Dear Kaushal Kumar Sarswat ,

These links might be useful for you to explore:

https://mcl.mse.utah.edu/heating-rate-in-dsc/

https://www.tainstruments.com/pdf/literature/TA039.pdf

In general, a low heating rate allows for the material to undergo that transition at a very slow pace, giving that data more "accuracy". In a polymeric material for example, or in an alloy, if the melting points of the components are different, these two might be observed nicely and separated, if you "give" the sample enough time to go through that transition (meaning you heat it slowly). If you increase the temperature fast, the structure of the material might not have enough time to show you every little effect that is happening.

Hope the links help!

The simplest answer is that the faster you heat the less time a sample spends at each temperature. Therefore, at faster heating rate it takes higher temperature to reach the same conversion, e.g., 50%. This can be demonstrated strictly by looking at the integrals of the basic rate eq:

da/dt=k(T)f(a) for different heating rates.

This explanation holds for any kinetic process, i.e., glass transition, crystallization, polymerization, decomposition, etc. Note, this shift is oftentimes explained (incorrectly) as some sort of "thermal lag". While real, the latter, however, must be eliminated via a proper calibration. In other words, one must not measure and report data obtained with “thermal lag”, i.e., without proper calibration. DSC must be calibrated at each heating rate by using melting point standards, typically pure metals. For your T-range Indium would work. If your DSC calibrated properly, the onset T for Indium melting should be independent of the heating rate. But calibration has its limits. For instance, if you use 10mg Indium for calibration it will most likely hold for 10mg sample of your glass, but will have “thermal lag” if you use, say, 50mg sample. However, since your sample crystallizes, it will melt on the next heating so you would be able to see if the onset T of its melting shifts with the heating rate. If the calibration holds, it will not.

Answers are given already. Organic materials are not highly thermally conducting. Therefore, the dynamic temperature shown by instrument is not the same as the temperature of the sample. If heating is faster the temperature shown by instrument goes much higher as compared to the sample. The thermal conductivity of organic material will not change much with temperature. Higher the heating rate the peak/base line shifting will shift to higher temperature as the temperature lag increases. If the heating rate is kept very low, say

In my opinion, Prof. Vyazovkin's explanation, at least on a qualitative level, is the clearest and most comprehensive. As for the extension to decomposition processes, I would like to add something from my experience. In the case of a chemical reaction, the shift of the entire temperature range to higher temperatures due to shortening the residence time of a substance at a certain temperature is characteristic of an autocatalytic reaction, whereas a reaction without self-acceleration starts at the same temperature regardless of the heating rate, but the temperature range is of course stretched out.