What is the difference between these two works ?

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The discussion clarifies the distinction between mechanical work in the work-energy theorem and thermodynamic work in the first law of thermodynamics. Mechanical work relates to changes in kinetic energy, while thermodynamic work involves internal energy changes due to heat transfer and work done on a system. The two concepts are not interchangeable; mechanical work assumes no state changes, while thermodynamic work accounts for changes in state. An example illustrates that internal energy can be expressed in terms of state variables, highlighting the differences in application. Understanding these distinctions is crucial for correctly applying the principles of physics in various scenarios.
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Some examples in textbook make me confused when these two works are discussed at the same time.

One of the works is the (mechanical) work in work-energy theorem:

<br /> \Delta K = \sum_iW_i,<br />

where K is the kinetic energy and W_i was the work done by the i-th force.

The other is the (thermodynamical) work in the first law of thermodynamics:

<br /> \Delta U = Q + W,<br />

where U is the internal energy of the system, Q is the heat transfered, and W is the work done on the system by surroundings.

Are the two works the same when we want to use work-energy theorem and the first law of thermodynamics at the same time?

Can anyone give some criterion to distinguish these two works ?

Thank you .
 
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Hi Variation:

You might want to refer to the overall energy balance equation:

dPE + dKE + dU = Q - W

Now the first equation have certain assumptions, that the change in potential energy is zero (such as it is on a flat surface) and the change in internal energy is zero (no state changes, or temperature and pressure changes). This is your classic pulling a block on a flat surface problem.

The second equation, you have thermodynamic work. You have state changes, and this assumes that no mechanical energy, potential energy or kinetic energy. So basically on a flat surface and not moving. A classic problem is the cylinder with the piston. You put it near the flame and it would expand. Flame provides heat (Q) and expansion is work (W)

I hope this helps.
 
This two works are barely the same. Let me give you the simplest example. Internal energy is
U = U(S, V, N)
with
dU = TdS - PdV + \sum \mu dN
-PdV = dW
being the elementary mechanical work.
You get precisely the same elementary work with mechanics of continuous media (consider the simplest case of a diagonal stress tensor \sigma_{ij} = -P \delta_{ij}).
 
Thank you all.
 

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