How does radiation transfer of heat fit in with the Clausius statement of second law of thermodynamics?

The Clausius statement of the second law of thermodynamics says that heat flows from a hotter body to a colder body. Heat can flow in many different mechanisms. In the mechanism of radiation for transferring heat, the body emits radiation though there may not be a temperature difference between it and outside. A simple example: consider a body in a complete vacuum, the vacuum has no defined temperature (acc. to this stack) but it still should emit radiations.

And this radiation that is emitted may travel through space and strike another body which may have hotter temperature than the body emitted it and then cause it to heat up. So, this seems like a violation of the second law.

A possible resolution: The lightwave radiated out by the body will redirect itself (somehow?) to strike only bodies colder than it... but this almost seems ridiculous to think about.

• If radiation from the first body can reach the second, then couldn't more radiation from the hotter second body reach the first, conserving the second law? Oct 4, 2020 at 17:41
• That's some deep thinking... I'll have to think about it Oct 4, 2020 at 17:43

A simple example: consider a body in a complete vacuum, the vacuum has no defined temperature (acc. to this stack) but it still should emit radiations.

That is correct, and the radiation emitted by a body is given by

$$\dot Q=εσAT^4$$

And this radiation that is emitted may travel through space and strike another body which may have hotter temperature than the body emitted it and then cause it to heat up. So, this seems like a violation of the second law.

It depends on what you mean by "heat up". If you mean there will be a net transfer of energy from the low-temperature body to the high-temperature body so that the temperature of the higher temperature body increases, that would be a violation. However, at the microscopic level energy can transfer from the lower temperature body to the higher temperature body as long as there is not a net transfer of energy from the low-temperature body to the high-temperature body.

At the microscopic level, some particles of the higher temperature body can have a lower translational kinetic energy than the average kinetic energy, owing to the distribution of speeds of the particles about the average (Stephan-Boltzmann distribution). When energy is exchanged between the two bodies, some of the lower kinetic energy particles of the higher temperature body may increase, meaning there can be a transfer of energy from the low to high-temperature body at the individual particle level. That does not violate the second law, because at the macroscopic level the net transfer of energy involving all the particles will be from the high to low-temperature body.

Hope this helps.

• I don't understand this part "translational kinetic energy than the average kinetic energy," Oct 4, 2020 at 18:47
• @Buraian What don't you understand about it? Oct 4, 2020 at 18:54
• I did not know there were two kinds of kinetic energy.. What exactly is difference between translational and average? Oct 4, 2020 at 18:55
• @Buraian There are three kinds of kinetic energy, translational, rotational, and vibrational. Translational normally means random speeds of particles in 3 dimensions. "Average" is not a kind of kinetic energy. The kinetic temperature of an object is associated with the average speed of all the particles or the average kinetic energy moving around. Some particles have speeds lower than the average, some higher. Look up the Stephan Boltzmann distribution. Oct 4, 2020 at 19:06
• So, is second law of thermodynamics a macroscopic principle rather than a microscopic one? Oct 4, 2020 at 19:09

Clausius's differential inequality $$dS\ge \frac{\delta Q}{T}$$ can also be written as an inequality between rates as follows $$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA \tag{1}\label{1}.$$ In $$\eqref{1}$$ $$\mathcal B$$ is the body of the system receiving heat through its boundary $$\partial \mathcal B$$ at a rate $$\dot q$$ and the temperature of the surface element $$dA$$ is $$T=T(dA)$$. As written this inequality has only "surface heat sources" but it can be generalized to include "volume heat sources"; Truesdell calls it the Clausius-Duhem inequality[1] : $$\frac{dS}{dt} = \dot S \ge \oint_{\partial \mathcal B} \frac{\dot q}{T} dA + \int_{\mathcal B} \frac{\dot s}{T} dm\tag{2}\label{2}.$$ In $$\eqref{2}$$ the quantity $$\dot s$$ represents the heat supply per unit mass $$dm$$ and per unit time (it is a rate) at temperature $$T=T(dm)$$. When the process including the heat transfer is reversible one has equality in $$\eqref{2}$$.

This is a very natural generalization of Clausius's inequality and it also includes radiation that is absorbed "bodily". Just as with $$\dot q$$ the sign of $$\dot s$$ tells you in what direction does "heat", i.e., energy and entropy may flow; more specifically when $$\dot s$$ is the radiated heat supply between two bodies then depending on their relative temperatures one body may be the source while the other the sink, or vice versa. Of course, if they have the same temperature then there is no net flow between them, for whatever one absorbs it will also radiate it out.

[1] Truesdell: Rational Thermodynamics, page 117

• So, you're saying that even in radiation, you need another body which is 'cold' for a hot body to radiate it's heat into? Secondly, I'm not very sure of the reference. Here is my ignorant take on it after reading it's wiki It seems so that this book was a 'different kind of formulation ' to thermodynamics. So, how accurate is this formulation? Is it widely adopted? Oct 4, 2020 at 20:26
• You need one warm body to radiate into space (the latter has no positive temperature), but you need another body to absorb some of that radiation. "Rational Thermodynamics" is one of several formulations of thermo-dynamics. What you usually learn in school is thermo-statics (so-called "equilibrium thermodynamics") but the Clausius -Duhem inequality is not controversial among the various formulations. Oct 4, 2020 at 20:47
• and despite what the en.wikipedia.org/wiki/Rational_thermodynamics says rational therrmodynamics has nothing to do with statistical thermodynamics, this axiomatically phenomenological and macroscopic theory originated by Coleman and Noll some 60 years ago. There are several non-equivalent schools or formulations of irreversible thermodynamics, rational thermodynamics is one of them; I am not competent to judge their relative merits. Truesdell's (Coleman&Noll) is the most explicitly mathematical and axiomatic in appearance and practice, all in the spirit of Euler and Lagrange. Oct 4, 2020 at 20:54