Question

Arrange the following transition temperatures given by Differential Scanning Calorimetry (DSC) in ascending order.
A. Melting temperature
B. Crystallization temperature
C. Degradation temperature
D. Glass transition temperature

• A. D, B, A, C
• B. A, C, B, D
• C. B, A, D, C
• D. C, B, D, A"

Hint:

Think of the states of a semi-crystalline polymer as you heat it: 1. Glassy Solid (Amorphous part is frozen). 2. Heat past Tg \(\rightarrow\) Rubbery Solid (Amorphous part is mobile). 3. (Sometimes) Heat further \(\rightarrow\) Crystallization (Tc) (Mobile chains organize). 4. Heat past Tm \(\rightarrow\) Liquid Melt. 5. Heat much further \(\rightarrow\) Degradation (Td) (Burning/Decomposition). So, Tg < (Tc) < Tm < Td.

Answer 1

The Correct Option is A

Step 1: Identify thermal transitions in a semi-crystalline polymer as temperature rises.
- D. Glass transition temperature (Tg): The temperature at which amorphous polymer regions shift from a rigid, glassy state to a flexible, rubbery state. This is the initial significant transition during heating.
- B. Crystallization temperature (Tc): Above Tg, polymer chains gain mobility and can form ordered, crystalline structures. This crystallization process is observed as an exothermic peak on a DSC scan during cooling (heat released) or, for an amorphous sample, after Tg during heating. It occurs between Tg and melting.
- A. Melting temperature (Tm): The temperature at which crystalline regions melt, and the material becomes a viscous liquid. Tm is always higher than Tg due to the greater energy requirement.
- C. Degradation temperature (Td): The temperature at which the polymer's chemical bonds break, leading to decomposition. This chemical change happens at a significantly higher temperature than melting.

Step 2: Order these temperatures in ascending order. The sequence is: Glass Transition \(\rightarrow\) Crystallization \(\rightarrow\) Melting \(\rightarrow\) Degradation.
This corresponds to the order D, B, A, C.