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The laws of entropy says entropy can only increase. On the other hand, if I take a hot object, it will naturally convert its heat into EM radiation.

How is this possible? Does EM radiation count as entropy?

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    $\begingroup$ "radiation" isn't entropy - neither is "heat". Heat transferred at a given temperature will increase / decrease the entropy of a target system (depending on the direction of heat transfer and the temperature). The "universe" system increases in entropy when you add "radiated heat" to it. $\endgroup$ – Floris Aug 25 '15 at 19:28
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    $\begingroup$ The second law of thermodynamics ranks right up there with relativity and quantum mechanics in the "most misunderstood" sweepstakes. And almost every attempt to construct a pop-sci statement makes things worse, not better. The version that you have (or think you have here) is badly mangled and completely incorrect. $\endgroup$ – dmckee --- ex-moderator kitten Aug 25 '15 at 19:49
  • $\begingroup$ @dmckee That is why I am asking. I need guidance. $\endgroup$ – PyRulez Aug 25 '15 at 20:04
  • $\begingroup$ @dmckee Then what do you think of this pop-sci attempt? (You did say "almost every"). I've got to say, I'm pretty impressed by this one. $\endgroup$ – WetSavannaAnimal Sep 5 '15 at 7:54
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The connection of heat to entropy in thermodynamics is through:

second law

where S is entropy Q is heat T is temperature, and it is through differential changes.

This in no way means that heat is entropy .

The easiest way to acquire an intuition of entropy is to read up on the statistical definition which can be proven to be the same as the thermodynamic definition.

All matter that we know is countable, and composed of particles/atoms/molecules and radiation in the form of photons. Therefore statistical methods can be applied from which the thermodynamic equations emerge.

Specifically, entropy is a logarithmic measure of the number of states with significant probability of being occupied:

In contrast to the macrostate, which characterizes plainly observable average quantities, a microstate specifies all molecular details about the system including the position and velocity of every molecule. The more such states available to the system with appreciable probability, the greater the entropy. In statistical mechanics, entropy is a measure of the number of ways in which a system may be arranged, often taken to be a measure of "disorder" (the higher the entropy, the higher the disorder)

statentropy

where kB is the Boltzmann constant, equal to 1.38065×10^−23 J/K. The summation is over all the possible microstates of the system, and "p" is the probability that the system is in the i-th microstate.

So entropy comes as a number from an ensemble of particles which can be in microstates, including photons. It is not an attribute of particles/photons.

Heat is also a collective phenomenon of all the particles constituting matter, that emerges from the statistical mechanics of the ensemble.

In the kinetic theory, heat is explained in terms of the microscopic motions and interactions of constituent particles, such as electrons, atoms, and molecules. Heat transfer arises from temperature gradients or differences, through the diffuse exchange of microscopic kinetic and potential particle energy, by particle collisions and other interactions.

EM radiation is composed of photons and adds to the entropy of the system by increasing the microstates and the probabilities associated, in addition to diminishing the temperature of an object . Heat is energy in transfer and radiation is one way of transferring energy.

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There are 3 forms of heat transfer: radiative, convective, and conductive heat transfer. All 3 forms of heat transfer are normally operating at the same time.

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The correct version of the statement you grope for in your question's title is that the heat of a closed system at thermodynamic equilibirum cannot spontaneously transform into other forms of energy: this follows from the second law because the definition of a closed system at thermodynamic equilibrium is one which has reached its maximum entropy (most likely) distribution of energy amongst its microstates. In particular, all macroscopic parts of a closed system at thermodynamic equilibrium must be at the same temperature.

Systems not in this state, e.g. hot things in contact with cold things, can spontaneously convert some of their heat into work, as does a heat engine drawing heat from a hot reservoir and expelling some of it into a cold reservoir, converting the unexpelled leftover to work. The maximum fraction of heat drawn from the hot reservoir that can be converted thus is the Carnot efficiency $1-\frac{T_{cold}}{T_{hot}}$, with $T_{cold},\,T_{hot}$ the temperatures of the cold and hot reservoirs respectively.

A hot object radiating belongs to a system not in thermodynamic equilibrium. For simplicity, think of the object in deep space. Deep space itself is at a temperature of about $3{\rm K}$. So the hot object radiates into space, thus cools down; at the same time, the radiation in deep space around the object is being absorbed by the hot object, albeit at a much lower rate. Eventually, the rate of radiation and absorption will be equal, and the object will be at thermodynamic equilibrium with its surroundings. That is, it too will be at $3{\rm K}$.

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  • $\begingroup$ If a system was in equilibrium, would there be any radiation? $\endgroup$ – PyRulez Sep 5 '15 at 11:21
  • $\begingroup$ @PyRulez There would be no radiation lost from the system. In my example here, collisions / vibrations within the object at $3{\rm K}$ still emit radiation (since they involve accelerated charge), but the radiation emitted is balanced by that absorbed by the object from the CMBR. $\endgroup$ – WetSavannaAnimal Sep 5 '15 at 11:53

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