A question about thermal conductivity and heat transfer

The sensation of something feeling "hot" is a function of the rate of heat transfer to the skin and the duration of the transfer. The product of the rate and duration of heat transfer is the amount of heat transferred to the skin.

The rate of heat transfer of the aluminum foil, being a metal, is relatively high because of its high thermal conductivity. But the amount of heat available is low because it is very thin making its mass per unit area and its heat capacity very low, all other things being equal. This is why you can take the foil directly out of the oven without an oven mitt and not get burned. If, on the other hand, you attempted to take the metal rack out of the oven without a mitt you could get burned.


A widely cited thermal injury study (Alice Stoll et al) determined the theoretical combinations of absorbed energy rate and duration of exposure for thresholds of blister and pain. It was based on experiments conducted on human subjects.

For an absorbed rate of 5 w/cm$^2$ and duration of exposure of 1 s, a theoretical total absorbed energy of 5 J/cm$^2$ is predicted for threshold of blister and 2.5 J/cm$^2$ for threshold of pain.

The amount of thermal energy available to transfer to the finger, bringing the temperature of the aluminum to the temperature of the finger, from a 1 cm$^2$ piece of aluminum foil (approximate surface area of finger tip), 0.02 mm thick (average for foil) at a temperature of 200 C (392 F) is about 0.8 J, making it more than 5 times less than the blister threshold and more than 2.5 times less than the pain threshold.

Just a curious thought, what should we do to get burn by the aluminium foil? I mean what modification should we make to the foil? And may you please provide formula/methodology for the calculations?

In order to increase the potential for pain and burn from the aluminum, one would have to sufficiently increase its thickness and/or its temperature.

The formula used to calculate the theoretical amount of heat $Q$ available per cm$^3$ of foil relative to, say, the finger, and assuming the foil does not transfer heat to anything else, was

$$Q=\rho c (T_{Al}-T_{f})$$


$\rho$ = the density of aluminum, taken as 2.71 g/cm$^3$

$c$ = the specific heat of aluminum, taken as 0.872 J/g$^o$C

$\rho c$ = the volumetric heat capacity of aluminum, 2.43 J/cm$^{3}$ $^o$C

$T_{Al}$ = temperature of aluminum, taken as 200$^0$C (upper temperature of a moderately hot oven)

$T_{f}$ = temperature of the skin of finger, taken as 32.5$^o$C.

From the above data the available thermal energy is about 400 J/cm$^3$. Multiplying this by the thickness of the foil, 0.002 cm, gives you 0.8 J/cm$^2$ of foil.

As indicated previously, the theoretical threshold for blister and pain for 1 sec contact is 5 J/cm$^2$ and 2.5 J/cm$^2$. To approach those levels the foil thickness would need to be theoretically increased by a factor of 5 and 2.5 respectively. Alternatively, the oven temperature (= temperature of the foil prior to touch) would have to theoretically be increased to about 870 $^o$C and 450 $^0$C, respectively. These temperatures are well above those attainable in household ovens.

It should be noted that there were a number of assumptions made in the above, most of which err on the side of caution. Perhaps the most significant is the theoretical thresholds of pain and burn. They were based on experiments on very thin (most vulnerable) skin (the inside surface of forearm and back of finger (middle phalanges)). The skin on the finger pads is thicker making them less vulnerable. So the thresholds for the finger pads would be higher than those used here. This should err on the side of caution.

Hope this helps.

The foil is so thin that it loses its heat energy very quickly. Even though it may initially be as hot or hotter than the pizza when removed from the oven, its heat will cool very quickly. When you touch it, it reaches thermal equilibrium with your skin before it can heat your skin enough to burn you.