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

Physics Asked on March 1, 2021

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.

2 Answers

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.

Answered by Bob D on March 1, 2021

Clausius's differential inequality $dSge frac{delta Q}{T}$ 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] : $$frac{dS}{dt} = dot S ge oint_{partial mathcal B} frac{dot q}{T} dA + int_{mathcal B} frac{dot s}{T} dmtag{2}label{2}.$$ 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

Answered by hyportnex on March 1, 2021

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