# Heat energy of two systems

I am confused from the time I saw the tile and carpet problem.Here it is.. In winter, when you step into your bathroom (with tiles) your legs can feel the coldness but when you step out from there and on the rug(carpet) you feel warm.Why? And this was the answer:

"The carpet or the rug is able to absorb the heat better than the tile floor.

That is why the tile floor feels colder to your bare feet than a rug does even though they are the same temperature."

1. Carpet is not a very good conductor.but why is it able to absorb heat?
2. How can both have the same temperature?

Can heat energy be different in two systems when it has same temperature?

• take a piece of cloth and a tile, place both in water of temperature $37C$, wait a sufficient time so that both are of the same temp; take them out of the water and test with your finger which one feels warmer, the cloth, the tile or the water? – hyportnex Feb 13 '18 at 14:42
• "The carpet or the rug is able to absorb the heat better than the tile floor." That's the opposite of true, and the opposite of what happens. I can see why this confused you. – JMac Feb 13 '18 at 17:09

If you place two media at different temperatures in direct contact with one another, the interface between the two media will come to a temperature intermediate between those of the two media; and a temperature profile will develop within each medium, with the temperature varying between the interface temperature, to the original temperature of each medium far from the interface.

The temperature at the interface can be obtained by solving the transient heat conduction problem involved. The predicted temperature at the interface satisfies the following equation: $$\frac{(T_h-T_i)}{(T_i-T_c)}=\frac{\sqrt{k_c\rho_cc_c}}{\sqrt{k_h\rho_hc_h}}$$ where $T_i$ is the interface temperature, k is thermal conductivity of a material, $\rho$ is the density of the material, and c is the heat capacity of the material. So, the product $k\rho c$ of each material is the key physical property parameter involved. If the product $k\rho c$ is much higher for the cold medium than the hot medium, the interface temperature will be much closer to that of the cold medium, and, if the product is much higher for the hot medium than the cold medium, the interface will be much closer to that of the hot medium.

In the case of the tile, the product is much higher for the tile than for your flesh, while, in the case of the rug, the effective product (effective because the carpet is a fabric medium) is much lower for the rug than for your flesh.

The following is the derivation of the equation I presented: If I have a semi-infinite medium (running from x = 0 to x = infinity) at an initial temperature $T_0$ and, at time t = 0, I suddenly change the boundary temperature at x = 0 from $T_0$ to $T_1$, the temperature as a function of time and position within the medium is given by the so-called complementary error function similarity solution to the transient heat conduction equation: $$T=T_0+(T_1-T_0)\left[1-\frac{2}{\sqrt{\pi}}\int_0^{\frac{x}{2\sqrt{\alpha t}}}{e^{-\xi^2}d\xi}\right]$$where $\alpha=\frac{k}{\rho c}$ is the thermal diffusivity of the medium. It follows from this solution that the temperature gradient and heat flux q at the boundary x = 0 of the medium are given, respectively, by:$$\left(\frac{\partial T}{\partial x}\right)_{x=0}=-\frac{(T_1-T_0)}{\sqrt{\pi \alpha t}}$$and$$q=-k\left(\frac{\partial T}{\partial x}\right)_{x=0}=k\frac{(T_1-T_0)}{\sqrt{\pi \alpha t}}=\sqrt{k\rho c}\frac{(T_1-T_0)}{\sqrt{\pi t}}$$
If we have two semi-infinite media, one at an initial temperature $T_h$ and the other at an initial temperature $T_c$ and, at time t = 0, and we place the two media into direct contact with one another at time t = 0, the heat flux at their interface must match at all times. This means that $$q=\sqrt{(k\rho c)_h}\frac{(T_h-T_i)}{\sqrt{\pi t}}=\sqrt{(k\rho c)_c}\frac{(T_i-T_c)}{\sqrt{\pi t}}$$where $T_i$ is the temperature at their interface (which, according to this relationship is constant for all times). So, from this equation, it follows that: $$\frac{(T_h-T_i)}{(T_i-T_c)}=\frac{\sqrt{k_c\rho_cc_c}}{\sqrt{k_h\rho_hc_h}}$$ or, equivalently, $$T_i=\frac{\sqrt{k_h\rho_hc_h}}{\sqrt{k_h\rho_hc_h}+\sqrt{k_c\rho_cc_c}}T_h+\frac{\sqrt{k_c\rho_cc_c}}{\sqrt{k_h\rho_hc_h}+\sqrt{k_c\rho_cc_c}}T_c$$