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Difference between work and heat in thermodynamics

  1. May 8, 2010 #1
    In the equation [tex]\Delta U = \Delta Q + \Delta W[/tex] I'm trying to conceptualize what this means but I don't really know whats meant by heat and work. Would thinking of it as the change in heat results from the work done to or by the system while the work could be anything from beaming x-rays through the system to sending seismic waves through it? Or would a better way to look at it be heat is the energy used in changing the temperature of the system while work is the energy used to change the velocity of the gas particles. The latter doesn't make much sense to me since temperature is pretty much a result of the kinetic energy of the gas particles isn't it.
    Last edited: May 8, 2010
  2. jcsd
  3. May 8, 2010 #2
    I wouldn't recommend using the equation as you have posted it. The delta attached to the internal energy, U is correct, but the deltas attached to the heat and work can be misleading.

    Let us take an example. Consider a partly inflated balloon.

    The balloon contains internal energy, U. We can demonstrate this by letting out the inflating gas over a fan blade and making the fan spin.
    We can take the ballon somewhere else and still do this. The internal energy is permanent, until we let it out.
    We do not have to let it all out, so there could be some left after partial deflation. This is why it is correct to talk of a change in internal energy [tex]\Delta[/tex]U.

    We can also add heat to the balloon with say a hair dryer. As you have pointed out, this will raise the temperature of the balloon and its contents. We will have added a specific amount of heat, Q and when the dryer stops the heat input stops. It did not have this heat before so this is why I say using [tex]\Delta[/tex]Q is inappropriate.
    If we try the experiment with the fan we find that we can make the fan spin faster/longer than before so we conclude that the internal energy has increase by the amount of heat added.

    [tex]\Delta[/tex]U = Q

    This is positive since heat added to the system increases the internal energy

    However we are not finished yet. We could also mechanically squeeze the balloon, increasing the pressure inside instead of heating it.
    If we tested with out fan we would see that this also had the effect of increase spin speed so we would conclude that the work, W we had done in squeezing also increased the internal energy.

    [tex]\Delta[/tex]U = W

    So if we both heat and squeeze we get your equation

    [tex]\Delta[/tex]U = Q + W

    But there is more
    We know that when we heat the balloon it will expand somewhat. When it expands it pushes back the rest of the universe and does work on the rest of the universe. the work represented by this expansion is negative since it is done by the balloon, not on it.

    This work is the product of the pressure and the change in volume

    W = -P[tex]\Delta[/tex]V

    So the overall equation for this system (balloon) is

    [tex]\Delta[/tex]U = Q+W -P[tex]\Delta[/tex]V
  4. May 8, 2010 #3
    Thanks a lot. That squeezing the balloon example cleared up the confusion I had about all this. In college we learned that work is "force over a distance". This made no sense to me but thinking of it as the force you have to work against to move a mass a distance it makes sense to me. I'm guessing the work done in squeezing a balloon is moving the gas molecules against the force of the random collisions between other gas molecules.
  5. May 9, 2010 #4
    OK so now to answer your second question.
    You have stated a simple version of the first law of thermodynamics, sometimes called the no-flow version.
    In principle all forms of energy exchange need to be taken into account since this is really a statement of the principle of conservation of energy.

    In thermodynamics the common forms encountered are

    Internal energy = u
    Flow energy = energy kinetic energy of movement
    Displacement energy = pressure volume work
    Potential energy = Z
    Heat received or rejected = Q
    Other work done on or by the system eg turning a turbine blade = W
    Electrical energy received = E

    If we consider the steam in a steam turbine electricity generator we would encounter all of these so the first law energy balance might look like this.

    [tex]{u_1} + {p_1}{v_1} + \frac{{{V_1}^2}}{2} + g{Z_1} = {u_2} + {p_2}{v_2} + \frac{{{V_2}^2}}{2} + g{Z_2} + Q + W + E[/tex]
  6. Jun 11, 2010 #5
    I have a question with regards to this subject. I noticed that some of my engineering textbooks (with regards to internal energy) refer to internal energy by the definition ΔU = q + w. Other times I see it as ΔU = w - q. I know it has something to do with the internalization of energy from systemic standpoints and Macrocosmic (outside the system; surroundings) standpoints, but why do they do this? Why would they feel the need to make it so confusing?
    Last edited: Jun 11, 2010
  7. Jun 11, 2010 #6
    Hello Joe.

    I don't know what branch of engineering you might end up in but pretty well all branches have examples of this phenomenon. It is to do with sign conventions.
    Sign conventions are all about what we regard as positive or negative eg current direction in a circuit, clockwise or anticlockwise, hogging or sagging or in the case of thermodynamics in or out. They arise when we have a choice of definition and sometimes the original one may seem less obvious in hindsight.

    You should realise that thermodynamics was developed in the heyday of steam locomotive engineering. Steam engineers talked about "so much heat in" resulting in "so much expansion work out". Since both were real world quantities both were reckoned positive. So the energy change was one minus the other.
    Later wisdom suggests that an overall convention energy in = positive; energy out = negative was preferable to cover the many new applications of thermodynamics but the former had become entrenched by then.

    Sorry but you will always need to beware of the sign convention adopted by a particular author and check it out before using any formulae he provides.
  8. Jun 11, 2010 #7
    The simplest example is a cylinder full of gas, closed by a piston, where you can add energy by two ways:

    1. Compressing the gas with pushing the piston (this is adding work)
    2. Putting the cylinder into contact with a body with higher temperature (this is adding heat)

    In the first case the transfered energy is easily evaluated, since it equals the force on the piston times displacement (= -p*dV). In the second case energy is transfered by colisions of molecules: this energy can't be evaluated directly, so heat is calculated as difference between energy change and work, where energy change between two states can be calculated by measuring how much work is needed to reach state 2 from state 1 when the system is thermaly isolated (no heat transfer).

    Check http://en.wikipedia.org/wiki/Mechanical_equivalent_of_heat
    to see how the same change of state can be achieved by adding work instead of heat (which allows calculation of the change of internal energy).
    Last edited: Jun 11, 2010
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