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Homework Help: Basic Thermo Help!

  1. Feb 1, 2006 #1
    Im having trouble finding the reasoning here. Work done by the system is called positive??????

    For example, a gas pushing up on a piston and causing the piston to rise.

    Well, it seems to me that as the gas pushes up on the piston, the gas does work on the piston, so the gas should LOOSE energy because it did work in raising the piston. However, the sign convention says that the work should be positive???

    What? Why!
  2. jcsd
  3. Feb 1, 2006 #2
    because work is defined [tex] \int_{V_i}^{V_f} pdv [/tex]

    for work to be positive the final volume is larger than the initial
    for work to be negative the final volume is smaller than the initial

    I don't know if this is true for all cases but this is how I always looked at it
  4. Feb 1, 2006 #3

    Doc Al

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    sign convention

    The sign of the work depends on what force you are considering: the force that the gas exerts on the piston (work done by the gas) or the force that the piston exerts on the gas (the work done on the gas).

    If you use a convention in which work means work done by the system, then the first law would be written as [itex]\Delta U = Q - W[/itex]. (Yes, positive work then means that the gas loses energy--all else being equal.)

    But an equally good convention (and one becoming more popular) is to use work to mean the work done on the system. In that case, positive work adds energy to the system and the first law is written as [itex]\Delta U = Q + W[/itex].
  5. Feb 1, 2006 #4
    I just do it as my book seems to, which makes more sense to me. Keep track of the energy. Call it Q in- W-in, and Q-out W-out. If the gas moves the piston, it used energy up, so thats W-out. Also, if a heating element raised the temp of the gas, thats Q-in. Plain and simple. No need for all this stupid is it, by?, on?, is it plus? minus? nonsense. The change in energy of the system is simply Qin+Win - Qout - Wout. I cant see why this straight forward, intuitive method is not the one used? It eliminates all sources of confusion.
    Last edited: Feb 1, 2006
  6. Feb 1, 2006 #5
    I guess I am using your convention (1)

    But by calling [tex] Q = Q_{out} - Q_{in} [/tex]

    and [tex] W = W_{out} - W_{in} [/tex]

    I still think its rather silly not to use this in/out convention though.
  7. Feb 2, 2006 #6

    Doc Al

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    I'm not sure what you mean by "out" minus "in". There's just Q and W. (Depending upon how you define W, it will be plus or minus. It seems like you are implicitly assuming that W means the work done by the system.)

    Personally, I like the "newer" convention that always talks about work done on the system. After all, it's the system that I care about. But it doesn't matter, as long as you know which convention you are using.
  8. Feb 2, 2006 #7
    Ok, let me give an example of what I mean. I see what is being done here by the convention, but I still find it unnecessary:

    Example: Say you have a closed system of gas inside a cylinder. And assume that there is an electric heating element in the system. The heating element will cause a temperature rise and expand the gas, raising the pistion.

    So what im saying is, simply keep track of the energy of the system.

    [tex] \delta Q + \delta W = dE [/tex]

    or in our case:

    [tex]( Q_{out} - Q_{in} ) + (W_{out} - W_{in}) = d E = 0 [/tex]

    Where [tex] Q_{out} [/tex] is the amount of heat that flows out of the system, [tex] Q_{in} [/tex] is the amount of heat that flows in (in this case due to the electric heater), [tex] W_{out} [/tex] is the energy loss because the gas makes the cylinder rise, thus decreasing the overall energy of the system, and [tex] W_{in} [/tex] is the work done to increase the energy of the system, zero in our case.

    So you just keep track of if the system's overall energy is going up or down depending on what its doing. You dont care if its a positive or a negative anymore.

    In a way, im taking [tex] Q-W = \Delta E [/tex] and making it more intuitive, by asking whats going on to the energy of the GAS in this process.
    Last edited: Feb 2, 2006
  9. Feb 3, 2006 #8


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    Why is dE=0?That is not true.
    Just to make it absolutley clear, The first law says that, the change in internal energy in a process (when the state of the system goes from A to B), is equal to the net heat supplied into the system minus the work done by the system. That is, in the form that you have written,

    [tex] \Delta E = Q - W [/tex]

    Where, Q, the net heat exchanged, is positive if heat is supplied into the system and Q is negative if heat goes out of the system. W is positive if work is done by the system on the surroundings and negative if work is done on the system.

    Also, when there is more than 1 source of heat or work,
    [tex]( Q_{out} - Q_{in} ) + (W_{out} - W_{in}) = d E [/tex]
    is exactly what you do. How else will you find the net heat (or net work due to various sources) entering or leaving the system? The internal energy change will not be 0.
    Last edited: Feb 3, 2006
  10. Feb 3, 2006 #9
    Because I have a closed system.

    What I have done is eliminate the need for a sign convention. Im keeping track of energy contained within the system.
  11. Feb 3, 2006 #10


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    Even if it is a closed system, why is dE=0?. A closed system can exchange heat and work with the surroundings and as a result, the internal energy can change.

    For example, if I have a a closed cylinder (a closed system) and I supply 100J of heat into the cylinder, what is the change in Internal energy? It will be 100J, as W=0
    and Q=100
    Last edited: Feb 3, 2006
  12. Feb 3, 2006 #11
    Hmmm, yes you are right. Its equal to zero in two cases. Case (1), the system converts all the work to heat, or case(2) it is a cycle. In my case it was zero because all the heat was converted to work, raising the piston.

    You are right that in general, dE is NOT zero, because if the walls are perfect insulators, then heat cant flow, and it will get trapped, thus increasing the energy of the system. (A net positive delta E ).

    Yes, quite right. But the walls have to be perfectly insulated.

    What im saying would by the generalized version would be:

    [tex] (Q_{out} - Q_{in} ) + (W_{out} - W_{in}) = E_{2} - E_{1} [/tex]

    Where [tex] E_{2} [/tex] means the internal energy of the system at some later state, and [tex] E_{1} [/tex] means the internal energy of the system at some initial state. A positive delta E means the sytem lost internal energy. A positive delta E means the system gained internal energy.
    Last edited: Feb 3, 2006
  13. Feb 3, 2006 #12


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    No, I still think that that is not true. How do you know that all the heat is converted to work?

    Now, heat and work interactions are recognised only at the boundary of the system. So, in your example, if you calculate the work done by the piston as it expands, and the heat supplied to the system from the resistor, then you can calculate the change in Internal Energy (which need not be 0).

    In fact, I say it is impossible for heat to be entirely converted to work, as that contradicts the second law.

    Case 2, is of course, correct as Internal energy is a state function, and the change is 0 for a cyclic process (as the initial and final states are same)

    Now, what does that mean? How can the internal energy of the system go out or come in? There are two ways a system can exchange energy with its surroundings, heat and work. As a result of this, the internal energy of the system changes. So, the internal energy of the system doesn't leave or enter the system.

    I agree with
    [tex] (Q_{out} - Q_{in} ) + (W_{out} - W_{in}) = E_{B} - E_{A} = \Delta E [/tex]

    where [itex] \Delta E [/itex] is the change in the internal energy of the system between states A and B.
    Last edited: Feb 3, 2006
  14. Feb 3, 2006 #13
    Simple, because the heat flow increases the internal energy of the gas. This causes the gas to expand and raise the piston and do work. Thus, zero change in internal energy. It is very possible. (The piston is free floating). Zero change internal energy processes occur in idealized cases where total energy conversion occurs. Another example would be an eletric fan. It converts all the electric energy into kinetic energy of the mass flow of the air on its blades. Hence, delta E of that system is zero as well. (A steady state process will have delta E of zero too.)

    I edited what I wrote because of that, but you caught it before I posted the fixed edit.

    Thanks for your post siddhart. That was very helpful. You are brilliant, marvelous.
    Last edited: Feb 3, 2006
  15. Feb 3, 2006 #14


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    But, total energy conversion is not possible in real life, so what I am saying is delta E will not be zero. Same way, for the electric fan, some of the electric energy will be lost as heat to the atmosphere (resistors heating up, friction, etc).

    I guess I should have waited for you to edit it. I knew that you wouldn't have let that slip.

    Glad to be of some help, but brilliant? You should see some of the other guys in my college!
  16. Feb 3, 2006 #15
    Yes, thats why I implied 100% efficiency. I assumed the work loss to be minimal. Its a valid engineering assumption, because of how small the loss is relative to the total energy of the system (in some but not all cases of course) :smile:

    Are you in UT Dehli? I bet you are you smart guy.
    Last edited: Feb 3, 2006
  17. Feb 3, 2006 #16


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    Ok, but I'm still not comfortable. Violating the second law takes courage :bugeye:!

    Never heard of UT Delhi.
    Last edited by a moderator: Mar 26, 2007
  18. Feb 3, 2006 #17
    Well, let’s say 100J of heat flows in, 99.99J of work are done. 0.01J of work is lost. Who cares? As an engineer, that is insignificant in my design process. Its the very important engineering assumption.
  19. Feb 3, 2006 #18


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    Ok, I agree with what you say. In such situations, where the efficiency is so high it would not matter. Point taken.

    I don't have experience in this, so, is it really possible to have such a high efficiency in an engineering process? For example, I read that efficient heat engines have an efficiency of about 45%.
    Last edited: Feb 3, 2006
  20. Feb 3, 2006 #19
    Probably not. But im fairly sure there are situations that arise when one can ignore the losses even if they are moderate. Say ~5%. A steady state process would be a good approx.

    Here is an example. A fan in a room. The fan will increase the air temp of the room. At the same time the rate of heat loss of the room increases until the rate of heat loss equals the rate of power consumption. At that point the temp remains constant, and the energy change becomes zero.
    Last edited: Feb 3, 2006
  21. Feb 3, 2006 #20
    I found a mistake in my equation. To be accurate, it sould read one of two ways, either:

    [tex] (Q_{in} - Q_{out} ) - (W_{out} - W_{in}) = E_{B} - E_{A} = \Delta E [/tex]


    [tex] (Q_{out} - Q_{in} ) + (W_{out} - W_{in}) = E_{A} - E_{B} = - \Delta E [/tex]

    Because based on my origional notation, [tex] (Q_{out} - Q_{in} ) [/tex] would mean a decrease in change in energy for an increase in heat flow IN. We let that one slip by the both of us :frown:
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