Question

In the first heating cycle of a differential scanning calorimetry (DSC) experiment, an as-spun Nylon 6 fibre showed an enthalpy change of 40 J/g during cold crystallization and an enthalpy change of 150 J/g during melting. If the heat of fusion of 100% crystalline Nylon 6 is 240 J/g, the percentage degree of crystallinity of as-spun Nylon 6 fibre (rounded off to 1 decimal place) is _________

Hint:

To calculate the degree of crystallinity, add the enthalpy change during cold crystallization and melting, then divide by the heat of fusion of 100% crystalline polymer, and multiply by 100.

Answer 1

Correct Answer: 45

Given: Enthalpy change during cold crystallization, $\Delta H_a = \SI{40}{\joule\per\gram}$ Enthalpy change during melting, $\Delta H_m = \SI{150}{\joule\per\gram}$ Heat of fusion for 100% crystalline Nylon 6, $\Delta H_x = \SI{240}{\joule\per\gram}$ Step 1: Calculate Net Enthalpy of Melting ($\Delta H_{net}$)
The net enthalpy of melting accounts for the heat absorbed during melting minus the heat released during cold crystallization: \[ \Delta H_{net} = \Delta H_m - \Delta H_a \] \[ \Delta H_{net} = \SI{150}{\joule\per\gram} - \SI{40}{\joule\per\gram} = \SI{110}{\joule\per\gram} \] Step 2: Calculate Degree of Crystallinity ($X$)
The degree of crystallinity is the ratio of the net enthalpy of melting to the heat of fusion of 100% crystalline Nylon 6: \[ X = \left( \frac{\Delta H_{net}}{\Delta H_x} \right) \times 100% \] \[ X = \left( \frac{\SI{110}{\joule\per\gram}}{\SI{240}{\joule\per\gram}} \right) \times 100% = \boxed{45.8%} \]