Contradiction in laws of thermodynamics

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Energy is defined as the ability to do work, and according to the first law of thermodynamics, it cannot be created or destroyed, only transformed. Work involves the transfer of energy, such as when a force accelerates an object, but this energy must originate from another source. The second law of thermodynamics states that entropy always increases, leading to thermal equilibrium where no usable energy remains, although energy itself is conserved in a non-usable form. At equilibrium, the kinetic energy of molecules is uniform, but individual particle velocities can vary. Understanding these principles clarifies misconceptions about energy and work in thermodynamic systems.
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following is what little I know about energy:Energy: its the ability to do the work.

First law says you cannot create nor destroy energy but can transform it to from one form to another. so its always 100 % conversation. but in a process you have 'work' done also. so does that mean 'work' doesn't consume energy.

In second law; entropy always increase but cannot decrease. In other words system always tends to achieve thermal equilibrium. Does it mean at its peak you left no usable energy, why? when you have 100 % conversation or has energy got destroyed.

Thanks for your explanations (in advance)
 
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If by "consume" you mean the energy ceases to exist, no.

For entropy, yes, it means temperature equilibrium.
 
* Work is energy flowing from one form to another. If for instance you push an object and it accelerates, its kinetic energy is increases because work has been done on it by the force that made it accelerate. But that energy must have come from elesewhere (The chemical energy in your mussels, for instance) so the total energy has not changed.

* Yes when entropy reaches its maximun, no usable energy is left. Energy is still conserved but it is in a format that canot be used any more.

Both your questions seem to be related to the stupid definition for energy given in introductory textbooks "Energy is the ability to do work". Dump that definition and you will be one step closer to understanding.
 
Energy is only 'available' to do work (a heat engine, for example) when there is a temperature difference. At equilibrium, with an ideal gas and a simple model, the Kinetic Energy of the molecules (temperature) is the same, everywhere so there is none available for conversion to mechanical energy (to do work).
I would not use the word "peak" in your original question. It's more of a limiting, maximum situation and will not decrease again.
 
dauto said:
* Yes when entropy reaches its maximun, no usable energy is left. Energy is still conserved but it is in a format that canot be used any more.
Entropy does not reach an absolute maximum. The ability to use internal (thermal) energy (ie the ability to use thermal energy to produce mechanical work) just requires a temperature difference.

Both your questions seem to be related to the stupid definition for energy given in introductory textbooks "Energy is the ability to do work". Dump that definition and you will be one step closer to understanding.
There is no need to change the definition of energy. Energy is the ability to do work. Work is the application of a force through a distance. Increasing the speed of a molecule requires work being done on that molecule. So it takes work to randomly increase the kinetic energy of molecules to increase temperature. This may not result in useful mechanical work on a macroscopic scale, but it still represents work being done at the microscopic level.

AM
 
sophiecentaur said:
At equilibrium, with an ideal gas and a simple model, the Kinetic Energy of the molecules (temperature) is the same

The mean kinetic energy uniform throughout the system in equilibrium, but the velocities of the particles will still be characterized by a distribution function.

Your argument remains the same, but I think it's worth being careful about these things.
 
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ZombieFeynman said:
The mean kinetic energy uniform throughout the system in equilibrium, but the velocities of the particles will still be characterized by a distribution function.

Your argument remains the same, but I think it's worth being careful about these things.

Agreed. I should have included the word 'mean' in there.
 
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