'll also notice that when the moist piece cooks, there is a great deal of moisture in the air of the oven. Also, Mercury has a high dipole moment as well, which is why a thermometer explodes inside of a microwave oven.
Comment #1 almost had it correctly -- his comment should read thermodynamic concepts of CONDUCTIVE AND CONVECTIVE temperature can't be usefully applied. Comment #5 was on the right track, in his observations of water in the microwave. Comment #9 was absolutely correct-- the 900MHz frequency was chosen because at that frequency water molecule dipole switching becomes resonant. Any faster, and the molecules do not have time to completely align to each new magnetic field, and any slower, some of the heat generated by the dipole switching is allowed to dissipate. Sorry comment #10, but you're applying conventional principles of EM radiation which do not apply-- there is no EM radiation, just magnetic field switching.
Posted by: Joel | August 18, 2008 9:45 AM
All,
Please excuse my ignorance of basic physics...
However, if a flat object was placed into orbit at the same radius as the Earth wouldn't one side heat to roughly 393K and the other side cool to near absolute zero? Assume the plane of the object is perpendicular to the suns rays and the object is made of some kind of reasonably good insulator (say asbestos).
My rough calculations appear to show that the sun facing side of the object would be radiating heat as fast as it was absorbed. Is this correct?
Posted by: Peter Schaeffer | August 19, 2008 5:13 PM
Joel: I'm afraid somebody has misinformed you as to the mechanism by which a microwave oven heats materials. The heating mechanism is dominantly due to the oscillating electric field acting on electric dipoles and mobile electric charges. Most foods consist of materials without a magnetic dipole moment, so magnetic field fluctuations don't directly force any mechanical motion, though a varying magnetic field will also drive free charges in a conductor by inductive coupling. There is no resonance in water at 900 MHz, check the graph in Jackson.
I prefer to talk about the radiation field in a microwave, rather than "EM radiation", because it's constrained to remain within a cavity and is operating in the near-field regime (where the physical dimensions of the components and spatial extent of the field are comparable to the wavelength of the radiation), but the fact remains that there are both electric and magnetic fields in a microwave oven, and there is a Poynting vector that, time averaged, points from the resonator to the cooking chamber, so there is a net flow of electromagnetic energy along there. The equations for electromagnetism in the near field are awkward because there are many mathematical simplifications that can no longer be applied, but they are still the correct ones to apply in this case.
Posted by: Winter Toad | August 19, 2008 5:20 PM
The blackbody temperature of space at earth's orbit has to be less than the average surface temperature of the earth, because the earth has been in that orbit and the sun has been emitting that level of power for long enough to bring the earth to at least that temperature.
Posted by: Anthony | August 23, 2008 2:31 AM
Anthony, there's a few complicating factors. The earth isn't a blackbody (I think it reflects something like 1/3 of incoming sunlight), and half of it is in shadow so "the temperature of space" is kind of hazy in this instance as discussed in the comments. And the earth generates its own heat from radioactive decay in the interior. It would be very interesting to take all these effects into account and try to derive the average surface temperature of the earth as accurately as possible.
Posted by: Matt Springer | August 23, 2008 2:33 PM
Geek talk?
Posted by: Anonymous | March 30, 2009 8:11 PM