It shows there are two components with two different melting point. However, check sometime the residual solvent showed a peak at the lower temperature.

You have done right way. However, I do suggest to follow the following steps if the first peak is for the solvent that could be avoided following these steps.

Step1: Equilibrate at 350C for 15 min and cool at 10C/min to 0C and isotherm for a min or 2 then heat at 10C/min to 350C; repeat those steps for another cycle.

Compare the melting peaks between two cycles and you will get the answer.

However, how do you know that there is no lower melting polymer.

To confirm it: you can extend one run to -70C as the sample is unknown.

Good luck.

It shows there are two components with two different melting point. However, check sometime the residual solvent showed a peak at the lower temperature.

You have done right way. However, I do suggest to follow the following steps if the first peak is for the solvent that could be avoided following these steps.

Step1: Equilibrate at 350C for 15 min and cool at 10C/min to 0C and isotherm for a min or 2 then heat at 10C/min to 350C; repeat those steps for another cycle.

Compare the melting peaks between two cycles and you will get the answer.

However, how do you know that there is no lower melting polymer.

To confirm it: you can extend one run to -70C as the sample is unknown.

Good luck.

I support the answer from Ahmed

In brif the highest peak represent the materials melting temperature and the lower represents a second phase or may be a structure change

It looks like a polyethylene/polypropylene blend to me, although I remember the recrystallization peaks as being at higher temperatures. You could try using a max temp of 200 C. I think you're degrading the polymers.

The themograms are quite similar to those of Linear Low Density Polyethylenes (LLDPE). LLDPE synthesized using regular multisite Ziegler-Natta catalytic systems usually are quite heterogeneous in terms of their chain length and architecture. That is, different catalytic active sites syntheize macromolecules differing in length (molecular weight), composition (comonomer content = short chain branching) and order (short chain branches distribution). These macromolecules differ significantly in their crystallizability and perfectness of crystallites they can form. Well-ordered long linear chains easily crystallize at higher temperatures and form crystallites with higher thickness lamellae (having higher melting temperatures), as compared with shorter poorly ordered chains with high branching content, resulting in double (sometimes multiple) crystallization and melting peaks on DSC thermograms. Prolonged heating of LLDPE above 200C, as David has already mentioned, can result in the polymer thermal destruction.

The graphs look peculiar. There is definitely more than one phase in the starting samples - judged by the double peaks. This may be caused either by a co-polymer that phase separates or that your sample consists of two or more blended phase separated polymers or a blend of a co-polymer and a homopolymer. It may be a blend of LDPE with e.g. LLDPE. Look for instance at: Article The application of DSC in identification of LDPE/LLDPE blend...

Is there any chance that you could run FTIR to identify the polymers of your samples?

Anders, I think it could be a LDPE/LLDPE blend because of the peaks temperatures, may be a 30/70 blend acording to that paper data. I have no idea either if that endotermic peak could be a reaction, I wonder why it only apear at constant tempeture of 350 and only in sample 1.

I'm working in a LDPE/LLDPE calibration curve too in order to confirm my hypotessis about that blend, but unfortunaly I can't access to a FTIR right now.

Thank you for your answer

German, Anders,

If the samples are heated in an inert gas, the decomposition is likely to be endothermic indeed, because you are generating gases (ethylene etc.) out of a condensed phase. The evaporating gases have more entropy, which comes at the expense of heat loss (or heat absorption from the surroundings). DG = DH - TDS. Just as you would expect a process that evaporates water out of a polymer (think polysaccharides) to be endothermic, the same should be true if you "evaporate" ethylene. An exothermic decomposition would happen if oxygen is present (burning) or if the polymer is intrinsically unstable (like nitrocellulose).

Now, why is the decomposition seen in sample 1 but not in sample 2 - that I don't know, but here it may be a kinetic effect - a catalyst (metal contamination?) may be present in one case that causes the decomposition to start earlier. It is hard to ensure that two polymer samples are essentially identical, and instabilities in an extrusion process, for instance, can cause a significant difference.

Thank you, Emanuel. I haven't consider that possibility, actually, I used Nitrogen as purge gas.

Thank you Emanuel Cooper for sorting out the thermodynamics behind decomposition under inert conditions. If I instead of decomposition think depolymerization then it obvious that it will be endothermic since the polymerization reaction of PE is highly exothermic.

Cheers, Anders