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How does time needed for cooking pasta (or any other food) change depending on stove power output (for simplicity, let's say flame size)? If it does... And why.

Starting with the point that water is boiling and continues in it (thus water temperature is constant, let's say, 100 degrees Celsius),

is there a difference in cooking time:

  • if I set the biggest fire flame (water is boiling strongly, many bubbles, water is really rough).
  • I set the smallest fire flame (but enough to keep water boiling) and water is quiet but still boiling.

EDIT: The simplest model I have in mind is:

1. Temperature is constant at 100°C;
2. The kind of pasta is irrelevant; for simplicity we can modelize it as a sphere;
3. Forget any kind of transport of water into pasta, amid compound into water, etc.

Now, I think the main issues are:

  1. how do we define cooked for this problem?
  2. does heat exchange coefficient change with fire size?
    larger fire -> greater heat goes into water -> much heat goes into pasta -> pasta cooked in less time
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This is a domain where experimental calibration is the correct approach. Simply because there are far too many variables to determine mathematically.

  • Pasta starts cooking at 80C, as starch starts absorbing water then. So the moment you add pasta matters.
  • The absorption speed largely depends on mechanical aid of stirring - as water boils, motion of the bubbles stirs the pasta. This process is far too complex to model mathematically.
  • Pasta of different cross-sections, composition, dryness will cook at different rate. This is again not something you determine through models, but experimentally.
  • there are other engineering factors to consider: splashing water (and pasta) when boiling too intensely, ratio of heat transfer into water and modelling temperature in the pot (hotter by bottom, cooler by surface); wasting energy through heat escaping through the sides, pasta sticking to the bottom of the pot or forming clumps if not stirred, is the water salted? (boiling point rise up to 4 Celsius degrees), is the pot covered (higher temperature near the top, which may speed up cooking but will build pressure and splash worse).
  • How do you define 'cooked' - the most accepted method involves taste test, that is experimental. For your model you'd need to develop a hardness function of pasta, determined through dedicated devices.
  • How uniformly will the pasta cook? If you cook thick noodles in very hot water shortly, they will be overcooked on the surface and hard inside.

In short, that all depends. Generally boiling stronger will speed the process up, but may not necessarily provide preferable results than gentle simmering.

Edit:

You've actually added some answerable questions.

  • how do we define cooked for this problem?

The 'correct' baseline is organoleptic, 'tastes just right'. If you want this to be more 'scientific' you can go with tests spread across many people, and do statistics. This is not necessary though, because you're not discovering a new field; pasta science is long known and established, so you can stand on the shoulders of giants: apply standard cooking conditions to pasta with known recipe, and it will result in 'cooked pasta'. So you have the baseline 'cooked pasta' template. Now for further tests you will need a more simple test, based on properties less cumbersome to measure - for example, mass. If the mass of drained pasta rose by a factor of 2.25, you have 'al dente'. With a factor of 2.4, you have it thoroughly cooked. It's an easy, reliable test usable in practice.

  • does heat exchange coefficient change with fire size?

Heat exchange coefficient is completely insignificant here. Pasta has a very low specific heat coefficient[citation needed] comparing to water, and with it being usually thin-walled, reaches the temperature of surrounding water within seconds after immersion; especially that it soaks the hot water rapidly. So, no, except for very low heat where you dump a lot of pasta into little water significantly reducing the water temperature (and resulting in lousy clumped cooked pasta) you're not changing regarding heat exchange coefficient.

Cooked pasta differs from raw, soggy pasta in content of water absorbed. (after all, if you cool down cooked pasta, it doesn't revert to uncooked...) And that is what changes with cooking intensity. Wet pasta secretes starch, which forms a weak barrier obstructing further absorption. Stirring washes it off, and allows water to penetrate deeper; motion moves water through pores and promotes penetrating cellular membranes. In short, water that boils hard stirs the pasta and lets it cook a little faster. Temperature and heat transfer have little influence here: if you stir a slow-simmering pasta with a ladle, you'll get it cooked just as fast as if you cook it on high fire.

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  • $\begingroup$ OK, you pointed out interesting aspects. updated OP $\endgroup$
    – mattia.b89
    Jun 8 '18 at 21:16
  • $\begingroup$ @mattia.b89: see my update. $\endgroup$
    – SF.
    Jun 9 '18 at 23:02
  • $\begingroup$ OK, so can we say it is just (approx.) a mass transport problem? How can we modelize it? $\endgroup$
    – mattia.b89
    Jun 11 '18 at 17:40
  • $\begingroup$ Could be helpful these two sources? sciencedirect.com/science/article/pii/S0735193398000608 citeseerx.ist.psu.edu/viewdoc/… $\endgroup$
    – mattia.b89
    Jun 11 '18 at 17:40
  • $\begingroup$ @mattia.b89: Somewhat. Extrapolating Fig2 from the second link, we have specific heat of dry starch at ~1J/gK (water is 4.1). The first link gives a ballpark of 0.2W/mK (water@100C: ~0.7) That means pasta won't deplete heat in immediate vicinity even without stirring, as water will draw it in faster than pasta draws it out; also, 1g of pasta will need ~100J to reach 100C from 0C. Heat transfer at 0.2W/mK for 1mm thick pasta and 100K temperature rise will be 20Watt or 20J/s. That gives us about 5 seconds until almost frozen pasta is at 100C. $\endgroup$
    – SF.
    Jun 11 '18 at 18:06
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If the temperature is constant 100 C that is all that matters.

Putting more heat into the pot does not put more heat into the pasta. That heat just goes to vaporize more water.

If anything a rapid boil would be longer as heat transfer from a vapor is less than heat transfer from a liquid (same temperature). Steam will also be 100 C.

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  • $\begingroup$ Could you expand you answer: why doesn't heat exchange coefficient matter? $\endgroup$
    – mattia.b89
    Jun 9 '18 at 15:28
  • $\begingroup$ @mattia.b89 Heat exchange is based on temperature (not heat). I did say the coefficient matters - it will be less in the vapor phase. $\endgroup$
    – paparazzo
    Jun 9 '18 at 15:37
  • $\begingroup$ with "If the temperature is constant 100 C that is all that matters. [...]" I understand temperature is the only variable, thus heat exchange coefficient doesn't matter... $\endgroup$
    – mattia.b89
    Jun 9 '18 at 15:40
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More dark or low frequency infrared radiation created from the eye of the stove is present in the pasta on full blast which heat the food more directly then bringing pasta to a boil. Normally the radiation from the stove eye heat the pot then the pot atoms heats the water atoms then heats the food molecules. A UV cooker and light ovens skips those mediums of heat transference cooking the food directly and that is why they are so much faster than a conventional stove. Chemistry.SE does cook and may want to ask there.

The OP does not mention the heating element. Cooking time differs because of how the food is warmed by what type of element and the amount of radiation produced. When you put food on that is not the same temperature as the stove eye it cools the eye. Depending on the efficiency and robustness of the stove to bring the eye back up to temperature matters in cooking time.

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  • $\begingroup$ So where is the stove eye on an induction stove? $\endgroup$
    – Solar Mike
    Jun 8 '18 at 16:22
  • $\begingroup$ What does a UV cooker have to do with a question on boiling pasta? $\endgroup$
    – paparazzo
    Jun 8 '18 at 23:15

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