Does adding heat to a material, thereby increasing electrical resistance in the material increase or decrease entropy?

Follow up questions:
Is there a situation were Heat flux ie. thermal flux, will change entropy?
Does increasing resistance to em transfer prevent work from being done?

  • $\begingroup$ @ronmaimon: updated question with follow ups if you are interested. $\endgroup$ – Argus Jul 12 '12 at 1:07
  • $\begingroup$ If resistance prevents work is there an equation that describes it yet if not may I get my name in there? lol $\endgroup$ – Argus Jul 12 '12 at 1:14
  • $\begingroup$ you can ask a separate question--- heat flux is really entropy flux, and it depends on where you dump the heat how it works in detail. Increasing "resistance to EM transfer" is unclear--- you probably mean increasing the resistance of a circuit, and this generally reduces the heat dumped, just because there is less current. The entropic considerations are not important in this, just the dumping of heat--- the entropy produced is the heat created over the temperature. $\endgroup$ – Ron Maimon Jul 12 '12 at 6:19

If by "heating" you mean "adding heat", then the answer is yes, except for the unusual situation where a material is at negative temperature. When you add heat to a system

$$ dS = {dQ\over T} $$

and this is always positive when T is positive. This is the definition of the thermodynamic temperature in the most fundamental way of looking at it, the partial derivative of S with respect to U at fixed volume and fixing all other conserved quantities is the reciprocal temperature $\beta$.

The only exception to this rule is for systems where there is a negative temperature. This occurs in spin-systems, where there is a maximum energy state and a minimum energy state. As you add heat energy to these systems, the entropy rises, then falls, which means that the inverse temperature smoothly goes down to zero, then turns negative. This corresponds to a temperature that goes to infinity and comes back out the negative side at negative infinity, going toward zero from the other diretion. Negative temperature systems are rare, since they require an upper bound on the energy, which means you are restricted to nuclear spins, which are decoupled from electron spins for a long time.

The question of whether entropy always increases with increasing temperature is a different question, and this has to do with the sign of the specific heat. For normal systems, the specific heat is always positive, so that the temperature increases with energy (beta decreases), and this is true for negative temperature systems too, as long as you define the specific heat properly as the change in beta with U. Even at negative inverse-temperature, the negative inverse-temperature becomes larger negative with increasing energy.

For neutral large black holes, as Carl Brannen points out, the specific heat is negative.

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    $\begingroup$ What about "smart materials" with "memory"? If you put them in hot water they restore the previous shape. So if I distort them at room temperature, I create chaos, but then putting them in hot water, increases their temperature and they become more ordered (take shape I programmed the material to be)? $\endgroup$ – bodacydo Jul 17 '12 at 0:51
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    $\begingroup$ @bodacydo: Any visible order or disorder on a macroscopic scale is completely insignificant compared to the atomic scale, by 20 orders of magnitude at least. Entropy is illustrated using macroscopic information or order, but it is fundamentally about the much larger microscopic state lack of information, which is all atomic scale motion. You aren't creating chaos on a molecular scale. You could say the same thing about melting ice that I've inscribed the contents of a book onto. $\endgroup$ – Ron Maimon Jul 17 '12 at 4:39

A higher temperature will cause the atoms in the material to vibrate more, increasing the number of microstates available to the material. Thus, the entropy also increases, since the (microscopic) definition of entropy is

$$ S = k \log \Omega $$

where $k$ is the Bolzmann constant, and $\Omega$ is the number of microstates.

See also the "Statistical Thermodynamics" section of the wikipedia entry on entropy.1

This does not directly have anything to do with the electrical resistance, although the electrons carry part of the entropy of the material.


The resistivity of some materials increases with temperature and with others it decreases. The way we manufacture zero temperature coefficient resistors (stable resistors) is by balancing these effects. But ignoring that, your question is also about entropy, does it always increase with temperature?

An assumption of statistical mechanics (see Callen, page 28) is that "The entropy is continuous and differentiable and is a monotonically increasing function of the energy." Thus if rising energy always causes a rise in temperature, your statement will be true. Such a material would have a negative heat capacity. Surprisingly, there are a lot of hits for "negative heat capacity" in google, and so I suppose one of these examples will be a contradiction to the assumption that entropy always increases with temperature.

The easiest one to explain has to do with black holes. With black holes, the entropy is proportional to the surface area as was famously discovered by Hawking and others. On the other hand, the temperature decreases as the black hole becomes larger. Consequently, raising the temperature of the black hole (which is done by making it smaller), causes a decrease in the system's entropy (as the area of the black hole gets smaller).

  • $\begingroup$ "heating" doesn't necessarily mean "increasing temperature". It means adding heat. Under this circumstance, all positive temperature systems increase their entropy, and all negative temperature systems decrease it. For black holes, heating reduces the temperature (negative specific heat) but increases the entropy, as always. This is definition of temperature. $\endgroup$ – Ron Maimon Jul 11 '12 at 5:48

Entropy always increases as U increases, that's it :)

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    $\begingroup$ Either one reads this statement as equivalent to the formula in Rons answer, or it's just not true (e.g. if one considers adiabatic compression). $\endgroup$ – Nikolaj-K Jul 11 '12 at 14:11

Mszep gave a very good answer; However I have some doubts if its in general true that entropy will increase with temperature(case 2 below), or (for an isolated system) with energy : -

Suppose $X$ is a given physical system. Here for definiteness we assume that microscopic description of $X$ is given in framework of quantum mechanics so that we have a simple definition of term "microstate" as an eigenvector of Hamiltonian $H$ of the system.

Case 1: Isolated system

If system is isolated and has energy $E$ then for computing entropy all that one need to do is work out following two steps :

  1. find number of eigenstates $\Omega_E$ of $H$ corresponding to energy $E$.

  2. Compute $S(E)$ as $S(E)=klog_e\Omega_E $.

If Hamiltonian of the system is such that $\Omega_E$ increases with $E$ then $S(E)$ will be an increasing function of $E$. But I am not sure if this condition of increasing $\Omega_E$ is always satisfied for all nontrivial many body physical Hamiltonians. Also it should be noted that there could be additional complications in that Hamiltonian itself may change with increase in energy. (e.g. as Mszep remarked, for an atomic system vibrational modes may come into play at higher energy; and so in this particular case more energetic system can be argued to have more microstates and hence more entropy)

Case 2: System in contact with a heat reservoir

When system is at some nonzero temperature $T$ then its total energy is not fixed, but we have a Boltzmann distribution of energies; and entropy is now given by following formula :

$S=-k\displaystyle\sum_i p_ilog_ep_i$

Here sum is over all microstates available to the system; and $p_i=exp(-E_i/kT)/Z(T)$, where $Z(T)=\sum exp(-E_i/kT)$ is partition function at temperature $T$.

Now to see if $S$ increases with temperature we need to compute its derivative wrt $T$ and check if its positive or not:

$\displaystyle\frac{dS}{dT}=-k\sum_i (\frac{dp_i}{dT}log_ep_i+\frac{dp_i}{dT}) $

Since net change in probability should be zero so $\displaystyle\sum_i\frac{dp_i}{dT}=0$.

So we are left with :

$\displaystyle\frac{dS}{dT}=-k\sum_i \frac{dp_i}{dT}log_ep_i \tag1$

$p_i$ tells probability and so should be less than 1, and so $log_ep_i$ should be less than (or equal to) zero for each $i$.

On the other hand,


and $\displaystyle\frac{dZ}{dT}=\frac{Z}{kT^2}<E>$ ($<E>$=expectation value of energy).

So $\displaystyle\frac{dp_i}{dT}=\frac{E_i-<E>}{kT^2Z}exp(-E_i/kT)\tag2$

This quantity is positive for states whose energy is greater than $<E>$ while its negative for states whose energy is less than $E$. Thus, to the extent this elementary analysis is valid, entropy may not necessarily increase with increase of temperature.

More calculations (Added on 09-05-13) :

Substituting (2) into (1) we get :

$\displaystyle\frac{dS}{dT}=-k\sum_i \frac{E_i-<E>}{kT^2}p_i log_ep_i $

$=-k\displaystyle\sum_i \frac{E_i-<E>}{kT^2}p_i log_ep_i$

$=-k\displaystyle\sum_i \frac{E_ip_i log_ep_i-<E>p_i log_ep_i}{kT^2}$

$=-k\displaystyle\sum_i \frac{E_ip_i (-\frac{E_i}{kT}-log_e Z)-<E>p_i (-\frac{E_i}{kT}-log_e Z)}{kT^2}$

$=-k\displaystyle \frac{ (-\frac{<E^2>}{kT}-<E>log_e Z)-<E> (-\frac{<E>}{kT}-log_e Z)}{kT^2}$

$=k\displaystyle \frac{ (\frac{<E^2>}{kT}+<E>log_e Z)-<E> (\frac{<E>}{kT}+log_e Z)}{kT^2}$

$=\displaystyle \frac{<E^2>-<E>^2}{kT^3}$

$=\displaystyle \frac{<(E_i-E)^2>}{kT^3}$

Thus for a system in equilibrium with another system at a (positive) temperature $T$ entropy will increase with increase in temperature.

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    $\begingroup$ You're making it far too complicated! The quantity we call internal energy ($U$) in thermodynamics is $<E>$, not $E_i$. Temperature is defined as $T = \partial U/\partial S$, so $1/T = \partial S/\partial U$. So S must increase as U increases, unless the system has a negative temperature (e.g. there's a population inversion). $\endgroup$ – Nathaniel Jul 11 '12 at 8:29
  • $\begingroup$ Now I understand where the confusion is. You are right in that internal energy U is defined as expectation value of energy, but the point is that for an isolated system with total energy $E$ expectation value of energy is same as $E$. $\endgroup$ – user10001 Jul 11 '12 at 11:31
  • $\begingroup$ Sure, but in that case $T$ is still defined such that $1/T = \partial S/\partial U$ (it's just that $S$ means $k\log \Omega$ instead of $-k\sum_i p_i \log p_i$. Or rather, the two expressions are equal for such an isolated system.) So it's still automatically true that entropy increases when energy is added, unless the temperature is negative. It just follows from the definition of temperature. $\endgroup$ – Nathaniel Jul 11 '12 at 14:03
  • $\begingroup$ Ya, but definition of temperature doesn't say anything "physical" in this case. By definition "Temperature is negative" is just another of saying that "entropy decreases with E". So your (second from the) last statement simply becomes - "Entropy increases when energy is added, unless entropy decreases when energy is added". $\endgroup$ – user10001 Jul 11 '12 at 21:45
  • $\begingroup$ Yes, absolutely, it is indeed a very simple tautology. That's why I said you were making it more complicated than it needed to be. The relevant physical fact is just that negative temperatures can exist in certain known systems, but are rare. $\endgroup$ – Nathaniel Jul 12 '12 at 8:51

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