You know that feeling when you open an oven quickly and you jump back as you feel as though you were being burnt? Well why aren't you getting burnt??
The main factors that I can think of for this off the top of my head are:
Temperature dissipation (probably not a big factor)
The rate of temperature transmission is probably a lot lower with air
The speed at which the air moves by your face probably also alters the feeling
The air is distributed on a relatively large surface (your face)
The air in the oven doesn't have that much energy total. An oven is quite small
Let's just start with some basics: Let's say that the temperature of the air in the oven will bee 200°C.
To simplify the model, I'm just going to assume that the air doesn't decrease temperatures very fast and that your face is actually getting the 200 degrees (if this approximation isn't suitable, then we can probably just assume that the actual oven temperature is 220° and it'll probably be closer to 200° at the end) . This is just to simplify the situation.
Now intuitively, the rate of transmission of temperature from air is relatively low. I think that between a 40°C room and 40°C water, you feel a lot more burnt from the water. Water simply transmits temperature a lot faster. This also explains why putting your hand in an oven (without touching it) would be uncomfortable but not terrible whereas touching a metal plaque would definitely burn you.
Let's start modeling!
From a quick search, I found this formula:
$P = h × A × ΔT$
where:
$P$ the power transmitted in $W$
$h$ the proportionality constant in $\frac{W}{m^2 \cdot K}$
$A$ the surface area in $m^2$
$ΔT$ the temperature difference between the two surfaces in $K$
We can apply this formula with some values k and face surface area. let's just say that a face is about $A = 0.02m^2$. The oven temperature is $200°C = 473.15 K$ and the temperature of a (living) human being is around $37°C = 310.15K$. In this case, $ΔT = 163K$. Let's take a value for the proportionality constant of $h = 10 W/(m^2 \cdot K)$.
When plugging these values into the formula, we get $P = 33W$. This means that we have 33W per second that gets added to the skin in the form of heat. $ $
| Thermal power per surface in $kW/(m^2)$ | how much you need to go to a hospital (if facing the power for 1s) |
|---|---|
| >1 | Pain |
| 10 | First degree burn |
| 20-40 | Second degree burn |
| 100+ | Third degree burn |
We have $P = 33W$ for a $0.02 m^2$ Surface for a human face. This means that $heatflux = h \cdot ΔT = 1630 W \cdot m^{-2} = 1.63 kW \cdot m^{-2}$. This seems to be bellow the pain threshold but an important threshold is being ignored: the value of $h$ changes a lot based on the speed of the air moving past you. This is why there's a difference between the "real" and "feels like" temperatures in a weather app. Higher relative speeds increases h. This means that if we take the rush of air towards you into account, we can get an $h$ closer to $h = 35 \frac{W}{m^2 \cdot K}$ which gives $heatflux = h \cdot ΔT = 5.7kW \cdot m^{-2}$. This gives us a value that would hurt and not burn based on the values above.
This is coherent with the real data (pain but no burn).
This graph shows exposure for 1s, but if we increase the amount of time, any of these temperatures can give really bad burns. The human body can dissipate heat at a certain rate which avoids burns by making a lot of the heat dissipate into the rest of the body rather than adding up to the temperature at the point of contact.
The time of exposure matters a lot as you could get a third degree burn by touching 68°C water for 1s or by touching 60°C water for 5s. This means that the initial temperature isn't the only thing that matters. The total amount of heat matters more.
The issue is that within something like 2-3s, $3912W \cdot m^{-2}$ will give you a first degree burn. This means that while 1s might not be enough to burn a face, 3s might be.
Unless you use a Heat Pack 24 Industrial Hot Air Blower to cook your lasagna, your oven probably can't output an ovenful of 200°C air every second continuously. This means that the amount of thermal energy that could be outputted in a second would be the thermal energy of the air in the oven. An average oven might have a volume of 60L or so. This gives an amount of energy of about $E = 13kJ$. This means that if all of the energy in the oven were discharged into your face in a single second, we would have $P = E * 1 = 12kW$. That's a lot and would amount to $heatflux = 12 \cdot 10^3 \cdot \frac{1}{0.02} = 600kW$