# Does a frying pan use radiation?

I have a project to prove that the colour black absorbs more light than the colour white in a creative way. Our first idea was to time a white frying pan and black pan and see which would melt the ice cube first, but is this proving radiation? Because I looked at saw pieces of metal above where the flame would go, making me guess pans use conduction. Will a flame ever give off enough radiation to be noticeable?

• When cooking using a pan over a flame, most of the heat doesn't come from the light of the flame (radiation), but rather the hot gases created by combustion in the flame. – MaxW Mar 28 '16 at 16:45
• True, most of the heat is carried by convection. That being said, flame does produce a good deal of thermal radiation, more than enough to be noticeable. You have to devise some ingenious way to separate that from convection. – Ivan Neretin Mar 28 '16 at 17:26
• A light mill proves that black absorbs more light than the color white. – aventurin Mar 28 '16 at 20:28
• Isn't the heat the pan is putting into the food almost purely conduction? The heat gets to the pan through all three modes of heat transfer, but once it's there, I don't think you would get much cooking going if you suspended a steak an inch above the pan. Sounds like a good way to warm some milk before bed... if you want to wait an hour. – SendersReagent Mar 28 '16 at 21:54

You may estimate the contribution of thermal radiation by using the Stefan–Boltzmann law, which describes the power $P$ radiated from a black body in terms of its temperature $T$:

$$P=\sigma\cdot A\cdot T^4$$

where $\sigma=5.670\,367(13)\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}$ is the Stefan–Boltzmann constant[reference] and $A$ is the surface area.

The formula can be corrected for real materials by introducing the emissivity $\varepsilon$:

$$P=\varepsilon\cdot\sigma\cdot A\cdot T^4$$

For an ideal black body, the emissivity is $\varepsilon=1$. For polished metal surfaces, the values are significantly lower.

For example, for a hot black frying pan with $\varepsilon\approx1$, $A\approx600\ \mathrm{cm^2}=0.06\ \mathrm{m^2}$, and $T\approx230\ \mathrm{^\circ C}\approx500\ \mathrm K$, the power may be estimated as

\begin{align} P&=\varepsilon\cdot\sigma\cdot A\cdot T^4\\ &=1\times5.670\,367\times10^{-8}\ \mathrm{W\ m^{-2}\ K^{-4}}\times0.06\ \mathrm{m^2}\times\left(500\ \mathrm K\right)^4\\ &\approx200\ \mathrm W \end{align}

However, in addition to thermal radiation, heat transfer also occurs via thermal convection and thermal conduction. You could exclude these modes by placing the frying pan in a vacuum.

You could also estimate the total heat transfer due to all three modes from the electrical power of the stove that is required to keep the frying pan at a constant temperature.