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Assume you have an oven chamber of fixed volume (let's say $1 \text{ m}^{3}$) which also releases any excess pressure. At the bottom is a tray that contains boiling water. You heat up the oven until the temperature inside reaches $350 {}^\circ\text{F}$.

  1. How much water can the air hold at that temperature? My guess is that because the pressure is constant at $1 \text{ atm}$, this would be the saturated vapor density of $598 \text{ g/m}^{3}$. (http://hyperphysics.phy-astr.gsu.edu/hbase/Kinetic/watvap.html#c1). But then it's also at $350 {}^\circ\text{F}$ - so could it be superheated?

  2. Why can the oven temperature reach $350 {}^\circ\text{F}$ when the maximum temperature of steam at $1 \text{ atm}$ is $212{}^\circ\text{F}$? Is it because the air molecules are able to heat up beyond what the water molecules can? Or are both at the same temperature?

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350 °F are 176 °C, which is well above the boiling point of water; Liquid water can not get hotter than 100 °C at 1 atm. However, gaseous water (Steam) can become much hotter. The oven can only hold around 1 atm; at the point where it exceeds this pressure in the inside, the mixture of gases will stream out of it in the respective proportions.

So what will happen is that during the heating process, air will exit the oven. as more and more water evaporates, this air will contain more and more water (which you will probably see condensing above the oven) according to the increasing amount of gaseous water in the oven.

When the 350°F are reached and the water amount in the air will reach your 598 g/m³, there will be less air and water leaving the oven, as some of the water condenses back into the puddle on the tray, but as long as you keep the pressure constant by allowing the water to escape, this will be vanishingly less and most of it will escape the oven. As long as there is liquid water on the tray, this water will have 100 °C, the air/water mixture in the oven will have 350 °F and the amount of gaseous water in the oven will be around 598g/m³. The water molecules in the air will approximately have the same kinetic energy as the air molecules; as most of the air molecules are heavier, the water will be accordingly faster. This will increase the heat conductivity in the oven, heating the food faster; if you shove your sourdough into the oven, water will condense on its surface too, giving the energy which was used to evaporate the water (which is a lot compared to similar substances!) to the loaf and heating it even faster, yielding a delicious crust. Enjoy it ;)


Edit; The density of 598g/m³ seems to be the density of pure steam at 100°C. Due to thermal expansion, I would expect it to be less at 180°C.

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  • $\begingroup$ Ha, thanks a bunch! So to summarize, $\endgroup$ Feb 14 at 23:04
  • $\begingroup$ 1) 598 g/m³ is the correct amount of water 2) The steam / air mixture can indeed increase its temperature past 100°C. Both molecules will have the same kinetic energy (why?). Can we compute how fast the water on the tray will evaporate? $\endgroup$ Feb 14 at 23:18
  • $\begingroup$ 1) The value for the amount of water in the air for this temperature is something I took from you without checking. I could compute/validate it if needed. 2) yes, the air/water mixture will get hotter than 100 °C, as the water in the air will not longer be liquid. 3) the same kinetic energy (distribution) is due to the constant collision between the molecules, which will result in the same energy distribution. $\endgroup$ Feb 14 at 23:39
  • $\begingroup$ 4) an estimation of how fast the water could evaporate will depend on; how fast the water can escape the oven, how much heat the oven can deposit in the water per second, the surface area of the water and the ventilation of the oven and the pressure in the oven (no pressure difference between the oven and the outside may be ok for a first estimation but will not necessarily be true). The effect will likely be dominated by how fast the water can escape or how fast your oven can bring the heat to the water. $\endgroup$ Feb 14 at 23:40
  • $\begingroup$ you will have the fastest evaporation with a door opened a centimeter and a combination of convection and heat from underneath with the tray standing at the bottom and the slowest evaporation for a tight closed door and no convection with the tray far away from the heat source. $\endgroup$ Feb 14 at 23:40

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