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Difference between ΔH and q?

  1. Sep 18, 2012 #1

    Can someone please explain the difference between ΔH and q? I know that ΔH represents enthalpy of heat whereas q represents quantity of heat. Surely, there must be a difference between them since quantity and enthalpy are different measurements.

    The thing that I'm confused about is why did my lecturer tell me that ΔH = q?
  2. jcsd
  3. Sep 18, 2012 #2


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    [tex]Q = E-w[/tex]
    [tex]\Delta H=\Delta E+\Delta PV[/tex]

    Heat is the Energy transfer due to thermal interactions, like, heating water. Enthalpy is the TOTAL energy of the thermodynamic system, including Internal Energy and its Pressure-Volume Energy (Work system does to displace the environment to do its work)

    For many processes involving solutions and not gases, it is safe to assume that ΔPV=0 and ΔV=0; making ΔH=q.
  4. Sep 19, 2012 #3
    Make sure you're not confusing a single state with a change in state for a system.

    Also make sure you're not confusing heat and energy. Heat, by definition, is a transfer of energy (due to a temperature differential). H represents the enthalpy of the system, which is more or less a measure of how much energy it contains. ΔH represents a change in that amount of energy, and would usually be represented by a change in temperature.

    ΔH = q only in certain circumstances. If there is work, then your enthalpy change does not equal your heat.
  5. Sep 19, 2012 #4
    I am not sure if the previous answers are exactly correct.
    The internal energy E of a system is the sum of the kinetic energies of all of the particles in a system and all of the potential energies that arise from chemical bonding situations or intermolecular attractions or repulsions.

    The first law of thermodynamics says that ΔE = q – w

    ΔE is the change in the internal energy of the system
    q is the heat that has flowed into the system
    w is the work that has been done by the system

    It is fairly easy to see, when it is put in this way, that the first law is just a restatement of the law of conservation of energy.

    So why do chemists, engineers, physicists, and geologists work with enthalpy H instead of internal energy E?

    Well Enthalpy is defined by ΔH = ΔE + Δ(pV)

    When a chemist does a reaction that produces a gas, or when a kettle is boiled in an open room, the system actually does work: the gas has to push back the surrounding air against its pressure to make room for itself. So for work at constant pressure, work done by the system = pΔV

    We can put the two equations together for constant pressure conditions:
    ΔE = q – w; w = pΔV = Δ(pV), and we will get ΔE + Δ(pV) = q, or ΔH = q

    So, ΔH = q for the conditions that a chemist usually works with: a constant pressure environment (does not matter which specific pressure), and the system does no other work and has no other work done on it than the work of expansion.

    ΔH is a modified internal energy designed to take account of work of expansion for the volume change of a system.

    If a reaction is carried out at constant volume, then w = 0, and ΔE = q.
    ΔH = ΔE + Δ(pV) will be rather greater than q if a gas is given off, because there will be a large increase in pressure in the closed vessel.
  6. Sep 20, 2012 #5
    I think what I said was correct, but I do still confuse the "w" and the Δ(pV).

    Would it be correct to say that "w" is shaft work (work from some external source), and Δ(pV) is work due to change in atmospheric conditions? How do we separate the two?

    Are you saying that they're the w = Δ(pV) = 0 at constant pressure?
  7. Sep 20, 2012 #6
    Perhaps it should be noted that H is a state variable or state function, and therefore a propetry of the sytem, whereas q is not a property of the system at all.

    H is defined by the equation

    H = E + PV

    Where H is the enthalpy, E is the internal energy , P is the equilibrium pressure of a sytem whose volume is V.

    q is defined as the heat transferred between the system and its surroundings during some thermodynamic process and is therefore a property of the process, not the system.

    differentiating the first equation yields

    dH = dE + PdV + VdP

    which by the first law is

    dH = q - w + PdV + VdP

    But PdV is the first law work

    dH = q + VdP

    Now others are correct in saying that most chemical experiments occur at constant pressure so dP = 0 and under these conditions

    dH = q

    Now consider adding heat (q) two systems

    System 1
    is a block of metal standing on the bench and we add heat, raising its temperature from T1 to t2, all taking place at atmospheric pressure

    There is no (significant) change of volume involved so no work is done and all the heat added goes to raising the temeprature

    q = specific heat (also called heat capacity) x Temperature change

    q = Cp (T2 - T1)

    However system 2 is a balloon of gas which expands considerably when we heat it from T1 to T2

    So the heat added not only raises the temperature but also does work

    Both of these contribute to the internal energy (E) of the system

    This time we have

    q = Cv (T2 - T1)

    which includes the work of expansion.

    We should consider the 'twins' ( specific heat at constant pressure and constant volume )

    {\left( {\frac{{\partial H}}{{\partial T}}} \right)_p} = {C_p} \\
    {\left( {\frac{{\partial E}}{{\partial T}}} \right)_v} = {C_v} \\
  8. Sep 20, 2012 #7
    Why is Cv used? Isn't volume changing?
  9. Sep 20, 2012 #8


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    Not if you heat something in a rigid tank.

    Sure, no such thing as a perfectly rigid tank. But many tanks (I suppose even PET bottles) are rigid enough if the temperature/pressure changes are not too large.
  10. Sep 20, 2012 #9
    Sorry if it was not clear, but the original question was

    What is the difference between heat and enthalpy, not what is the difference between enthalpy and internal energy.

    Ok so why use Cv?

    Well what is internal energy?
    It is the sum of all the energies input to a system. So it can be increased by adding heat alone or mechanical work alone or some other form of energy or by some combination of these.
    If they do not leave the system they all end up as internal energy.

    So the first law says E = heat in minus work out.

    So the heat in is q which is the product of the specific heat and the temperature change and the work is the energy associated with the change of volume.

    Consider E is some function of pressure, volume and temperature E = f(P,V,T)

    But P,V,T are not all independent and can be linked by an equation of state. That is to say we only need know two to determine the third. So since we can express pressure in terms of volume and temperature we can say E = f(V,T)

    So differentiating

    [tex]dE = {\left( {\frac{{\partial E}}{{\partial T}}} \right)_v}dT + {\left( {\frac{{\partial E}}{{\partial V}}} \right)_T}dV[/tex]

    So if you like

    [tex]{\left( {\frac{{\partial E}}{{\partial T}}} \right)_v}[/tex]

    is another definition of Cv and the first product is the heat added

    [tex]{\left( {\frac{{\partial E}}{{\partial V}}} \right)_T}[/tex]

    is another definition of P and the second product is the mechancial work added

    and we can recognise the first law dE = q + (-w)

    This is quite a mathematical approach.
    Here is a more practical one.

    We cannot (easily) experimentally measure the internal energy change directly.

    However we can easily measure, by calorimetry, the heat required to make a specified change. This does not tell us the distribution between heating and work achieved. That is some of the heat we put in passes out of the system as work done.

    So the total heat input = change in internal energy plus the work done

    'Total heat' or 'heat content' are old names for enthalpy .

    For instance say we have a kettle of water at 0° C and we boil off 1kg of the water (convert it to steam).
    We find in International Steam Tables (eg Callenders) that the latent heat (enthalpy) of steam at 0° C is listed as 2495 kJ/kg, the vapour pressure of steam is 615 Pa (at 0° C) and the specific volume 1 kg of water is 204.5 kg/m3

    So we have to do 615 * 204.5 of work per kg = 125.8 kJ/kg

    So the internal energy edit : change of the 1kg of steam we generate is (2495 - 126) = 2369 kJ/kg
    Last edited: Sep 21, 2012
  11. Sep 20, 2012 #10
    (1) A "system" needs to be defined before we start operating on it, and the thermodynamics we are talking about refers to "closed systems", i.e. those which can exchange heat and energy with their surroundings (outside the system), but not matter.

    (2) A system starts out with an initial internal energy, so that everything we are talking about arises from the processes we are applying to the system. We will almost certainly get less confused if we talk about the change in internal energy rather than the internal energy as such,

    that is ΔE = q - w

    and the last sentence of the previous post should read "So the increase in the internal energy of the 1 kg of water we have turned to steam is ..."

    (3) It is usually very easy to directly measure an internal energy change -- we do it in a bomb calorimeter. The most accurate thermochemical measurements are usually made in bomb calorimeters (constant volume), and mathematical routines are used to convert from ΔE to ΔH. We normally arrange for some of the "system" in our bomb to be gaseous so that there is decent compressibility, and no pressure rise that threatens the integrity of the bomb. Systems involving gases can easily be studied at constant volume.

    (4) ΔH = q whenever a process is carried out at constant pressure and no work is done other than work of expansion. ΔE = q when a process is carried out at constant volume, and this is not an impossible or impractical experiment for many systems.

    (5) In general neither ΔH nor ΔE is equal to q, and this applies, most importantly, to the fact that in a cyclic situation where a system is brought back to its initial state, both ΔH and ΔE are zero, and work must be done on the system to compensate for any heat that it gives out, or heat must be fed into the system to produce any useful work it is made to do in the process. In either of these cases q = w, and both are clearly path functions that depend on the machinations of the operator rather than properties of state of the system.
  12. Sep 20, 2012 #11

    1) I agree that you need to define a system, but you also need to define the system boundary and the system process. Where was it specified that the system had to be closed?

    2) This is just not true and at variance with your post#4. Heat and work are transferred across the boundary, they are not system variables.
    Thank you for the correction to 'change in internal energy', it shows how one should not rush these things.

    3) Callender did not use bomb calorimeters to prepare steam tables, although they have their uses. Why do you think you think chemists and engineers work in terms of bond enthalpies and prepare tables of them?

    4) & 5) and post#4 I think we are agreed, except that I did not say impossible to measure, just more difficult than simple measuring the total heat input.
  13. Sep 20, 2012 #12
    (1) Agree about system process; "closed system" is simply a tacit understanding -- If material is going to flow out of a system, there is no way of including, for example, the chemical energy of the bonds in the exiting material in either q or w, so that a different form of the first law would be needed.

    (2) Mea culpa. I used the expression "everything we are talking about" very loosely indeed.

    (3) Chemists use enthalpy tables so that they can estimate exotherms in practical laboratory situations from standard enthalpy of formation, when they are working in a constant pressure environment; they obtain standard enthalpy of formation for various materials for the tables in various ways, but deduction from internal energy of reaction in a bomb calorimeter study is a very common, and one of the more accurate ways. I have much less idea about how engineers work; I am a physical chemist.
  14. Sep 21, 2012 #13
    Hello JohnRC, I think we are largely on the same wavelength. In particular I am impressed by your use of q and w for heat and work respectively, rather than some form of delta as many loosely use. It emphasises that these are not differential quantities but actual values and that the whole of q or w are transferred.

    In the same vein it is worth pointing out the difference between this usual form of the first law (heat in, work out) and that used by many others with both q and w measured in the same direction. It can be very consfusing to come across two versions of the first law.

    As regards closed systems, how about precipitation reactions in solution? Or reactions that evolve a gas?

    go well
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