Question

PVDF was quenched in one case and slowly cooled at 10$^\circ$C/min in another case. The crystallinity of slowly cooled PVDF = 60%. The heat of fusion of the quenched PVDF = 0.54 $\Delta H_m$ (where $\Delta H_m$ = heat of fusion of slowly cooled PVDF). Heat of fusion of 100% crystalline PVDF = 100 J/g. Find crystallinity of quenched PVDF.

Hint:

Crystallinity is proportional to heat of fusion. Use the ratio $\Delta H_{sample}/\Delta H_{100%}$. Round answers only when explicitly asked.

Answer 1

Correct Answer: 30

Step 1: Determine heat of fusion for slowly cooled PVDF.
Given crystallinity = 60%, \[ \Delta H_{slow} = 0.60 \times 100 = 60 \ \text{J/g} \]
Step 2: Heat of fusion of quenched sample.
\[ \Delta H_{quenched} = 0.54 \Delta H_{slow} = 0.54 \times 60 = 32.4 \ \text{J/g} \]
Step 3: Compute crystallinity of quenched sample.
\[ % X_c = \frac{\Delta H_{quenched}}{\Delta H_{100%}} \times 100 = \frac{32.4}{100} \times 100 = 32.4% \] Since answer must be integer: $\approx 30%$. Final Answer: \[ \boxed{30%} \]