Let's solve **Question 4** step by step: --- ### **4) A glass has an annealing point of 516°C (viscosity of 10¹³.⁴ P) and a softening point of 686°C (viscosity of 10⁷.⁶ P).** #### **a) Calculate the activation energy (Q).** Given: - Annealing point, \( T_1 = 516°C = 789K \) - Softening point, \( T_2 = 686°C = 959K \) - Viscosity at \( T_1 \), \( \eta_1 = 10^{13.4} \) P - Viscosity at \( T_2 \), \( \eta_2 = 10^{7.6} \) P The viscosity equation: \[ \eta = \eta_0 e^{Q/(RT)} \] Take logarithms: \[ \ln \eta = \ln \eta_0 + \frac{Q}{R} \cdot \frac{1}{T} \] For two points: \[ \ln \eta_1 - \ln \eta_2 = \frac{Q}{R} \left(\frac{1}{T_1} - \frac{1}{T_2}\right) \] \[ Q = R \cdot \frac{\ln (\eta_1/\eta_2)}{\left(\frac{1}{T_1} - \frac{1}{T_2}\right)} \] Plug in the values: - \( R = 8.314 \) J/mol·K \[ \ln (\eta_1/\eta_2) = \ln (10^{13.4}/10^{7.6}) = \ln (10^{5.8}) = 5.8 \ln 10 = 5.8 \times 2.303 = 13.357 \] \[ \frac{1}{T_1} - \frac{1}{T_2} = \frac{1}{789} - \frac{1}{959} \] \[ = \frac{959 - 789}{789 \times 959} \] \[ = \frac{170}{756051} \approx 0.000225 \] Now plug into Q: \[ Q = 8.314 \times \frac{13.357}{0.000225} \] \[ Q \approx 8.314 \times 59364.44 \] \[ Q \approx 493,606 \ \text{J/mol} \] \[ Q \approx 494 \ \text{kJ/mol} \] --- #### **b) Calculate the working temperatures if the working range is \( 2 \times 10^4 - 10^8 \) P.** Let \( \eta_1 = 2 \times 10^4 \) P, \( \eta_2 = 10^8 \) P. Again use: \[ \ln \eta = \ln \eta_0 + \frac{Q}{R} \cdot \frac{1}{T} \] Set up two equations: \[ \ln \eta_1 = \ln \eta_0 + \frac{Q}{R} \cdot \frac{1}{T_1} \] \[ \ln \eta_2 = \ln \eta_0 + \frac{Q}{R} \cdot \frac{1}{T_2} \] Subtract: \[ \ln (\eta_2/\eta_1) = \frac{Q}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] \[ \ln (10^8/2 \times 10^4) = \ln (5 \times 10^3) = \ln 5 + \ln 10^3 = 1.609 + 6.908 = 8.517 \] \[ 8.517 = \frac{Q}{R} \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] \[ \frac{Q}{R} = \frac{493,606}{8.314} = 59,400 \] So, \[ 8.517 = 59,400 \left( \frac{1}{T_2} - \frac{1}{T_1} \right) \] \[ \frac{1}{T_2} - \frac{1}{T_1} = \frac{8.517}{59,400} = 0.0001434 \] Now, you need one more equation to solve for \( T_1 \) or \( T_2 \). Use the viscosity equation for one known point from part (a): Let’s use \( \eta_2 = 10^8 \) at \( T_2 \): \[ \ln 10^8 = \ln \eta_0 + \frac{Q}{R} \frac{1}{T_2} \] But we don't know \( \eta_0 \). To estimate, use the softening point (\( T_s = 959 K, \eta_s = 10^{7.6} \)): \[ \ln 10^{7.6} = \ln \eta_0 + \frac{Q}{R} \frac{1}{959} \] \[ 7.6 \times 2.303 = \ln \eta_0 + 59,400 \times \frac{1}{959} \] \[ 17.499 = \ln \eta_0 + 61.92 \] \[ \ln \eta_0 = 17.499 - 61.92 = -44.421 \] Now use for \( T_2 \): \[ \ln 10^8 = -44.421 + 59,400 \frac{1}{T_2} \] \[ 18.420 = -44.421 + 59,400 \frac{1}{T_2} \] \[ 18.420 + 44.421 = 59,400 \frac{1}{T_2} \] \[ 62.841 = 59,400 \frac{1}{T_2} \] \[ \frac{1}{T_2} = \frac{62.841}{59,400} = 0.001058 \] \[ T_2 = \frac{1}{0.001058} = 945 \text{ K} \] Now for \( T_1 \): \[ \frac{1}{T_1} = \frac{1}{T_2} - 0.0001434 = 0.001058 - 0.0001434 = 0.0009146 \] \[ T_1 = \frac{1}{0.0009146} = 1093 \text{ K} \] --- ### **Final Answers** **a) Activation energy:** \[ Q \approx \boxed{494 \ \text{kJ/mol}} \] **b) Working temperature range:** \[ \boxed{T_1 = 1093 \ \text{K} \ (820^\circ \text{C}), \quad T_2 = 945 \ \text{K} \ (672^\circ \text{C})} \] --- **Let me know if you'd like to see the solutions for 5 and 6 as well!**