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Kitchen Notes: Heat Transfer and Cooking
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ELEM
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PostPosted: Mon May 18, 2015 3:51 pm    Post subject: Does water stay hotter longer depending on how it was heated Reply with quote

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?
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Dilbert



Joined: 19 Oct 2007
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Location: central PA

PostPosted: Mon May 18, 2015 5:59 pm    Post subject: Reply with quote

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.
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ELEM
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PostPosted: Mon May 18, 2015 6:47 pm    Post subject: Natural gas vs. Electric heating and heat retention Reply with quote

Dilbert -

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).

Best,
ELEM
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Dilbert



Joined: 19 Oct 2007
Posts: 1193
Location: central PA

PostPosted: Tue May 19, 2015 6:10 pm    Post subject: Reply with quote

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.....
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JonR0
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PostPosted: Fri Jan 08, 2016 8:04 pm    Post subject: Heat Capacity of Foods (and combinations thereof) Reply with quote

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.
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Simon
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PostPosted: Thu Nov 17, 2016 4:34 am    Post subject: Thank you very much for this wonderful piece Reply with quote

Hi Michael,

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
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PaulT
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PostPosted: Fri Dec 02, 2016 5:34 pm    Post subject: reheating food Reply with quote

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.
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Z-man
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PostPosted: Mon Jul 23, 2018 6:40 pm    Post subject: Baking different size objects Reply with quote

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.

Thanks

Z
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Dilbert



Joined: 19 Oct 2007
Posts: 1193
Location: central PA

PostPosted: Mon Jul 23, 2018 8:41 pm    Post subject: Reply with quote

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 . . .
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Z-man
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PostPosted: Tue Jul 24, 2018 5:54 pm    Post subject: Reply Reply with quote Delete this post

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.

Z
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