Understanding the Relationship Between Work and Heat in Thermodynamics

[dEdt]

1) If work is done on a system, then W = ΔE. But how can we calculate the work done by the system on its surroundings? Intuitively, I would think the answer is -ΔE, but I can't prove it :(. When I try, I run into contradictions (two planets at rest, gravitationally attracted to each other; if one planet is considered as the system, then its energy will increase. Wext = -ΔE, then the other planet shouldn't speed up, but it does).

2) The work article says that the total work done in an isolated system is independent of the frame of reference. What's the significant/implications of this?

 

[jambaugh]

Use the equal and opposite force law. Remember work = force times displacement but that is a "vector" equation so it can be negative if force is applied in the opposite direction.

As far as independence of frame of reference note that it is not true for non-isolated systems if you consider moving frames of reference. Example: You drop a rock and gravity does work on it as it falls a certain distance increasing its Kinetic energy from 0 to 1/2 m v^2.

Now I'm parachuting down at constant speed v/2. I see the rock initially moving up at speed v/2 then gravity does first negative work then an equal amount of positve work as I see the rock slow then begin to fall again at speed v/2. From my perspective 0 total work is done.

But then again this is not an isolated system as gravity is affecting it from outside. What does the system being isolated tell you?

 

[dEdt]

For when using Newton's third law to get W_ext = -W, doesn't that imply that the two forces act over the same distance? But if the two forces act on different systems/points, why would that be true?

If the system is isolated, then the total momentum is zero...?

 

[jambaugh]

For when using Newton's third law to get W_ext = -W, doesn't that imply that the two forces act over the same distance? But if the two forces act on different systems/points, why would that be true?

If the force is applied by contact then it is true. I push you with my hand. You slide backward pushing against the floor. I do work on you, you do negative work on me. You do work on floor (heating it up) the floor does negative work on you (so you don't accelerate but slide at constant rate).

Now in the case of say gravity it is a bit trickier. I jump out of a plane (with a parachute. We don't want me to go splat). The Earth via gravity does work on me (increasing my kinetic energy). There is an equal and opposite force of gravity by me on the Earth but the Earth doesn't move nearly as much as I do. So it would seem to invalidate this thesis. However we treat the gravitational field itself as an intermediate actor in this scene. It goes rather like this:

The gravitational field acts on me doing positive work. I do equal negative work on the the field decreasing the potential energy of the me-Earth system. The field does (less) work on the Earth pulling it slightly toward me (in proportion to our relative masses so the center of mass remains fixed). The Earth does equal and opposite negative work on the gravitational field likewise decreasing slightly potential energy.

This appears to be how forces of all kinds work. One can in fact do work on the gravitational field around you which then propagates as a gravity wave.

If the system is isolated, then the total momentum is zero...?[/QUOTE]
Not zero, but constant. Similarly with energy and angular momentum. These will be conserved.

 

[dEdt]

Got it, thanks.