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Entropy: measure of amount of energy unavailable to do work?

  1. Aug 24, 2016 #1
    Dear All,
    I am trying to understand the concept of "entropy" in thermal engineering point of view.
    I have basic idea about "entropy" i.e. the change in entropy is used to determine the direction in which a given process will proceed.

    However following two more definition (of entropy generally available in text books) are confusing and not allowing me to really understand it.
    1. “Entropy can also be explained as a measure of the unavailability of heat to perform useful work”. Or “Entropy quantifies the energy of a substance that is no longer available to perform useful work”.

    2. “Entropy is sometimes referred to as a measure of the inability to do work for a given heat transferred.”
    Can anyone explain me above definitions rather entropy concept with simple examples with mathematical explanations…
  2. jcsd
  3. Aug 25, 2016 #2
    All definitions of entropy are interconnected...the definition that you posted is a direct result of the 2nd law of thermodynamics which essentially states that heat energy(low grade) cannot be converted to any other form of energy (say mech. work) with 100% efficiency .i.e. there will always be some heat rejected to the sink.

    now, consider an irreversible isothermal cycle, in which wc(compression work) > we(expansion work) and this extra compression work is done by some external agent ( lets generalize external agent as "the universe") to ensure the process is cyclic. The above considerations imply that the universe invests high grade energy(work) and gets back low grade energy(as heat from the compressed gas) and from the 2nd law the universe has lost something and it is essentially the ability to do useful work( this is because, it has dq in place of dw which is not interconvertible with 100% efficiency).

    therefore, with every irreversible cycle, universe's accumulates dq which potentially cannot be converted into any other form of energy with 100% efficiency and if ds measures this inefficiency then we have
    dq <= ds T (where T is the proportionality constant) and for the process at hand, dq = dw(isothermal) and therefore dw <= ds T where dw is the unavailable energy which essentially means the enegry that cannot do work.

    note : 1. for an irreversible cycle, expanding gas does less work than when it is contracting. This is essentially because of initial conditions which you can understand easily.
    2. the inequality in T ds>= dq is due to the lack of information on the temperature T, if you observe, T is not uniquely defined for an irreversible process.
    Last edited: Aug 25, 2016
  4. Aug 25, 2016 #3

    Andrew Mason

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    Entropy is a thermodynamic state function that is defined in terms of a process between two states. So it is bound to be confusing. There are a variety of ways you can look at it. It provides a quantitative basis for the second law. With the development of atomic and kinetic theory, it was realized (ie. Boltzmann) that the second law is essentially a statistical law.

    A change in entropy of a body is defined in classical thermodynamics as the integral of heat flow divided by temperature along a reversible path between two thermodynamic states: ΔS = ∫dQrev/T. Since heat will not flow between two bodies unless there is a temperature difference and it will only flow from higher to lower temperature, in any real process in which heat flow occurs between two bodies, the entropy increase of the cooler body will always be greater than the entropy loss of the hotter body. They will approach equality as the temperature difference approaches 0. That is just a consequence of ΔS = ∫dQrev/T.

    If you want to picture what entropy represents, try this: Entropy is a measure of the extent to which a thermodynamic process can be completely reversed without the addition of mechanical work from outside the system. The change in entropy is a measure of how far from the original states the system and surroundings will be after that reversal. For example, if you store the mechanical energy output of a Carnot engine and then reverse the cycle using the stored energy to cause heat to flow from the cold reservoir back to the hot, the system and surroundings will be in their original thermodynamic states after all the stored energy is exhausted. Total change in entropy is 0 representing 0 difference between the final and initial states of both system and surroundings after reversal. For a real engine, that reversal will always result in the system and surroundings not making it back to their original states (ΔSsys + ΔSsurr > 0). You will never have a process in which the system and surroundings reach and exceed their original states using just the energy gained from the mechanical work generated in the forward process (ΔSsys + ΔSsurr < 0).

  5. Aug 29, 2016 #4
    Entropy is a measure of disorder. Not useful energy or non-useful energy. In order to obtain useful energies we use Gibbs and Helmholtz Free energies as A(Helmholtz)=U-T*S and G=H-T*S where U is internal energy, H is enthalpy and the TS is non-useful energy. I hope that helps.
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