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Calculating enthalpy from other data

  1. Mar 18, 2009 #1
    Hi everyone:

    I am not really a thermo person, but need to use thermo data in some other calculations. I have enthalpy, heat capacity, temperature, and density, as well as other data.

    I can't calculate/reproduce the enthalpy data using the other parameters and would really appreciate some clues as to where I'm going wrong (because I need to know if I should use the enthalpy data or the other data in my new calculations, or if there's some other problem). My understanding is that dH/dT=Cp. The data table is attached.

    The thermo data are from multiphase solidification calculations. Could it be that for phase change, the equation above doesn't account for latent heat? If this is the case, then my problem is solved and I will use the enthalpy values in my new calculations.

    Here's another question. Sorry--it is probably really basic. The enthalpy values from this solidification calculation are all negative. Is this OK? I mean, can I use these negative values in my other transport calculations? My understanding is that a system can be defined so that enthalpy is positive or negative. But if I remove the negative, then the values decrease with increasing temperature and this can't be right. Can someone point me in the right direction for resolving this? I may not have explained myself very well here. Thanks a lot.

    Attached Files:

  2. jcsd
  3. Mar 18, 2009 #2


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    The equation

    [tex]C_P=\left(\frac{\partial H}{\partial T}\right)_P[/tex]

    definitely is not useful for a phase change (because it would predict a heat capacity of positive or negative infinity).

    Enthalpy, like energy, is relative; it's not useful to consider the magnitude unless the reference point is clearly stated.
  4. Mar 18, 2009 #3
    Thanks a lot Mapes! Really appreciate it.
  5. Mar 19, 2009 #4
    Usually the standard thermodynamic reference is based on standard temperature and pressure conditions (25°C, 1atm) and elements in a reference state. In these reference conditions, the enthalpy is usually defined to be zero. Typically, the enthalpy of O2 gas, H2 gas, N2 gas , ..., Fe solid, ... will be zero by convention at 25°C and 1 atm.

    In standard temperature and pressure conditions, most substances will have a negative enthalpy. For example liquid water has an enthalpy of about 16 MJ/kg. This enthalpy represents then the heat of formation from the elements of this substances in standard conditions of temperature and pressure. (i.e. the heat of reaction for H2+0.5O2 -> H2O)

    For entropy data, an absolute scale is used based on the Nerst theorem.

    I suggest you to spend some time on the http://webbook.nist.gov/chemistry/" [Broken].
    Last edited by a moderator: May 4, 2017
  6. Mar 19, 2009 #5
    So here is my next question.

    If I understand correctly, it is OK for the enthalpies from multiphase solidification to be negative.

    But here is my issue: for multiphase solidification, my enthalpies are negative. The container around the solidifying stuff is being heated up and may melt. I have SEPARATE multiphase thermo results that describe how heating up and melting happens in the container walls. The enthapies from that calculation are also negative. So negative enthalpies for the solidifying fluid and for the walls.

    Now I want to take those results and put them into my new calculations (that describe transport--length and time scales associated with solidification of the fluid/heating of the walls.

    In my new calculations, I incorporate the thermo results by defining enthalpy versus temperature tables. Can I simply take the negative enthalpies of formation of (1) the fluid and (2) the walls from each separate thermo calculation and input them?
  7. Mar 19, 2009 #6


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    I'm having a problem with the idea of heating up and melting (for the container walls) being exothermic. How do you interpret this?
  8. Mar 19, 2009 #7
    Hi Mapes:

    I don't know if you're interested, but I've attached examples of melting and solidification thermo calculations. Can you see, by looking at the results for a single T step (i.e., see system H at bottom of a T step), how enthalpy is calculated? You mentioned that the Cp equation is not relevant here. So using the other thermodynamic output, how else can enthalpy be calculated?

    Attached Files:

    Last edited: Mar 19, 2009
  9. Mar 19, 2009 #8
    Sorry, I didn't see your question. Please see the examples I attached to my previous post. I think it's because in the thermo model, the system is (1) the container walls or (2) the solidifying stuff, but not both together. There's a separate calculation for each material (solidifying or melting) and I think the thermo software always adjusts so that H is negative. You have identified my problem (here's where my weak thermo background comes in): putting the two separate calculation results together. Can the negative enthalpies of formation from both separate calculations be used in the same calculation?

    In my transport calculations, the software requires that enthalpy versus T is positive. So I can't simply remove the negative signs from either one of the thermo calculations (H vs. T will end up positive). The thermo results describe the properties of 1) the melting or 2) the solidifying substance at a given T. They don't describe heat flow into or out of the melting or solidifying system.
    Last edited: Mar 19, 2009
  10. Mar 19, 2009 #9


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    It looks like enthalpy vs. T is positive; that is, enthalpy always increases with temperature, as it should. Am I missing something?

    Enthalpy is only defined up to a constant; only changes in enthalpies are meaningful. You can add 109 J mol-1 to all your values if you want; it shouldn't make a difference. If it does, there's something wrong with the calculations.
  11. Mar 19, 2009 #10
    Hi Mapes:

    I think you have solved one of the problems. I finally understand what it means that only changes in H are significant...That we can ADD to our values. Thanks! Sorry for being so dim.

    In these results files, enthalpy vs. T is positive. But we have other results for other conditions where it doesn't. There are certain situations in multiphase systems where it can decrease, if my understanding is correct.
  12. Mar 19, 2009 #11
    So one more dumb question, just to be sure. It doesn't matter then, that the enthalpies for each separate thermo calculation are both NEGATIVE (for solidification and melting)? Because only changes in enthalpy matter?
  13. Mar 19, 2009 #12

    I think you should describe you experiment a little bit more:

    For my first answer I had not noticed the "multiphase solidification" and I answered as for the enthalpies of substance. Apparently you are dealing with differences of enthalpies, probably measured by calorimetry of DTA or something alse, I don't know.

    If you are considering something like the solidification of ice, then the difference of enthaply (liquid-solid) should be positive.

    To use your data correctly, we need to know what they represent exactly.
  14. Mar 19, 2009 #13
    Hi Lalbatros:

    They are enthalpies of formation. Mapes has explained that it does not matter that they are negative in the calculation output (see files attached to my post above to Mapes). Mapes explained that I can add some constant to all the enthalpies to make them positive because it is only the changes that matter.

    I do still have a question though, that I have asked Mapes about. The thermo software always output negative enthalpies for any calculation. So my calculation for melting of a container results in negative values and my other calculation for solidification also outputs negative values. My question is, can you confirm that it does not matter that the container (heating up) and the solidifying stuff (cooling down) both have the same sign? Sorry, I am being extra careful because I want to be sure to build a solid foundation for my transport calculations. Thank you so much for your interest and help!
  15. Mar 19, 2009 #14

    I looked at your excel files (test.xls).
    Are you a geologist, or are you a cement chemist or engineer?
    I guess a geologist, considering the kbar seen in your files.
    But the temperatures are typical of the cement industry althoug a bit low!
    Could you comment a little bit on what you are doing?
    The excel files look like output from a chemical equilibrium software or database, at bit difficult to read without more information.
    Could you eventually tell me which software it is, if it is not home-made.

    As a hobby a few years ago I worked on phase calculations for the cement system (CaO, SiO2, Al2O3, Fe2O3), working only from (paid) published litterature data, and without any other software than a programming language. I found it very difficult to get consistent thermodynamic data from the litterature and to match known experimental facts from the industry (like the amout of liquid phase and the melting temperatures). In the end it was only good as an exercice.

  16. Mar 19, 2009 #15
    bumblebee77: "They are enthalpies of formation."

    Lal: That's ok then, and indeed they should increase with temperature. You should check the conventions used in your thermodynamic data source. I guess, it is possible to calculate order of magnitudes, based on chemical bounds enthalpies accounting. That may however need some time, and I suggest that only in case you have time and taste for that.
  17. Mar 19, 2009 #16
    Hi Lalbatros:

    I am a geologist. I'm using software from a friend for the thermo. It is calibrated on experimental data for magma. The EXCEL files that I attached here are summary files. Each calculation produces multiple files--one for each phase, but I thought it would be too confusing to include them all here. If you are interested, I can answer any questions about interpreting the files.

    The thermo calculations model magma solidification or melting of solid rock (two attached files provide an example of each). What I want to do is use these two files to find out how long it would take magma to solidify and how much of the surrounding crust would heat up and melt as a result. I am very rusty on thermo. So I need to make sure that I understand the results! You guys are wonderful for being so patient. It is so valuable to be able to ask about this stuff.

    My understanding is that in an isobaric multiphase system, enthalpy vs. T may go negative because of density changes.
  18. Mar 19, 2009 #17
    I just realized that I do have a problem. Mapes, the enthalpy values need to be in volumetric units (e.g., J/m^3) in my transport software. So I can't add a constant to them to make them not negative (changes final result). The software does not like the negative values though. Think I am stuck!
  19. Mar 19, 2009 #18


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    How about dividing by density and multiplying by atomic weight to get to J/mol?
  20. Mar 19, 2009 #19
    Hi Mapes:

    Great idea! I will give it a go.
  21. Mar 20, 2009 #20
    Hi bumblebee77,

    From your last posts, I think it would be careful to check the definitions assumed in the transport code you are using. It could well be that it is using non-standard conventions.
    For example, the reference state could be the solid phase of each substance.
    This would make no problem as long as there are only phase changes without chemical transformations. This is in principle ok for a pure transport code.

    Could you give us more information about this transport code?

    Besides, are you dealing with melting of pure substances, or do you also consider chemical solutions?

  22. Mar 22, 2009 #21
    Hi Michel:

    Thanks for your suggestions. I am expecting to hear from the transport code people tomorrow (will let you know), but I'm pretty sure that the transport code only cares about the H versus T relationship and other boundary conditions that I define.

    I am dealing with multiphase, multicomponent (i.e., liquid and solid) solutions. But the transport code doesn't care about chemical changes. As far as the code is concerned, there is a solidifying substance with solidification behaviour defined by the H versus T data from the thermo calculation.

    I did as you and Mapes suggested and changed the enthalpy output from the thermo calculations to make it positive (the transport code may only like positive enthalpy data) by adding a constant. In fact, I did this for several constant values. However, when I ran the transport code using the recalculated enthalpy values, I did not get identical results. I am trying to figure out why. The weird thing is that for recalculated + original enthalpy values that are negative, the transport results are identical. But where the recalculated enthalpy values are positive, the results are different from each other and from the results based on negative enthalpy. I am going to ask the transport code people why this might happen.

    A question: Below is output from a thermo calculation. I should be able to calculate enthalpy from the Cp and density and T (1150 deg. C) right?

    System mass = 100.15 (gm) density = 2.37 (gm/cc)
    G = -1648722.72 (J) H = -1232991.00 (J) S = 292.12 (J/K) V = 42.21 (cc) Cp = 148.47 (J/K)
  23. Mar 23, 2009 #22
    Hi bumblebee77,

    Let's assume the data you have provided are for temperature To.
    Then, as long as T is not too far from To, you can use:

    H(T) = H(To) + Cp(To)*(T-To)

    where the units used must be consistent, for example: H[kJ/kg], Cp[kJ/kg/K] .
  24. Mar 23, 2009 #23
    Hi Michel:

    Thanks a lot for being so patient with me! I am learning so much from being able to communicate with you and Mapes. It is really nice of you guys to spend so much time helping people out.

    So I'm not feeling very confident about the transport code I'm using. I was told today to ignore the error messages I get when I input enthalpy values with a negative sign. There are some other weird things going on too; for example, when I recalculate the enthalpy values to make them positive, the code doesn't always produce the same results. I think I'll try writing my own code in Matlab. At least that way, I'll know what it's doing!
  25. Mar 24, 2009 #24
    Hi bumblebee77,

    It all depends on the complexity of the problem you want to solve.
    However, in any case you would benefit by considering first a simple problem.
    You could then try to solve it by your own means analytically or by simple numerical methods.
    Then you could eventually compare with your transport code.
    It will also be easier to eventually discuss your Matlab code here.
    Personally I don't use Matlab very much (I use more often VB, java and C#) .
    I have done some radiation heat transfer calculations in Matlab in the past and a few other things.

    In any case, you need indeed to know what is going on.
    Last edited: Mar 24, 2009
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