Why is it more efficient to get thermal energy from a higher temperature source?

In summary: This is incorrect. Entropy is not disorder, it is the amount of energy in a system that is not in one particular state.
  • #1
fog37
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Hello,

Internal energy is the sum of the total KE and total PE of the system. Thermal energy is essentially representing the KE of the system and depends on temperature.

When studying heat engines, we learn that it is better, in the name of efficiency, to get the same 1000 J of thermal energy from a source with ##T_1## than from a source with ##T_2 <T_1##, even if the energy amount is the same...I am not sure why...The molecules in the higher T are moving faster clearly. Let's assume the sink has temperature ##T_3 <T_2<T_1##...

Thanks.
 
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  • #2
fog37 said:
Thermal energy is essentially representing the KE of the system and depends on temperature.
This is not correct in general. This is only correct for an ideal monoatomic gas. For all other substances some of the thermal energy is in the form of other degrees of freedom, such as the internal potential energy held in molecular bonds under tension or torsion. The case of ideal monoatomic gas is covered thoroughly in most classes because it is simple, not because it represents all materials.

fog37 said:
When studying heat engines, we learn that it is better, in the name of efficiency, to get the same 1000 J of thermal energy from a source with T1T1T_1 than from a source with T2<T1T2<T1T_2
For the same change in energy the change in entropy is less at a higher temperature.
 
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  • #3
thank you Dale. I am not at entropy yet.

could you give me an intuitive reason of why there is a lower change in entropy and what that is good for the efficiency of an engine whose job is to convert thermal energy into mechanical work?

Also, it is always about mechanical work, not electrical work, correct?
 
  • #4
fog37 said:
could you give me an intuitive reason of why there is a lower change in entropy and what that is good for the efficiency of an engine whose job is to convert thermal energy into mechanical work?
I'm not sure there is a way to look at it withou entropy that is really meaningful. However, there are practical considerations as well, that have a big impact on real world systems: high grade energy is easier to move than low grade energy. Moving the same amount of heat across a lower delta-t requires a larger heat exchanger and more thermal fluid flow. Hydraulic power is similar (lower pressure requires more flow) and electricity as well (lower voltage needs more current).

I don't think entropy explains these, but I'm not certain.
Also, it is always about mechanical work, not electrical work, correct?
I'm not sure if there is a relevant difference...
 
  • #5
fog37 said:
could you give me an intuitive reason of why there is a lower change in entropy
Well, if you haven’t studied entropy then I am not sure how intuitive this will be, but it basically follows directly from the 2nd law of thermodynamics. This essentially says that the entropy of an isolated system increases. Since energy moves from hot to cold in an isolated system that means that the entropy lost by the hot side must be less than the entropy gained by the cold side.

fog37 said:
it is always about mechanical work, not electrical work, correct?
It is about any work, electrical, mechanical, chemical, whatever.
 
  • #6
Thank you!
 
  • #7
Hello again,

When thermal energy passes from a high temperature substance to a lower temperature, that same energy becomes "lower quality" in the sense it is now in a lower temperature reservoir and it is more difficult to extract work from it (need to find a lower temperature reservoir).

Entropy means disorder, multiplicity, etc. seems to identify how energy is stored. The spontaneous direction of a transformation corresponds to an increase in entropy and also reduces the quality of the thermal energy...
Still a little foggy on what entropy really means. I know it is ##S=\frac {Q}{T}## which means that the higher ##T## the lower ##S## assuming ##Q## is constant...
 
  • #8
fog37 said:
I know it is ##S=\frac {Q}{T}## which means...

The correct equation is ##d S =\delta q/T## : the entropy change ##dS## of a closed system during a reversible change where ##\delta q## is an infinitesimal quantity of energy transferred by heat at a portion of the boundary where the thermodynamic temperature is ##T##.

Have a look at "THERMODYNAMICS AND CHEMISTRY" by Howard DeVoe
http://www2.chem.umd.edu/thermobook/
 
  • #9
fog37 said:
Still a little foggy on what entropy really means
Are you taking a class on thermodynamics? If so you may want to wait until the entropy lecture.

Entropy is the log of the number of micro states that correspond to the macro state. Sometimes it is described as “disorder” but the problem with that is that the word “disorder” is not well defined. Often people will put their own idea about messy rooms or dirty laundry into the idea of disorder and then make mistakes in the physics. I think that it is best to keep to the rigorous meaning.
 
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  • #10
Hello,

As far as heat passing from a high temperature ##T_h## reservoir to a lower temperature ##T_c## reservoir, the hot reservoir loses a quantity ##Q## while the cold reservoir gains a quantity ##Q##.

  • Qualitatively, I understand that having thermal energy stored at a higher temperature is "better". Some ability to do work is lost after the heat transfer. Thermal equilibrium means that there are no temperature differences and no possibility to do work.
  • The more disordered the system the higher its entropy. As the degree of order of the system decreases, so does its ability to do work. I guess the higher temperature reservoir loses entropy, even if its temperature is higher. But isn't a higher temperature reservoir more disordered from the beginning since the temperature is higher? A higher temperature system seems more "chaotic" and have more entropy than a lower temperature system. Clearly, we are concern with changes in entropy. The lower temperature reservoir gains heat and its entropy increases (which makes sense).
  • Useful work can be extracted from highly ordered systems...I read this in a book but I am not sure how to correctly interpret it...
Thank you!
 
  • #11
fog37 said:
But isn't a higher temperature reservoir more disordered from the beginning since the temperature is higher? A higher temperature system seems more "chaotic" and have more entropy than a lower temperature system.
The informal definition of entropy as “disorder” can be misleading. To see what’s going on here you’ll want to use the formal mathematical definitions of temperature and entropy based on the number of microstates corresponding to a given macrostate.

I learned this stuff from a college-level statistical mechanics text, but I expect that some of our members will be able to point you to a more accessible high-quality source somewhere on the internet.
 
  • #12
fog37 said:
But isn't a higher temperature reservoir more disordered from the beginning since the temperature is higher? A higher temperature system seems more "chaotic" and have more entropy than a lower temperature system
I find that this concept of entropy as “disorder” is unhelpful and very rarely useful to students. I would recommend abandoning it entirely.

Until you get to a statistical mechanics class it is far more useful to think of entropy as simply the relationship between temperature and energy: $$dS=\frac{\delta Q}{T}$$
 
  • #13
Thank you. I will refrain from order/disorder and entropy until I have a better grasp.

I was also reading/studying that a heat engine, which works in cycles, cannot completely convert a certain amount of absorbed heat into mechanical work (see Carnot and 2nd law of thermodynamic). Some of the initial heat must be exhausted into a colder reservoir. But this is true only when discussing a thermodynamic cycle, i.e. a transformation that returns the system to its original state. Apparently, it is POSSIBLE to convert a certain amount of heat completely into work if we are not talking about a cycle...What would be an example of that?

Obviously, work can be easily converted completely into heat. But I was unaware that the opposite was also true when not concerned with a cycle...
 
  • #14
Nugatory said:
The informal definition of entropy as “disorder” can be misleading.

Indeed! See, for example:

Frank L. Lambert: “Entropy Is Simple — If We Avoid The Briar Patches!” (http://entropysimple.oxy.edu/content.htm#increase)
Frank L. Lambert: “Entropy Sites — A Guide” (http://entropysite.oxy.edu/)
Harvey S. Leff: “All About Energy & Entropy” (https://energyandentropy.com/)
Harvey S. Leff: “Removing the Mystery of Entropy and Thermodynamics – Part V” ( [PDF]
Removing the Mystery of Entropy and Thermodynamics .. )
 
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  • #15
fog37 said:
. Apparently, it is POSSIBLE to convert a certain amount of heat completely into work if we are not talking about a cycle...What would be an example of that?
Gas pushes a piston, doing one Joule of mechanical work, losing one Joule of heat energy.
 
  • #16
Ok. Thanks.

So expression of the 2nd law of thermodynamics states that

"It is impossible for a heat engine working in a cycle to produce no other effect than that of extracting from ma reservoir and performing an equivalent amount of work"

The word "cycle" is key here because, as the textbook says, it is possible to convert heat completely into work in a noncyclical process. For example, an ideal gas undergoing an isothermal expansion: the gas starts in a state and ends in a different thermodynamic state (expanded volume). I guess the entirety of the absorbed heat is completely converted into work during this process...
 
  • #17
fog37 said:
For example, an ideal gas undergoing an isothermal expansion: the gas starts in a state and ends in a different thermodynamic state (expanded volume). I guess the entirety of the absorbed heat is completely converted into work during this process...

Here's just something I thought, it's kind of funny:

Let's say we have hot ideal gas with high entropy inside a cylinder. This gas can suck neither heat nor entropy from a warm source.

Now that gas expands adiabatically. It becomes cool gas with the same high entropy as originally. This gas can suck heat energy, and also entropy from a warm source.If we zoomed into a small area where the high entropy gas and the low entropy warm stuff are in contact, we would see some apparently low entropy gas becoming more entropic.
 
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FAQ: Why is it more efficient to get thermal energy from a higher temperature source?

What is higher grade thermal energy?

Higher grade thermal energy refers to thermal energy that is at a higher temperature and therefore has a greater potential for doing work. It is typically generated through processes such as combustion, nuclear reactions, or solar energy.

How is higher grade thermal energy different from lower grade thermal energy?

Higher grade thermal energy has a higher temperature and is therefore able to do more work compared to lower grade thermal energy. Lower grade thermal energy is typically waste heat that is released during energy production processes and has a lower temperature and less potential for doing work.

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Some examples of higher grade thermal energy sources include fossil fuels such as coal, natural gas, and oil, nuclear energy, and solar energy. These sources are able to generate high temperatures and produce significant amounts of thermal energy for various purposes.

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Higher grade thermal energy is used in a variety of ways, including heating buildings, generating electricity, powering industrial processes, and transportation. It can also be converted into mechanical energy through steam turbines or used directly for cooking and other household tasks.

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