I had to transfer the comment James made here for reference so I don’t have to back and forth to read and re-read. I finished testing. I used up all of the 440 grams of fuel to boil 17 litres of water.
I quote for reference:
Dan, yes.
“James, that radiation is from the outside of the pot, the water inside is not radiating heat outward, the water is still absorbing….correct? yes or no :-)”
Dan, Not to be a pain in the butt, but Yes and No. Metal absorbs say 99.9999999999999999999999999% of IR radiation (not a correct value.) Some IR manages to slip through directly, though depending on the thickness, density, blah, blah, blah… it is undetectable, generally.
IR is a radiation or enrgy. It absorbed by almost any mass. (Though some is actually transparent to it.) For metals, in general, what happens is it is absorbed, increasing the “heat” state. Heat can be interpreted in various ways: molecular vibratiion, lengthening of electron shells (exitatation), and/or even loss of electrons (ionization) and in extreme heat plasma. Anyway, the heat is retransmitted in a 360 degree probability sphere…some will go back the way it came, some will escape to the next atom. Some of the IR is duplicated as the atom “relaxes” and electrons drop back to their non-excited states. This releases the IR photon to the next atom with the same probability’s all over again. Sometimes it will only release part of the energy as a lower order photon and keeping part as “excitation”. Sometimes it will simply convect to another atom as Brownian movement or molecular “bumping.” Generally we say the pot gets hot. Anyway, there is no set formula for determining which will happen, only probabilities of possible dispositions for the added additional energy. The transfer of heat from water to metal to the outside is NOT instantaneous, though. Conduction is a measure for the SPEED at which this happens. (Metals form a different structure, but still the rules apply.) So looking at the total pot structure, you can envision the heat as being radiated from the outer surface. But it may have been produced from the second or third or fourth or fifth or some other layer of atoms UNDER the surface, not necessarily from the surface layer of atoms. But, assuming the pot has several thousand layers in its thickness, it will be difficult to know the difference. And, the IR will likely be at a different frequency than the IR released from the water, or the IR released from the flame. This is just basic laser physics as far as the frequency and it is based on the composition of the material releasing the photon…skip it…
Anyway, you can see that water and the pot act the same, except, you deal with two different materials. The water still radiates heat. It may be picked up by the other water molecules, and/or the pot. This “starts” the above heat transfers. There is no actual start, though (at least above absolute zero,) just an increase in what is already happening either in the plus (heating) direction or minus (cooling) direction. Newtonian views are usually enough, but they can fail for the last 10-20% of efficiencies. My example with the paint is one such, even though it is only 7-10%.
Anyway, the actual thermodynamics says heat flows in both directions at all times. The water radiates heat and it also absorbs heat. It radiates to the pot as it absorbs from the pot. The pot absorbes from the flame and radiates to the flame. The hotter something is the more it radiates.The difference between the input and output indicates the plus or minus direction of the heat flow (not correct terminology, but easily understood.) Most of the heat produced by a stove starts as IR. It pays to understand it at least in general terms. Is that better, Dan?
I’ll be back with final boil times and comments.
.
or RGH
