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Is it possible to extract the molecular kinetic energy from a system directly (without the use of a heat engine / temperature gradient) and convert that to another form of energy, such as electricity, or perform useful work?

Similar questions have been asked before:

However, I find the answers slightly lacking. They rest on one of two assumptions:

  • that the asker wants to transfer the heat away through a heat engine; or
  • that the Second Law of Thermodynamics applies to all systems in all situations — my understanding of thermodynamics (albeit rather limited) says that these laws are statistical properties that are true en masse, but not absolute laws of nature in the sense that they apply to all systems for arbitrarily short amounts of time.

Even if my understanding of the second law is incorrect, I don't understand why we can't extract heat energy from an object without a temperature gradient by placing it under certain conditions. For example, heat transfer via infrared radiation could be extracted from a gas of any temperature, placed in a glass sphere and isolated from the environment via a vacuum chamber:

diagram

The gas would slowly radiate its heat through the glass to the ambient container housing the vacuum, and solar panels lining this surface could feasibly collect this energy.

Note that this isn't a question about efficiency; I'm not concerned with how efficient this particular setup would be. It seems that if it works at all, one would be extracting thermal energy from an object without a heat engine.

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    $\begingroup$ In your diagram, what you don't show is the photovoltaic cells radiating their own IR radiation back at the gas at the center. If the cells are colder than the gas, then the net energy transfer will be from the gas to the cells. If the cells are hotter than the gas, then the net energy transfer will be from the cells to the gas. $\endgroup$ – The Photon Jul 29 '14 at 4:42
  • $\begingroup$ @ThePhoton I understand that the net energy transfer will be from the cells to the gas, but in the case that the cells are hotter than the gas, will they still be able to convert the thermal energy of the gas to electricity? That is the part I am most confused with here, as if you can extract thermal energy into electricity in that way, it still seems that you are extracting thermal energy from an isolated system (the cells and the gas, to make it isolated) into electricity and making the system colder as a result, without transferring any heat to a colder system. $\endgroup$ – Matthew McDermott Jul 30 '14 at 0:51
  • $\begingroup$ @MatthewMcDermott the photovoltaics are using the gradient between the hot incoming photons and the cold substrate to extract energy. The radiating body cools and the substrate warms. If there is no cooler body to reject the heat, then the temperature of the substrate rises and the reaction loses efficiency. $\endgroup$ – BowlOfRed Aug 19 '14 at 6:39
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"The gas would slowly radiate its heat through the glass to the ambient container housing the vacuum, and solar panels lining this surface could feasibly collect this energy."

No. If we assume the gas inside and the cells outside are both at temperature $T$, then no (thermal) energy can be extracted. They will be in thermal equilibrium. Whatever mechanism you want to come up with will be unable to extract energy.

If we assume some pathway that interacts with the IR radiation to complete a chemical reaction, then since the material is at that temperature, the reaction is just as likely to run in reverse and return the same radiation to the interior.

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  • $\begingroup$ Does this mean that there is no radiation from the gas when the gas and cells are in equilibrium? What is it about the temperature gradient that permits the actual transfer of energy via radiation? $\endgroup$ – jtbandes Aug 19 '14 at 6:44
  • $\begingroup$ Thanks so much for your answer. I'm a little confused, though. Are you saying that if the photovoltaic cells are hotter than the gas, they will not be able to absorb any thermal energy from it? Why is this? It seems like they should be able to, as the photovoltaics have no way of knowing the temperature of the gas, and the gas will radiate thermal radiation of a variety of wavelengths. $\endgroup$ – Matthew McDermott Aug 19 '14 at 6:49
  • $\begingroup$ There is radiation in both directions, but you can't extract work from it. $\endgroup$ – BowlOfRed Aug 19 '14 at 6:59
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    $\begingroup$ I don't know the specific reactions involved in photovoltaics, but if the components are all brought up to a temperature of the incoming radiation, then they are going to run backward (taking in electrons and emitting radiation) just as often as they run forward (absorbing radiation and producing electrons). There will be no net production of energy. $\endgroup$ – BowlOfRed Aug 19 '14 at 7:02
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    $\begingroup$ Yes, and a turbines behavior changes quantitatively when the source water is at the same level as the destination pool. There's no net flow of water, so the turbine stops turning. It does this without knowing the height of the water. The voltaic cell simply behaves differently at high temperatures than it does at low temperatures. Passive filtering will reduce the efficiency, not increase it. $\endgroup$ – BowlOfRed Aug 28 '14 at 7:12
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In your question you already have the answer, photovoltaic cells. They are not heat engines but change the black body electromagnetic radiation of the sun to electricity. Of course this depends on the fact that the black body temperatures of the cells are much smaller than the black body temperatures coming from the sun.

As @ThePhoton says , note the direction of the temperature differences. For objects at the temperature range of earth the system will be completely inefficient. It is better to boil water and make a steam engine than wait for the black body radiation to transfer energy to a solar panel.

All motors also are not heat engines changing one form of energy to another ( kinetic to electrical, or electrical to kinetic).

Hydroelectric engines change potential gravitational to electric.

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If you work through the equations of a solar cell in detail, you find that the maximum possible efficiency of a solar cell is a function of the cell's own temperature and of the radiation brightness temperature of the light hitting it. If the two temperatures are the same, as they would be in your drawing, you find that the maximum possible efficiency is 0. The cell can convert 0% of the light energy hitting it into electrical energy. This is of course what you expect from the Second Law of Thermodynamics. It is consistent with conservation of energy too, because the solar cell emits its own thermal radiation which is equal to the amount it receives.

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There has been a new approach that you may consider. Your question involves the energy conversion between heat energy and Gibbs free energy. Since the two both are non- conserved quantities, the changes in heat energy and both Gibbs free energy can be divided into the two parts: one is the fluxes, the other one is the productions. For heat energy, $\delta Q$ is the heat flux, and another term, the heat production comes from energy conversion. Using $d_iq $ denotes the heat production, $d_eq =\delta Q$ denotes the heat flux, a new state function can be defined by

\begin{align}dq=d_eq+d_iq.\end{align}

Where $q$ denotes the heat energy within the system. Similarly. we have

\begin{align}dG=d_eG+d_iG. \end{align}

Where $d_eG$ is Gibbs free energy flux. and $d_iG$ is Gibbs free energy production.

In your case, we have

\begin{align}d_iG=\sum_j\Delta \mu_jdN_j=-d_iq. \end{align}

For energy conversion, the first law does not involve the temperature gradient.

The entropy production

\begin{align}d_iS=\Delta \left(\frac{1}{T}\right)dq+ \sum_j\frac{1}{T}\Delta \mu_jdN_j+\Delta \left(\frac{p}{T}\right)dV\ge0.\end{align}

The driving force of the process is $\Delta(1/T)$, so for the second law, the temperature gradient must be considered.

More details can be seen in the paper

http://arxiv.org/pdf/1201.4284v5.pdf

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Bowlofred's answer is correct. You can get some insight by studying Feynman's Brownian ratchet - an attempt to get energy out of heat using statistical fluctuations (This also relates to your second point - that the Second Law is statistical in nature.) It consists of a tiny paddle wheel and a ratchet, and appears to be an example of a Maxwell's demon, able to extract useful work from random fluctuations in a system at thermal equilibrium in violation of the second law of thermodynamics. Detailed analysis shows why it cannot actually do this. A photovoltaic cell is rather similar - it only works if the electrons are mostly bound to start with, so that they can then be ejected by a photon.

The exact mechanism may be complicated, but the result is always the same - you cannot extract energy from heat without a heat engine that rejects heat at a lower temperature.

Of course, the random nature of heat means that you could be lucky and get a net flow of a handful of electrons in the direction you want through your external circuit in a given period of time. If you want to detect these electrons, though, you will need an amplifier which uses a lot of energy.

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There's no easy answer to your question as you don't define the nature of the gaz nor the nature of the cells.

"Even if my understanding of the second law is incorrect, I don't understand why we can't extract heat energy from an object without a temperature gradient by placing it under certain conditions. For example, heat transfer via infrared radiation could be extracted from a gas of any temperature, placed in a glass sphere and isolated from the environment via a vacuum chamber:"

You are actually right. To be able to consider this, you have to avoid to use the black body model for your system, where the temperature of each body are the only parameters defining your radiation energy exchange. In that case, the equilibrium is only a function of relative exchange surfaces and the initial temperature of the bodies.

Now if you consider for instance a grey body model, you have to pay attention to the absorption, reflection and transmission coefficients, which are a function of the wavelength (or frequency) and of the material considered (and also of the temperature...). In that case you can actually take energy from your gas and transform it into electricity, you just have to choose a correct combination of materials and correct intitial temperatures, according to the previous coefficients.

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