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  1. If the temperature of boiling water is about 100 ºC (depending on air pressure), why are boiled potatoes on a high stove flame cooked faster?

  2. From the English Wikipedia:

Acrylamide was discovered in foods in April 2002 by Eritrean scientist Eden Tareke in Sweden when she found the chemical in starchy foods, such as potato chips (potato crisps), French fries (chips), and bread that had been heated higher than 120 °C (248 °F) (production of acrylamide in the heating process was shown to be temperature-dependent). It was not found in food that had been boiled or in foods that were not heated.

If I boil potatoes on the highest stove flame, will the potatoes not reach the temperature of 120 ºC?

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  • $\begingroup$ Are you asking how fire (from a stove) can be hotter than boiling water? $\endgroup$
    – Arturo don Juan
    Dec 2, 2017 at 17:43

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Whether your stove is on 'high' or 'low' doesn't change the temperature of the water, i.e. the boiling point remains $100\ \mathrm{C}$. So the potatoes can NEVER reach $120\ \mathrm{C}$.

But objects being heated by hot water follow Newton's Law of Cooling (and Heating), which describes the heat transfer per unit of time, $\frac{dQ}{dt}$:

$$\frac{dQ}{dt}=hA[T_{BP}-T_{pot}(t)]$$

We can derive from this, for potatoes plunged into boiling hot water at $t=0$, with $T_0$ their ambient (and initial) temperature and $T_e$ their end temperature, their temperature evolution in time $t$:

$$T_e=T_{BP}-(T_{BP}-T_0)e^{-\frac{hA}{mc_p}t}$$

And the heating time:

$$t=\frac{mc_p}{hA}\times \ln\Big({\frac{T_{BP}-T_0}{T_{BP}-T_e}}\Big)$$

($m$ is mass of potatoes, $c_p$ their specific heat capacity and $A$ their surface area)

The factor $h$ is the heat transfer coefficient (in $\mathrm{Wm^{-2}K}$). It is obvious that when $h$ goes up, $t$ goes down.

In fluids, like water, $h$ is known to be somewhat dependent on the Reynolds Number, which is higher in turbulent conditions. More turbulent conditions favour higher heat transfer rates.

For that reason there may be a slight increase in heat transfer (and thus a slight reduction in cooking time) when comparing a vigorous boil (high heat) to a gentle simmer (low heat).

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  • $\begingroup$ What about cooking devices that put the water under pressure? en.wikipedia.org/wiki/Pressure_cooking $\endgroup$
    – Anedar
    Dec 2, 2017 at 18:40
  • $\begingroup$ Pressure cookers elevate the BP by a 10 or 15 C. I doubt if any could reach 120 C. $\endgroup$
    – Gert
    Dec 2, 2017 at 18:42
  • $\begingroup$ @Gert in a stable system (continuous boiling) isn't the temperature of both the potatoes and the water constant and equal, i.e. dQ/dt=0. If so, how does h (heat transfer coefficient) matter? Also could you provide an intuitive explanation of this phenomenon, beyond the equations (which are a mathematical representation of some attributes of the actual system) $\endgroup$
    – rapt
    Dec 2, 2017 at 19:24
  • $\begingroup$ @rapt: well, no. The potatoes start at ambient temperature and then slowly heat up till $100\ \mathrm{C}$. So $h$ matters GREATLY in heating/cooling processes. What Newton says is that the heating rate is proportional to the difference $T_{BP}-T_{pot}$ (in this case). I find that quite intuitive because heating is about transferring kinetic energy from the 'hot molecules' to the 'cold molecules'. Temperature is a measure of the average kinetic energy of the particles that make up an object. $\endgroup$
    – Gert
    Dec 2, 2017 at 19:54
  • $\begingroup$ @Gert I do not know how long it takes for the potatoes to reach 100 ºC. But are you saying that (1) I can accelerate the heating of the potatoes themselves up to 100 ºC by starting the cooking with the highest flame and (2) once the potatoes reach the water temperature (100 ºC), I can lower the flame (as long as it's high enough to keep the whole mass of water boiling), save energy, and still have the potatoes cooked at the minimum total time? This is not exactly what I experience when cooking beans - constant high flame significantly reduces the total cooking time. Could you explain? $\endgroup$
    – rapt
    Dec 2, 2017 at 20:59

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