Those who demonstrate barefoot stepping on hot (glowing) rocks (or charcoal), rely on high porosity materials (e.g., volcanic rocks and charred wood) with a very small effective conductivity \left\langle k\right\rangle and a very small effective volume-specific heat capacity \left\langle \rho c_{p}\right\rangle [i.e., a small thermal effusivity (\rho c_{p} k)^{1/2}]. This is rendered in Figure (a).
When two solids initially at two different temperatures T_{1}(t = 0) and T_{2}(t = 0), are brought together with a zero contact resistance, the interface temperature will remain constant at T_{s}. Upon using the continuity of temperature T_{i}=T_{j}\equiv T_{ij}, A_{ij}=A_{ji}, and applying the interfacial energy equation A_{ij}[-k_{i}(\triangledown T.s_{n})_{i}-k_{i}(\triangledown T.s_{n})_{j}+(\rho c_{p}Tu.s_{n})_{j}+(q_{r}.s_{n})_{i}+(q_{r}.s_{n})_{j}]\equiv Q\mid _{A}=Q\mid _{A_{ij}}+Q\mid _{A_{ji}}=\dot{S}, with no convection, radiation, and energy conversion, and using q_{\rho ck}(t)=q_{s}(t)=-k[T_{s}-T(t=0)]\frac{\delta T^{\ast }}{\delta x}\mid _{x=0}= -k[T_{s}-T(t=0)]=-\frac{k}{(\pi \alpha t)^{1/2}}[T_{s}-T(t=0)]\frac{\delta \eta }{\delta x}\frac{dT^{\ast }}{d\eta }=-\frac{(\rho c_{p}k)^{1/2}}{\pi ^{1/2}t^{1/2}}[T_{s}-T(t=0)] for the interfacial conduction heat flux for each medium, it can be shown that q_{\rho ck}(t)=q_{s}(t)=-k[T_{s}-T(t=0)]\frac{\delta T^{\ast }}{\delta x}\mid _{x=0}= -k[T_{s}-T(t=0)]=-\frac{k}{(\pi \alpha t)^{1/2}}[T_{s}-T(t=0)]\frac{\delta \eta }{\delta x}\frac{dT^{\ast }}{d\eta }=-\frac{(\rho c_{p}k)^{1/2}}{\pi ^{1/2}t^{1/2}}[T_{s}-T(t=0)]
T_{12}=\frac{(\rho c_{p}k)^{1/2}_{1}T_{1}(t=0)+(\rho c_{p}k)^{1/2}_{2}T_{2}(t=0)}{(\rho c_{p}k)^{1/2}_{1}+(\rho c_{p}k)^{1/2}_{2}} =\frac{T_{1}(t=0)+\frac{(\rho c_{p}k)^{1/2}_{2}}{(\rho c_{p}k)^{1/2}_{1}}T_{2}(t=0) }{1+\frac{(\rho c_{p}k)^{1/2}_{2}}{(\rho c_{p}k)^{1/2}_{1}} } interfacial temperature for two semi-infinite solids of different temperatures suddenly brought into contact.
From the above, when a solid, say material 1 with high values for k and \rho c_{p} and uniform temperature T_{1}(t = 0) = T_{s} , suddenly comes in contact with another solid, say material 2 with a much lower k and \rho c_{p} and at a uniform temperature T_{2}(t = 0) = T (t = 0) , then the contact surface will be nearly at T_{s} . This is the result of setting the thermal effusivity ratio (\rho c_{p} k)^{ 1/2}_{ 2} /(\rho c_{p} k)^{ 1/2}_{ 1} equal to zero in the above equation. Physically, this is because the material with a much higher \rho c_{p} will not have much change in its temperature. The solid with the lower values for k and \rho c_{p} will have a transient temperature distribution given by T(x,t)=T(t=0)+[T_{s}-T(t=0)]\left\{1-erf\left[\frac{x}{2(\alpha t)^{1/2}} \right] \right\} and the surface heat flux given by q_{\rho ck}(t)=q_{s}(t)=-k[T_{s}-T(t=0)]\frac{\delta T^{\ast }}{\delta x}\mid _{x=0}= -k[T_{s}-T(t=0)]=-\frac{k}{(\pi \alpha t)^{1/2}}[T_{s}-T(t=0)]\frac{\delta \eta }{\delta x}\frac{dT^{\ast }}{d\eta }=-\frac{(\rho c_{p}k)^{1/2}}{\pi ^{1/2}t^{1/2}}[T_{s}-T(t=0)] .
The human body has much higher k and \rho c_{p} than the high porosity rock. Therefore, the rock surface in contact with the sole of a foot will be nearly at T_{s}. To avoid damage to the living cells under the dead skin cells, the heat flow rate to the body (i.e., the sole of foot) should be small enough such that it can be removed by the interstitial blood flow
Consider a rock with a porosity \epsilon = 0.9, with the solid properties being that of silicon dioxide. Use q_{k}=-\left\langle k\right\rangle \triangledown T ,\frac{\left\langle k\right\rangle }{k_{f}} =\left(\frac{k_{s}}{k_{f}} \right)^{0.280-0.757 log(\epsilon )-0.057 log(k_{s}/k_{f})}0.2\lt \epsilon \leq 0.6 for \left\langle k\right\rangle and \left\langle\rho c_{p}\right\rangle _{V}=\left\langle\rho c_{p}\right\rangle \equiv \sum\limits_{i}{\rho _{i}c_{p,i}} =\sum\limits_{i}{\epsilon _{i}\rho _{i}c_{p,i}} for \left\langle \rho c_{p}\right\rangle and air as the fluid filling the pores. Using T_{s} = 32^{\circ }C and T (t = 0) = 500^{\circ }C, determine the heat flow rate to the surface after (i) 1, (ii) 10, and (iii) 100 s elapsed times.
\rho _{s} = 2,200 kg/m^{3}, c_{p,s} = 745 J/kg-^{\circ}C, k_{s} = 1.38 W/m-K, \rho _{f} = 1.161 kg/m^{3}, c_{p,f} = 1,007 J/kg-K, and k_{f} = 0.0263 W/m-K.
