# Do conventional ovens heat by thermal conduction or radiation?

I am curious which energy transfer mechanism dominates in a conventional oven, thermal conduction or thermal radiation?

A naive guess would be that early on, when the food is still cold, the heat transfer rate would be higher for radiative than conductive transfer due to the dependence on $\Delta T^{4}$ for the former compared to $\Delta T$ for the latter. However, it is not clear to me if the oven elements/burners heat the air and the air transfers the heat through conduction or the oven walls radiate in infrared and heat through radiative transfer.

If we are heating a solid food (e.g., beef/meat) and we assume the air inside the oven is a constant, uniform temperature then we only have conductive and radiative heat processes, correct? Meaning, I need not worry about convection.

If these assumptions are correct, then what heat transfer mechanism will dominate and why?

Let's assume the oven walls radiate like perfect black body radiators and the air inside the oven is uniform throughout (ignore the thin layer near the cooler meat). Assume I want to cook a prime rib (chosen because it can be bought in a cut that has a cylindrical shape), which is just beef and fat. I can look up the thermal emissivity and thermal conductivity of beef to find:

• $\epsilon$ ~ 0.74–0.78, depending on fat content; and
• $\kappa$ ~ 0.504–0.561 W m-1 K-1, depending on fat content.

We know the heat transfer rate for radiative heating can be approximated by: $$\dot{Q}_{rad} \approx \sigma \ \epsilon \ A \ \left( T_{oven}^{4} - T_{beef}^{4} \right) \tag{1}$$ where $\sigma$ is the Stefan-Boltzmann constant, $\epsilon$ is the radiative emissivity, $A$ is the surface area of the absorber, and $T_{j}$ is the temperature of the jth object.

Similarly, we know heat transfer rate for conductive heating can be approximated by: $$\dot{Q}_{con} \approx \frac{ \kappa \ A \left( T_{oven} - T_{beef} \right) }{ L } \tag{2}$$ where $\kappa$ is the thermal conductivity, $A$ is the surface area of the absorber, $L$ is the thickness of the absorber, and again $T_{j}$ is the temperature of the jth object.

Let's assume we buy a prime rib that is ~1.5 ft (~0.46 m) in length and ~5 in (~0.13 m) in diameter and approximate it as a cylinder. Then the absorbing surface area and thickness would be $A$ ~ 0.42 m2 and $L$ ~ 0.13 m. If the meat starts at 32 oF (273.15 K) and the oven is at 350 oF (449.82 K), the rates from Equations 1 and 2 when the meat first enters the oven are:

• $\dot{Q}_{rad}$ ~ 630–664 W; and
• $\dot{Q}_{con}$ ~ 297–331 W.

Obviously this is a function of time since the $\Delta T$ will change with time. The constant factors for each are $\sim 1.8 \times 10^{-8}$ W K-4 for $\dot{Q}_{rad}$ and $\sim 1.8$ W K-1 for $\dot{Q}_{con}$. One can go ahead and plot the transfer rates (see figure below) and find that radiative transfer always dominates.

If these assumptions are correct, then what heat transfer mechanism will dominate and why?

So if we can truly ignore convection, then it appears that radiation dominates over conduction.

• That does assume the meat is effectively in contact with the oven walls - by ignoring the low heat capacity of air. A better model of the oven would be to assume k for air with a length of the half the width of the oven – Martin Beckett Jan 5 '18 at 17:07
• @MartinBeckett - I knew there was something I was missing but was not sure whether the air would matter that much. I partly posted the question and answer because I have seen so many articles that argue for one or the other but none of them seem very definitive/conclusive. In many of the articles the dominant heating process is stated anecdotally but not actually shown to be true. So I thought perhaps I could provoke a better, more quantitative answer. – honeste_vivere Jan 5 '18 at 17:38
• @honeste_vivere Most convincing articles I'd seen gave radiation as the larger heat source in a conventional oven. Quite honestly though, I think choosing to use $T_{oven}$ for conduction works in your favour here. I can't imagine that adding the insulating effect of the air would increase the conductive heat transfer. By giving a really high estimate on conduction, I think it does a pretty convincing job showing that even any natural convection would still not be sufficient to dominate over the radiant heat transfer here. – JMac Jan 5 '18 at 18:37