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The First Law of Thermodynamics and Mechanical Work

  1. May 10, 2014 #1
    So I have a doubt that's been bugging me for a very long time, and my teacher's response to it was vague. And I would be very grateful if someone would help me clear the misconceptions I have. Now, first of all, according to the first law of thermodynamics, [itex]\Delta[/itex]U = Q + W, where U is the internal energy of the closed system, Q is the heat transferred to the system, and W is the work done on the system. Here comes the part that I don't fully comprehend; what "kind" of work done are we referring to here? For instance, say we have a trolley at rest on a smooth surface; would doing work on the trolley (pushing it) increase it's internal energy? I mean, what am I missing out here? Is internal energy defined relative to a stationary system? So we don't consider work to be done "on" the system unless the system is actually "deformed"?
     
  2. jcsd
  3. May 10, 2014 #2

    Doc Al

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    Staff: Mentor

    Work done on a system can increase both macroscopic energy as well as internal energy. In a thermodynamics setting, one generally looks at processes where work is done without adding to the macroscopic energy. Such as compressing a gas.

    So to answer your question, if you exert a net force on that trolley as you push it along you increase its total energy, but mainly its macroscopic kinetic energy not its internal energy. (Unless you deform it, as you point out.)
     
  4. May 10, 2014 #3
    If you have a system boundary, energy as either heat or work can cross the boundary to change the state of the system. You have to be able to distinguish which is which for thermodynamics.

    Consider the following cases for a tank of fluid such as water, where for each case the temperature of the water increases by the same amount from T1 to T2.
    1a. A resistor of an electrical circuit is outside the system boundary and is at a higher temperature than the fluid and transfers heat Q.
    1b. The resistor is inside the system boundary and transfers the same Q as before.

    2a. A paddle wheel connected to a shaft is inside the system boundary and the shaft work is W.
    2b. A paddle wheel and its ideal motor is inside the system boundary, and spins as before with the same voltage V and amperage I for the same amount of time t.

    1a is evidently heat transfer and 2a is evidently work transfer.
    What about 1b and 2b? Can you yourself state whether that is work or heat crossing the boundary?

    All 4 tanks are having their delta U increase by the same amount with energy crossing the boundary as either heat or work, and you can see sometimes the placement of the system boundary will determine whether you can call the energy transfer as either heat, or work. What if instead of the paddle wheel, we replaced it with friction pad? and included the friction pad either inside, or outside the system boundary? Heat? or Work?

    In neither of the cases did deformation take place. And none of the work was a product of pressure and volume changes ( which is to some degree a work W = force F times distance x type of calculation and easy to spot ). At other times the electrical, mechanical, magnetic, gravitational,... work can be difficult to see.

    Changes in KE and PE are not considered for changes in U, but are part of the overall energy E of the system from E1 to E2, although some exceptions probably do apply.
     
    Last edited: May 10, 2014
  5. May 10, 2014 #4
    But what is the difference between this work transfer, and between pushing the entire system? Also, isn't the total energy of the system U? The sum of all microscopic kinetic and potential energies of the constituent particles of that system?
     
  6. May 10, 2014 #5
    So, the kinetic energies of the particles are defined relative to the system? Like how a ball moving at 3 m/s on a train moving at 50 m/s is moving at 3 m/s relative to the train, but not to an observer outside the train?
     
  7. May 10, 2014 #6

    Doc Al

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    Well it's certainly true that the macroscopic translational KE of the ball depends on the reference frame.
     
  8. May 10, 2014 #7
    The equation ΔU = Q + W, usually neglects the changes in Kinetic energy and Potential energy of the system. The more general form of the equation includes these:

    ΔU+ (KE) + Δ(PE)= Q + W

    See Introduction to Chemical Engineering Thermodynamics by Smith and Van Ness.

    Chet
     
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