# Why work can't be extracted from thermal energy?

As I understand it, thermal energy is simply molecular agitation, and therefore nanoscopic mechanical energy.

So why can't this form of energy be used to extract work?

I consider here an isolated system, i.e., without a cold source to create work from a temperature difference.

• What do you mean by an "insulating system:? Nov 10, 2021 at 15:33
• It can be used. Such devices are called heat engines
– Dale
Nov 10, 2021 at 15:36
• I can recommend Feynman's ratchet and pawl for a better explanation than I can give Nov 10, 2021 at 16:15
• I corrected the typo, sorry for that, and thank you for the link Nov 10, 2021 at 17:42

As I understand it, heat is simply molecular agitation, and therefore nanoscopic mechanical energy.

I think you are confusing heat with internal energy.

Heat is the transfer of energy due solely to temperature difference. The kinetic energy at the atomic and molecular level (microscopic kinetic energy), what you call "molecular agitation", is properly called part of the internal energy of an object or substance, not heat. Heat transfers that kinetic energy from a hot to a cold source due to the temperature difference.

You can think of the operation of a heat engine in a cycle as roughly being the result of extracting some of the heat from the hot source performing work and passing the remaining heat to the cold source. For a heat engine to perform net work in a cycle, a cold source is always required per the Kelvin-Planck statement of the second law which is:

No heat engine can operate in a cycle while transferring heat with a single heat reservoir.

UPDATE:

Regarding your edits, the proper term for what you are now referring to as "thermal energy" is still "internal energy", not heat. You still can't equate internal energy with heat simply by referring to internal energy as "thermal energy".

That said, then of course you don't need a temperature difference to extract work from (reduce the amount of) internal energy. The first law for a closed system is

$$\Delta U=Q-W$$

where $$Q$$ is positive if heat is added to the system and $$W$$ is positive if work is done by the system (e.g., expansion of a gas).

For an insulated (adiabatic) system, $$Q=0$$ and $$\Delta U=-W$$. So you are extracting work from internal energy (what you call "thermal energy") without a cold reservoir.

Hope this helps.

• Hi Bob D. I'm a chemist who's probably been abusing heat/IE for 30 years, despite being an expert on heat transfer. So if I have an ice cube I should phrase that in terms of IE but if I add heat to it (to melt it e.g.) I should write that as heat transfer?
– Gert
Nov 10, 2021 at 17:08
• @Gert Hi Gert. Not sure what you mean by the "that" when you said "phrasing that in terms of IE". Nov 10, 2021 at 17:20
• I meant that in terms on energy content of a ice cube I should quantify its Internal Energy $U$ but when describe extraction of energy from it then it is Heat Transfer?
– Gert
Nov 10, 2021 at 17:24
• @Gert That is correct. Nov 10, 2021 at 17:27
• @Gert Describe heat transfer and internal energy transfer as what? Nov 10, 2021 at 17:41