Answer :
The time required to convert 100.0 grams of solid sodium at 20.0°C to sodium vapor at 1000.0°C, at a heating rate of 8.0 kJ/min, is approximately 521.5 minutes.
1. Energy to raise the temperature of solid sodium from 20.0°C to 97.8°C:
[tex]\[ Q_1 = m \times C_s \times \Delta T \][/tex]
[tex]\[ Q_1 = 100.0 \, \text{g} \times 28.2 \, \text{J/mol/K} \times (97.8 - 20.0)^\circ \text{C} \][/tex]
[tex]\[ Q_1 = 100.0 \, \text{g} \times 28.2 \, \text{J/mol/K} \times 77.8^\circ \text{C} \][/tex]
[tex]\[ Q_1 = 221,316 \, \text{J} \][/tex]
2. Energy for melting solid sodium:
[tex]\[ Q_2 = n \times \Delta H_{\text{fusion}} \][/tex]
[tex]\[ Q_2 = \frac{100.0 \, \text{g}}{22.99 \, \text{g/mol}} \times 2.60 \, \text{kJ/mol} \][/tex]
[tex]\[ Q_2 = 11,318 \, \text{J} \][/tex]
3. Energy to raise the temperature of liquid sodium from 97.8°C to 883°C:
[tex]\[ Q_3 = m \times C_l \times \Delta T \][/tex]
[tex]\[ Q_3 = 100.0 \, \text{g} \times 30.8 \, \text{J/mol/K} \times (883 - 97.8)^\circ \text{C} \][/tex]
[tex]\[ Q_3 = 100.0 \, \text{g} \times 30.8 \, \text{J/mol/K} \times 785.2^\circ \text{C} \][/tex]
[tex]\[ Q_3 = 242,446 \, \text{J} \][/tex]
4. Energy for vaporizing liquid sodium:
[tex]\[ Q_4 = n \times \Delta H_{\text{vaporization}} \][/tex]
[tex]\[ Q_4 = \frac{100.0 \, \text{g}}{22.99 \, \text{g/mol}} \times 97.4 \, \text{kJ/mol} \][/tex]
[tex]\[ Q_4 = 42,396 \, \text{J} \][/tex]
5. Energy to raise the temperature of sodium vapor from 883°C to 1000.0°C:
[tex]\[ Q_5 = m \times C_g \times \Delta T \][/tex]
[tex]\[ Q_5 = 100.0 \, \text{g} \times 20.8 \, \text{J/mol/K} \times (1000.0 - 883)^\circ \text{C} \][/tex]
[tex]\[ Q_5 = 100.0 \, \text{g} \times 20.8 \, \text{J/mol/K} \times 117^\circ \text{C} \][/tex]
[tex]\[ Q_5 = 244,416 \, \text{J} \][/tex]
6. Total energy:
[tex]\[ Q_{\text{total}} = Q_1 + Q_2 + Q_3 + Q_4 + Q_5 \][/tex]
[tex]\[ Q_{\text{total}} = 221,316 + 11,318 + 242,446 + 42,396 + 244,416 \][/tex]
[tex]\[ Q_{\text{total}} = 761,892 \, \text{J} \][/tex]
7. Time required:
[tex]\[ \text{Time} = \frac{Q_{\text{total}}}{\text{Heating rate}} \][/tex]
[tex]\[ \text{Time} = \frac{761,892 \, \text{J}}{8.0 \, \text{kJ/min}} \][/tex]
Now, we need to convert the heating rate to J/min:
[tex]\[ 1 \, \text{kJ} = 1000 \, \text{J} \][/tex]
[tex]\[ 8.0 \, \text{kJ/min} = 8.0 \times 1000 \, \text{J/min} = 8000 \, \text{J/min} \][/tex]
[tex]\[ \text{Time} = \frac{761,892 \, \text{J}}{8000 \, \text{J/min}} \][/tex]
[tex]\[ \text{Time} \approx 95.24 \, \text{min} \][/tex]
Therefore, it would take approximately 95.24 minutes to convert 100.0 grams of solid sodium at 20.0°C to sodium vapor at 1000.0°C at a heating rate of 8.0 kJ/min and a pressure of 1 atm.
