Yes, radiation has a temperature, too. The radiation in your oven is actually in thermal equilibrium with the walls of the oven.
Just look up a black body spectrum - it comes from assuming that the radiation of a black body in an oven is in thermal equilibrium with the radiation in the oven.
The thing about the microwave is that the radiation being used isn't from a thermal source, like the walls of an oven. So it's not that it doesn't have a temperature, it's just that the temperature of the microwaves themselves is actually pretty low. Kind of like how a fast enough blender can boil water.
I thought the spectrum of blackbody radiation was a function of the temperature of the *blackbody* -- not the radiation itself.
I thought the same thing. The black body radiation exists because of the temperature of the object.
Furthermore, the radiation of the "black body" walls of an oven is due to the fact that the oven walls have been heated by radiation, conduction and convection from the heat source of the oven, (gas, electric, wood or nuclear [just kidding about the nuclear]) and the walls are then radiating it back into the oven. That's why when you want to brown something, you should put it in a preheated oven-so the walls can radiate heat all the way around what you want browned.
A radiation in and of itself does not have a temperature. It has energy capable of heating other objects. DON'T confuse the heat of a flame with radiation. A flame has a temperature, and depending on the type of flame, a certain amount of heat that is also radiated.
Microwave radiation isn't a black body radiation in a microwave oven. It is a radiation at a frequency that is absorbed by molecules of water and to a lesser degree, fats and sugars, so they heat up. A microwave oven is simply a 2.45-2.50 GHz radio transmitter with its broadcasting antenna inside the box you place your food into to heat it up.
I think what everyone is posting here is basically correct; maybe just the way it's being phrased is at issue.
Every body acts like a black body in the sense that the body's temperature results in a characteristic emission spectrum. So there are definitely certain bands of radiation associated with materials of a given temperature. That said, "Temperature" is a measure of molecular vibrations, and so it doesn't really make sense to say radiation has a temperature itself. I prefer the way GaryProtein wrote it.
To say that radiation is 'in thermal equilibrium' with a warm body is somewhat correct in the sense that there is a relationship between temperature and radiation, but it is probably not strictly accurate in a thermodynamics sense.
Temperature is not just about molecular vibrations. There are two characteristics of any system that determine its temperature- the energy it stores internally at rest and the number of ways it can store that energy (this is tied in to the entropy). So, beyond vibrations temperature also ties in to rotations and unfettered random movement.
The simplest way to see that radiation has temperature is to think of it as a photon gas. The temperature is related to the random movement of the photons in the gas. This makes it, in fact, an ideal gas that happens to be ultra relativistic. It even obeys PV = nRT, if I recall correctly.
Even without quantum mechanics, though, if you get in to the nitty gritty of temperature's definition you'll find that radiation does have temperature because it presents another way for a system to store energy.
So, bottom line, radiation has a temperature like all other things. The cosmic microwave background, for instance, which does not have any known molecular origin in sight, has a temperature of 2.3 K
. Not to mention the fact that you can derive the temperature of the black body spectrum by finding the temperature of a body in thermal equilibrium with it without regards to any walls or cavity containing the photon gas.
Now, back to the article. The radiation in a microwave is fairly cold because the photons aren't really moving about very randomly, so they don't have a lot of different ways to store their energy. So, why does a microwave heat up food? Same reason the blade of a blender can boil water - friction, in a broad sense. Basically, when you use a microwave you're forcing the molecules to vibrate harder and they, by bouncing against each other and whatnot, turn that energy in to thermal energy.
I guess there's a lot about temperature -- especially regarding radiation and quantum effects -- that I don't know. Maybe Black Griffen would be willing to post an article about this -- or maybe you have a good link? This site requires authors to simplify somewhat so that articles are accessible to a broad range of readers, so maybe such an article isn't appropriate for this site. In any event, can you suggest some more reading on temperature?
Regarding radiation, there's some debate about microwaves and the back-and-forth motion and viscous heat generation versus excitation of vibrational modes. I have read from many sources what you have written, but others contradict it as well. Perhaps for this subject also you could suggest an authoritative source and post a link here? I'm not sure if there would be any peer-reviewed article on this, but that would be best if it existed.
Lesse, I don't know that a more detailed picture of temperature than what you give is necessary - I only provided more detail to justify what I felt was a necessary, but small, correction.
I would have liked to be able to link wikipedia's temperature article, but it contains the same error. I could recommend a few statistical mechanics/thermodynamics textbooks - on the undergrad level the only book I've ever used is Bowley and Sanchez
(p 166) and at the graduate level the two main standbys, Reif
(p 373 and following) and Landau & Lifshitz
(p 183 and following) have both served me well. This site
has a pretty good summary and mentions the phenomenon of negative absolute temperature, too. Having actually looked at these books, I have to retract a photon gas obeying PV = nRT - it obeys P = 4sigma * T^4 / 3 / c. Landau also shows that a gas of extremely fast moving electrons obeys the same law.
No need for quantum effects, actually, I chose to describe it as a photon gas because I figured that would be the fastest way to make my point apparent. This can all be done completely classically as long as one doesn't mind a few infinities popping up in the energy stored in short wavelength radiation
Perhaps the best way to explain temperature is the classical equipartition theorem - the thermal energy stored in a body is equal to kT/2 times the number of places it can store energy (strictly speaking, this is only true for "harmonic degrees of freedom" but just about everything is to some approximation). This includes random vibrations, movement in each dimension, energy stored in atomic and molecular bonds, etc. Reversed, temperature is a measure of how much random energy is available to each way a thing can store it. The definition of heat follows naturally from that - it is just a transfer of thermal energy. And heat is driven from hot things to cold things for no other reason than that the random movements inside of a thermal body tend to equalize the energy shared by the different energy stores, on average. Perhaps that's more technical than this site really needs... I dunno.
Re: the way microwaves heat things up. I haven't read any peer reviewed articles on the matter. Honestly, I can't rule out microwaves coupling to any particular degree of freedom over another, and I would guess it probably just depends on which molecule you're talking about. Regardless, when I talked about "friction" (I put in the scare quotes intentionally) I was just talking about how the molecules turn otherwise cold radiation into heat. The details could have included bumping, rubbing, or just good old fashioned re-emitting the radiation at a lower wavelength.
I went back to reread and potentially correct the wikipedia article. On closer examination, it seems to be fine. The entropy stuff under the second law definition of temperature is a little off, but overall it seems to be fine. Temperature
We now return you to your regularly scheduled cooking blog. :)
Both articles, the original Heat Transfer and Browning Foods and this one, omit one important thing. Something every American says every year in his Porsche dealership when buying 6 litre SUV for this year: "I only want something to get me from A to B".
In heat transfer, the heat must be also transferred from somewhere ("A") to somewhere else ("B"). For example, in the oven the "A" may be its walls for the radiation to the bun but the air surrounding the bun for the convection to it. And as regards the bun, "B" will be firstly just the bun surface where all those nice browning reactions are taking place. But in the moment we are curious about setting of the dough in the middle of the bun, the "B" shifts into the middle of the bun and the path of the heat gets one more step which works somehow in combination with the preceding ones. Etc. etc.
So, in my opinion matters are a bit more complicated (and different) than presented in both articles. Even the nice comparison to the baglayers crew can not help much. It definitely did not help me.
I have also enjoyed the stubbornness of second author about the "conduction by convection" or "convection being just conduction anyway" as if the convection was not the long decades accepted term for the heat transfer combined from the flow of the fluid in the vicinity of the wall and conduction from the fluid through the boundary layer to the wall.
Not to nitpick, but the point of a pressure cooker is to NOT steam food - or, rather, to not boil your water off. By increasing the pressure, water can be maintained in a liquid state at higher temperature than would be possible at atmospheric pressure. This means that you can cook things more quickly - higher temperature and greater heat transfer between your fluid and your food. If you've ever noted the high-altitude directions on some food packages, it's because water boils at less than 212 deg F, due to the low pressure. Of course, as a flatlander, it's only something I've read about...
Just stumbled upon your writing doing some research for a lecture.
Steam temperature effective range to 212F?
Your standard pressure cooker at 15psi uses a steam temp of 257F. In commercial units, I've cooked with steam at over 300F.
With regard to your simplified definition of cooking. Cooking does not have to use heat. It can also rely on an enzymatic process, or such of that with ceviche (acid).
All in all though, nice article.
Harry Otto, New York
Burr put the ** after the 212F to signify that the temperature is higher when pressure cooking. Sorry that this wasn't clear.
Clearly Burr doesn't think it's cooking if no heat is applied. I tend to agree with him on this as well. I've always referred to ceviche as "cooking" (always with quotes) because you're enzymatically tightening the proteins to produce a texture similar to that if you cooked it. Not the same thing. Now, since the scope of the website definitely includes food such as ceviche, sushi, and whipped cream toppings (I haven't written these articles yet, but I might!) we'll have to say there are at least two definitions to the word cooking that I use - 1. the act or preparing food for consumption, and 2. the act of physically altering the properties of food through the introduction of heat.
I rather liked both of these articles, though I think that the "Cooking Materials of Cookware" article did a bit better job of explaining thermal transfer, at least as it pertains to cookware.
Personally I think the sandbags analogy got stretched a bit thin and seemed to lose consistency as the article went on. I will admit that it's not the easiest thing to describe as an analogy. Most of the ones I can think of break down at some point.
The one I like best is to picture an object as a pool. The water in the pool is heat and the level of the water in the pool is temperature. The pool's width defines it's heat capacity. The wider the pool the more water it takes to fill it to a certain temperature. Radiation is easily described as simply shooting water into the pool from a hose, or even the pool filling from rain. Conduction requires connecting two pools adjacent to each other with pipes at the bottom of each pool. The thermal conductivity of a pool is basically how big the holes are to the pipes. Bigger holes allow water (heat) to travel at a faster rate into and out of a given pool (object). When two pools are connected if they are filled to different levels (temperature) water will flow from the higher pool to the lower pool until they are the same level (equilibrium). The water pressure is also higher depending on how much higher the high pool is, and thus will push water through the pipes even faster, slowing down as the two near equilibrium. In thermal conduction the mechanism is not the same but more heat is definitely transferred from a hotter object.
Experiment: Get two identical pans. Leave one pan at room temperature and heat another one. Place an identical ice cube in each pan and measure how long each one takes to melt. Intuitively the hotter pan will melt the ice faster even though both pans are significantly warmer than the ice. Conclusion, heat was able to transfer more quickly into the ice from the heated pan by conduction. Note: This experiment depends on neither pan actually reaching equilibrium with the ice before the end of the experiment. It would be more scientific to expose the ice cubes for exactly the same amount of time and then measure how much each melted by measuring either the cube (by weight) or melted water (weight or volume), but that also complicates it. heh
Obviously I'm not trying to describe a perfect mathematical analog, just making an analogy to explain the concepts.
BTW all the pools technically have the same depth... Absolute zero. I should also note that the width of the pool can be different at different levels as the properties of materials can change greatly depending on temperature (and pressure, which is not represented in this analogy along with many other things). Doing so is pushing the limits of this analogy, though.
That takes care of radiation and conduction. Convection is a bit more difficult and where the analogy tends to lose cohesion. To describe convection here you have to put lots of small "pools" on wheels. heh These pools can then move around in groups, transporting their contained water (heat) with them. That's a hell of an image... ;-/
I actually did like the author's stubbornly pointing out that the final stage of heat transfer into the food does not happen by convection.
It's theoretically possible... but it requires a food where your definition of what is the food and what is not are overlapped by an amorphous medium. Something like a stew where most of the liquid is actually strained off and not considered part of the food, maybe. Though I suppose some of this inevitably happens in deep frying or boiling anyway unless you keep a significant portion of the medium as your food the effect becomes negligible.
The author did make the point that he was referring to the final heat transfer into the food, not the entire system. Obviously a convection oven does use convection to heat food since convection is the mechanism of the oven it-self. I can also kinda see the other point of view, since convection as a mechanism for delivering heat to be conducted into the food does represent a different mode of cooking at the macro level, and as such that entire mode is generally referred to as "convection".
I think it boils down to point of view. Basically it’s the difference between asking how did the heat get to the food and how did the heat get into the food. To illustrate what I mean by that let me use a small analogy. Let’s say like millions of Americans you commute to work in a car. If I ask you, “How did you get to your home?” you would probably tell me you drove there. But if I ask, “How did you get into your home?” then you would probably tell me you opened the front door and walked in. In the very same sense convection can be the main mechanism that transports heat to your food, while it is conduction that gets the heat into your food.
The author alludes several times to a low Heat Capacity being a limiting factor on thermal transfer, which I think is partly in error. Heat Capacity is definitely part of the picture, but a low heat capacity doesn't equate to slow heat transfer. In fact a pan with high heat capacity takes longer to heat up, but it will also stay hot longer and maintain a more even temperature over time. High or low heat capacity may be desirable, depending upon what you are cooking. Heat transfer depends upon the heat conductivity of both your heating medium and the food as well as the difference in temperature. The quicker your heating medium heats up the sooner it will be heating your food. In fact aluminum is used in a lot of cookware specifically for this purpose.
Keep in mind that the purpose of putting heat into the food is to raise the temperature of the food it-self. The various chemical reactions we're trying to trigger happen at certain temperatures. The importance of controlling the rate of transfer is about controlling the distribution of temperature in your food over time. Time of course is the essential ingredient. Those chemical reactions take time. Increasing temperature generally increases the speed of those reactions, but also causes additional reactions, and can change the food in other ways. Also there are times when you want very uniform temperature gradients and times when you want very uneven ones, as noted.
The heat alone won't cook your food; you need to raise its temperature. If you cool the food at the same rate you heat it, simply passing the heat through the food without raising its temperature, you will achieve nothing.
I thought I'd mention a couple of points not really covered in any of these articles.
The first is surface area. This is somewhat covered, but not really directly. Surface area defines the size of the boundary we have to heat the food. Generally speaking the larger the surface area the faster the heating and thus cooking can occur. When you cut up a piece of meat you are increasing its surface area, at least relative to the air. This can greatly increase heat transfer, particularly from boiling, steaming, or baking. It also can expose more area to infrared radiation. It may or may not increase surface area relative to a pan, however. Here it depends partly on technique. If you cut up your meat and leave it sitting in the pan more is exposed to the hot air rising off the pan, but depending on how you cut it, you may not have exposed any more to the surface of the pan it-self. If you simply cut it but don't turn any of the chunks then the same surface area is in contact with the pan. Similarly if you bake cookies you should already be very familiar with the fact that smaller cookies will cook faster. As the author’s mentioned, cooking with oil fills in the gaps normally filled with air between your food and an otherwise dry pan. The oil acts like an extension of your pan, adhering to the contours of your food and creating a larger surface area through which to transmit heat.
I also wanted to bring up induction. It’s not really entirely within the scope of these articles. (I don’t know of any dishes that even could be cooked directly by induction…) But I thought I’d bring it up, mainly to refer to induction ranges that heat pans directly by induction instead of using a flame or heat coil. These ranges induce a current in the pan that converts directly to heat within the metal of the pan it-self instead of absorbing the head by conduction. The process results in less waste heat since there’s no burner or coil heating the air as well as your pan. (In fact the stovetop it-self generally stays cool to the touch until you put a pan on it and even then it’s only heated by conduction from the pan back to the stovetop.) Induction heating can also be very fast.
And now my disclaimer… All of this is just off the top of my head, so please don’t be too critical. I’ve done zero research since I’m merely posting this as a comment rather than writing an article. I believe it is reasonably accurate, though I’m sure that I’ve glossed over several things, and may even have a flat out error or few for all I know.
It is true that gravity moves mass, but, more generally, it is force that moves mass. It can be the force of gravity, but it can also be a number of other forces.
One thing I did not see was what the chefs/cooks refer to as "contintuation" cooking.
Once the medium used to introduce the heat transfer to the food is removed, cooking continues. The energy is still in the food until such time as it has had time to move from the higher energy state to its final state.
Cooking continues until the temperature that no longer changes the material is reached.
Meats are allowed to "rest" before serving because the liquids that move out return inside as the temperature cools.
Once food is cooked, it should be allowed to cool completely before being refrigerated or frozen. The rapid heat transfer of the outer edge of hot food creates a cold "barrier" to cooling inside. Bacteria can even grow before the temperature reaches the "magic" 40 F that impedes growth.
I just wanted to say thanks for helping me understand better how heat transfer in food effects the result. I was cooking a pork tenderloin roast
for my new wife and it came out perfect. I am a young air conditioning mechanic and know the basics of heat transfer, but you helped me broaden my knowledge. Again thank you for taking the time to share with the world!
Hi, I cooked my first ever meal this Christmas and have a query. The turkey instructions said it required 3 hours 20 mins but was not fully cooked in that time. I put the Turkey inside a large pot with a proper steel lid on in accordance with my wife's instructions. My mother would have put it in a tray with aluminium foil over it.
Question: Did the steal pot slow down the heat transfer from the oven to the Turkey in comparison to aluminium foil?
it's a bad idea to rely solely on "time to cook" by a pound chart.
in addition to the pan, the quirks of your specific oven, covered, uncovered, stuffed, unstuffed . . . .
a big variable is the temp of the bird when your start. completely thawing a large frozen bird - like a turkey - takes 4-5 days in the fridge.
even "fresh" birds often have ice crystals in their cavity - poultry and other meats do not freeze at 32'F/0'C because of the mineral/other contents of the water entrained in cells - so they can be kept below "water freezing" temperatures and still considered "fresh." any free water standing in the cavity will freeze however.
I've bought "fresh chickens" and had a devil of a time getting the giblets out - they're "frozen" inside the bird!
I'm not a fan of the pop-up "done" indicators - but if that's all you got . . .
best to invest in a decent thermometer.
Hi Dilbert, Thanks for responding. The question I meant to ask was this:
All other things being equal, will a Turkey, or anything else for that matter, cook quicker under circumstances A or B.
A. In a tray covered in aluminium foil.
B. In a steel pot covered with a lid.
Given the three hour time span would the heat transfer into the Turkey thourgh the aluminium or the steel pot be the same?
If not, why not?
most likely B "assuming" the usual sorts of roasting pans.
heat "transfers" convection, conduction and radiation.
heated air inside the oven makes up the convection part.
the pot/tray/whatever in contact with the oven rack does conduction.
exposed elements (if present) and the hot walls of the oven do the radiation part.
dull / dark objects absorb and emit radiant heat more readily than light colored / shiny objects.
the aluminum foil will reflect a lot of the radiant heat, so the typical dark color enameled roasting pan would be faster.
now, if it is a brand spanking new shiny polished stainless steel roasting pan . . . results will be different.
A couple responses to some posts over the years...
Hillman mentions "continuation cooking", but I believe it is more commonly called "carryover", and which I would consider to just be another example of conduction. The hot exterior of the food (because it was in contact with the pan, grill grates, etc.) is hotter than the interior, continues to move heat into the cooler interior. Even when removed from the cooking vessel, the exterior of the food is still hotter than the interior. For example, the outer surface of the steak is very hot from the grill, and thus will warm the cooler interior (pull your steaks off the grill when they're a little underdone!).
To Harvey and Dilbert, I agree with Dilbert's second answer. In an oven, the radiative mode is the most important, so a light-colored/shiny surface will slow cooking (aluminum) by reflecting the radiation, whereas a dark/dull black surface (like an enameled steel roasting pan) will absorb and re-emit heat more, and speed cooking. Another factor to consider is keeping warmth around the bird -- a close-fitting lid, or even cooking in a bag, will keep hot air / steam near the bird's surface, which will cook it faster than just the dry oven air. (It also inhibits browning, but that's not part of the question) The *best* advice to ensure doneness is to ignore time/weight guidelines and use a good probe thermometer. Stick it into the thick part of the thigh and pull the bird about 150-155F (it will carry over to 165F, which is a great level of doneness for poultry).
Could anyone give me the formula/relation used in calculating the cooking energy in joule or otherwise expended by cooking a particular meal when using a particular fuel? For instance, if I cooked 500 grams of rice using kerosene as my fuel (the source of heat) in say: 45 minutes, how would I calculate the cooking energy expended in performing the cooking task?
I read through a final year project by a university student who worked on evaluation of cooking energy for selected agricultural products and he stated somewhere that: ''Time (T) spent in cooking a food is directly proportional to the cooking energy (E) expended.'' I agree with this but in his analysis, he wrote that he spent 45 minutes in cooking a particular food using charcoal as the fuel; he then stated that:
Let 2 minutes of cooking = 1 Joule of cooking energy
then, 45 minutes of cooking will use up 45/2 Joule = 22.5 Joule of cooking energy.
Please, is this basis true?
He later stated in another analysis (with the view that cooking energy can be deduced from d amount of fuel consumed by a cooking process) that:
Let 1000 grams of fuelwood (solid biomass) = 1 Joule of cooking energy
then, 250 grams of fuelwood = 250/1000 Joule = 0.25 Joule of cooking energy
Is this argument right?
And lastly, he further stated that:
Let 1000 cubic centimetre of kerosene (liquid fossil fuel) = 1 Joule of -cooking energy
then, 50 cubic centimetre of kerosene = 50/1000 Joule = 0.05 Joule of -cooking energy
Is this argument equally valid?
These are my enquiries for the experts in the field of energy. I'll really appreciate it if someone could specify the most accurate method/formula/relation used in calculating cooking energy expended while cooking for a period of time using a solid or liquid fuel and also shed more light on the expressions stated above. Thank you all.
Abayomi Adewuyi, AMIMechE
I'm a bit puzzled about the term "cooking energy" - for example to cook 500 grams of rice, in water, one could calculate how much heat is needed to raise the water&rice&pot& lid to boiling from ambient. once it is "at cooking temperature" the pot&lid radiates heat, the water evaporates, taking more heat out of "the system" - so a continuing heat input is required to keep the pot at temperature over the cooking time.
how much fuel is burned determines how much heat is generated - each fuel contains "heat energy" which is released when burned.
for example a kilogram of coal would produce roughly 3.6x10^7 Joules
you can look up the energy content of various fuels - the fuel density can be used to convert between fluid / weigh for liquids.
the precise amount of heat released - especially for "natural" fuels - coal, wood, charcoal - will vary. petroleum liquids will also vary depending on how refined they are.
burning a kilogram of charcoal will produce X amount of heat energy whether there is a pot of rice over the charcoal or not. if there is forced air - the charcoal will burn faster - but excepting for the question of "complete combustion" - will not produce more or less absolute heat release - a kilogram is a kilogram - whether ir burns really fast or really slow.
same with kerosene or propane - a certain mass is burned and produces a specific amount of heat.
keep in mind that a large amount of heat released is not "absorbed" by the food being cooked - there is a _lot_ of "wasted" energy.
cooking appliances - like stoves - usually have knobs to regulate the amount of fuel being consumed. that means the rate at which fuel is being consumed varies - so going strictly by time will be inaccurate. if the fuel consumption rate varies, the heat energy produced in 2 minutes of cooking will not be the 45/2 ratio you mentioned for 45 minutes
for fuels like wood and charcoal, they will continue to burn after the food is finished cooking - they would have to be "extinguished" in order to "save" the remainder.
"Let 1000 grams of fuelwood (solid biomass) = 1 Joule of cooking energy
then, 250 grams of fuelwood = 250/1000 Joule = 0.25 Joule of cooking energy"
is simple math - but true only if the wood is very consistent.
your first question:
"For instance, if I coked 500 grams of rice using kerosene as my fuel (the source of heat) in say: 45 minutes, how would I calculate the cooking energy expended in performing the cooking task?"
easy - weigh how much fuel is consumed, kerosene contains about 46,300 Joules per gram.
Microwaves are for people who don't cook too.
These are the important facts.
Remember food does NOT heat form the in side out in a micro. Yes some otherwise intelligent people still believe this.
Reheating sketti sauce or other things that boil at lower temps than water means use 50-75% power depending on oven wattage or there will be cleaning to do. Since microwaves heat by moving the molecules, give your food a minute to stop dancing around before you eat it - just in case. We don't know how bodies actually react long term to food that moves.
When cooking on an electric stove :( remember cooking directions are typically for gas cooking where stove top heat gets turned off things cool rapidly. Electric coils keep the heat going longer and removing the pot/pan to another burner is advisable when temperature changes are critical to the recipe.
What method of heat transfer does "induction" fall under? Conduction, convection, radiation?
The heat is produced (induced) in the pot or pan. From there it's conduction into the material into the pan. There is no heat transfer from the cooktop to the pan - (the flow is actually from the pot to the cooktop).
It was suggested to me that oven cooking at 150C my beef curry would result in the curry reaching 150C. But I maintain it cannot exceed 100C unless it loses all the water I added as per the recipe for curries. Am I correct?
Yes you are correct - except for the top and edges which might dry out and allow the temperature to increase past 100C.
Many thanks for your reply Michael Best Regards Bill Edwards EH39 5BL Scotland UK
I am not an engineer, but I have a question for one. A natural gas utility advertises that "WITH NATURAL GAS, WATER HEATS UP TWICE AS FAST AND STAYS HOTTER LONGER."
The “heats up twice as fast” claim is obviously imprecise but makes some sense, given that a gas burner comes on instantly at full heat while an electric burner heats gradually. However “heats up” is pretty vague; “heats up” to what extent? Also, all else being equal the quantity of water should affect the relative times to a particular temperature because once both cooking methods hit full output the electric burner might at some cross-over point be more efficient with less heat escaping. At least it’s debatable.
But “stays hotter longer”? Could it be that water heated by a gas flame somehow stays hotter longer than water heated on an electric burner? Different molecular reaction? If the claim is somehow true, would the difference in heat maintenance disappear once the water hits the boiling point?
this is about stove tops I presume?
the 'heats up faster' bit is as you suggest fairly clear. not only is a gas flame 'instantly' hot - compared to the time lag for a spiral coil to heat up - but gas burners can deliver more BTU/time than the classic electric coils.
the big residential gas burner can put out about 22,000 BTU/hr - the (lossless) equivalent of near 30 amps at 220v nominal. so you have instant 'lots more heat' on the bottom plus combustion gases running up the sides of a pan.
it takes one BTU to raise one pound of water one Fahrenheit degree - so pint's a pound the world round,,,, two quarts of water for pasta, roughly 4 pounds, starting off at 75'F going to 212'F, delta T 137 F' x 4 pounds = 548 BTU needed - if the entire 22,000 BTU/hr gas burner were absorbed (it's not....) it would take 1.49 minutes to boil. lossless obviously does not apply - it takes longer - not all the heat is absorbed by the pan/water....
a large spiral electric coil will vary, but wattage wise they are in the 2100-3000 range. and the 3000 watts is roughly half the "power" of the 22,000 BTU/hr burner.
alternate heat sources - induction / radiant - alter the picture slightly - but "power in vs power out" still applies and gas will still win.
"stays hotter longer" - there must be some fine print there somewhere.
makes no sense - and certainly the fuel/heat source makes zip difference to how fast a pot cools down once removed from the heat.
Thank you for the quick and clear response.
There is no fine print in the ad on "stays hotter longer". I assumed from the context that the gas company was referring to stovetop cooking. On reflection, although it seems less likely, perhaps it was referring to hot water heaters. If so, there too a gas hot-water heater would presumably heat faster than an electric but I doubt that gas fueled hot water tanks are typically better insulated than electric ones to support "stays hotter longer."
The whole claim seems a good reason to take advertising with a few grains of salt (suggesting a discussion of boiling point elevation).
well, if the gas company is advertising their own hot water heaters, it is possible those devices are "better insulated" than the usual. but you're right, there's nothing more inherent about gas hot water heaters over electric to 'keep hotter longer'
with the minor exception of a pilot light. most gas appliance in the US have switched to electronic ignitions - no more pilot lights.....
I was seeking more about the heat capacity of foods, in fact having asked Siri for the heat capacity of a casserole. Burr's article does a decent job of describing heat transfer, and categorizing the speed and effect of various approaches to cooking. With graduate courses in Convection, Conduction and Radiation heat transfer under my belt, though, it didn't serve my needs that well. He felt he needed to use some metaphor to make heat transfer easier to understand. It may have worked for some, but clouded the topic for me.
Any help in finding an article on the heat capacity of foods would serve my interest better, and be appreciated. My hope is to determine how long to heat a leftover dish to serving temperature, knowing its mass, the mass of its container, and the desired serving temperature. This shouldn't be nearly as complex as determining how long to cook a dish in which phase and chemical changes occur.
thank you for making the effort to put this piece together. I was putting a piece of content together for my own website to explain the difference in pan materials used in Germany. Your article acted as a good entry point and I hope it was alright that I linked to your article as a source.
Greetings from Germany,
Simon from https://pfannenhelden.de
JonR0 said, in part:
>Any help in finding an article on the heat capacity of foods would serve my interest better, and be appreciated. My hope is to determine how long to heat a leftover dish to serving temperature, knowing its mass, the mass of its container, and the desired serving temperature. This shouldn't be nearly as complex as determining how long to cook a dish in which phase and chemical changes occur.<
This is what I would refer to as "the wrong question". Knowing the mass of food, and mass of container, is only somewhat useful. And you're missing the characteristics of the heating medium completely! Look at your desired end conditions: All of the food is at or above the desired reheat temperature. It's been decades since my last heat transfer course, but to get to your end conditions, look at the path of the heat from the medium to the middle of the food. Significant thermal resistance comes from the container (if it's ceramic/glass type - a thin metal foil tray would offer virtually no thermal resistance compared to the food) and then the food itself. The heating of the food would occur almost completely through conduction, unless it is a liquid that can move, or the food is stirred one or more times. Stirring can only be done for something like a pasta in sauce, or mashed potatoes, and not for something like a cake or quiche that has to have its structure retained.
Now look at the heat path through the food. A container that is cubical will have the longest path to the center (there aren't many spherical containers), while the same mass of food shaped in a rectangular solid with a depth that is much smaller will result in there being not only more surface area exposed to the heating medium but a shorter path to the center of the smallest dimension. If you assume that the medium has unlimited heat capacity (heats the surface of the container/food to its temperature instantly and maintains that temperature) then the heating of the food depends only on how long it takes to get the center up to temperature, and the shorter the path the quicker that will occur. A hotter medium will also accelerate this, up to the point that the outer layer is damaged from the heat before the center gets to temperature.
So knowing the mass of the food, and the container, is secondary. One must know/learn how fast the medium can transfer heat and its temperature, and how quickly the food conducts it based on its shape, constituents, and thickness. If you have numeric values for these, then it's a simple conduction problem.
It's almost never a real world question of "how much power/how many BTU's of gas are burned", the question is how long does the heating medium that has its characteristic energy consumption rate have to be applied to heat the variable container size of variable heat capacity food. This is in part because with the possible exception of microwave heating, the losses of energy to the environment are large in comparison to the amount of heat actually transferred to the food. One number that appears on the 'net is 1.8 J/g/degree C for pasta (referenced http://www.physicspages.com/2015/07/13/heat-capacity-of-pasta/) but this presumably refers to the uncooked hard noodle, not the final one with significant water absorbed, cheese and sauce added, etc.
All this said, IMO the best way to determine the time to reheat something is experience. Obtain a food thermometer, place its indicating point in the thickest part of the food, place all in the heating medium, and record the temperature over time.
For food which can be stirred, one can apply heat until the average temperature is the serving temperature, then stir to even out the temperature of the entire serving. This will be much sooner than just waiting until the coldest part has reached serving temperature. It also moves cooler food into contact with the hot container, increasing the average rate at which heat is absorbed, since the heat transfer is related to the temperature difference.
Now if you really want the heat capacity of the food for some reason, you can place a uniformly warmed portion of it into a known quantity of water, wait for the two to come to equilibrium, and calculate from the rule of mixtures. But I maintain that the more useful value in the real world is the rate of heat transfer within the food.
Hi. I'm trying to find an answer to what seems like a pretty straight-forward problem. Let's say I'm baking an item, like a sausage, a long hot-dog, a long log of dough, or even a steel bar. Imagine the length is very long compated to the diameter, such as a continuous length of rod being fed through an oven at a constant speed. Negating the heating at the ends, I'm only concerned with the radial heat penetration within the rod or "sausage" far from the ends. Also negating any convection that might occur within the "sausage", let's say that heat penetration within the rod occurs totally from conduction.
Ok, now, imagine that, through painstaking trial and error, I have figured out the perfect cooking time and temp for a "sausage" with a 2" diameter. I have measured the outer-wall temp (say 250 degrees) and the internal temp (say 150 degrees) at the center. Now I want to duplicate that same result with different diameter "sausages", say a 1", 3", or 4" sausage, so that both the internal and external temps match the original size.
I have figured out that a sausage with twice the diameter has twice the surface area, and will absorb heat at the surface faster than a small rod, so the heat must be set lower. However it also has four times the volume, so the cook time must also be longer. The inverse is true for the smaller rod. One would think that I could simply turn the heat down by 1/2 and raise the cook time by a factor of 4, but it doesn't seem to be that simple.
The question I have is: Is there a math formula I can use to calculate the cooking temps and times for different sized rods (sausages) to produce the exact same results as the known size? (Keep in mind that I'm no mathematician or engineer, so any long, mathematical explanation will be lost on me. I'm really just looking for a formula I can punch into my calculator that will get me into the ballpark (no pun intended).) Any help would be appreciated.
yes, of course one could set up all the necessary equations. then you do it by ear.
some of the factors: if cooking on a flat grill/pan, the contact area for heat transfer will be almost the same. the large diameter will pick up slightly more heat due to the larger area exposed to radiant heat from the flat. contact cooking will have huge variables of how often turned, crusting, casings, curl, etc. the commercial solution to that issue is the roller hot dog cooker . . .
cooking in water/air, the total surface area is more controlling.
cooking in water/air/uniformly surrounded, the consideration of how fast heat flows into the tube is delta-T inside (refrigerated temp?) to outside ("cooking" temp) plus the heat transfer coefficient - which changes with the temperature of the 'tube' - as the contents cooks, heat transfer slows down (1) because the consistency changes and (2) delta-T changes.
to achieve the same temperature gradient, you'll need to use a lower cooking temperature and more time. the variability is such that theoretical calculations can not be blindly relied on - so one cooks and takes notes . . .
Thank you very much for the quick reply. Consider that I'm "cooking" in air (baking). Also, I used the term "sausage" because this is more of a cooking website, but what I'm really cooking is more related to the steel bar than a sausage, in that the material is non-porous, contains almost no water, and has known thermal conductivity and heat capacity. (Thermal conductivity for instance is 0.2, or almost exactly that of water.) Likewise, my cooking temps are extremely high to set the proper gradient across the substrate (around 500 to 2000 degree F). To ensure even heating, the rod is rotated (like a hot dog on a roller) during heating, to prevent both uneven heating not only from any contact surface but also from the convection of the air, which of course heats the bottom more than the top due to boundary layer effects.
If it's just a matter of trial and error, I guess I can keep doing that, at least until I can get a baseline on a graph to help predict other sizes. I was hoping there would be some easier, mathematical way to calculate these predictions (even if only a ballpark range). Anyhow, thanks again for your reply.