Chapter 11.13, Problem 103AAP

To determine

The temperature of the soda-lime glass.

Answer to Problem 103AAP

The temperature of the soda-lime glass is $736.7°\text{C}$.

Explanation of Solution

Write the equation involving the viscosity to temperature for a viscous flow in the glass.

  $\eta ={\eta }_0{e}^{Q/RT}$                                                                                                            (I)

Here, viscosity of the glass is $\eta$, pre-exponential constant is ${\eta }_0$, molar activation energy for viscous flow is $Q$, universal gas constant is $R$ and absolute temperature is $T$.

Conclusion:

At the viscosity of ${\text{10}}^{\text{14}\text{.3}}\text{P}$, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_1={\eta }_0{e}^{Q/R{T}_1}$                                                                                                          (II)

Here, initial viscosity of the soda-lime glass is ${\eta }_1$ and initial temperature of the soda-lime glass is $T_1$.

At the viscosity of ${\text{10}}^{\text{9}\text{.9}}\text{P}$, write the equation involving the viscosity to temperature for a viscous flow in the glass.

  ${\eta }_2={\eta }_0{e}^{Q/R{T}_2}$                                                                                                         (III)

Here, final viscosity of the soda-lime glass is ${\eta }_2$ and final temperature of the soda-lime glass is $T_2$.

Divide Equation (II) by Equation (III).

  $\begin{array}{c}\frac{{\eta }_1}{{\eta }_2}=\frac{{\eta }_0{e}^{Q/R{T}_1}}{{\eta }_0{e}^{Q/R{T}_2}}\\ =\frac{e^{Q/R{T}_1}}{e^{Q/R{T}_2}}\\ =\exp \left[\frac{Q}{R}\left(\frac{1}{T_1}-\frac{1}{T_2}\right)\right]\end{array}$                                                                                  (IV)

Substitute ${10}^{14.3}\text{P}$ for ${\eta }_1$, ${10}^{9.9}\text{P}$ for ${\eta }_1$, $430\text{kJ}/\text{mol}$ for $Q$, $8.314\text{J}/\text{mol}\cdot \text{K}$ for $R$ and $570°\text{C}$ for $T_1$ in Equation (IV).

 $\begin{array}{l}\frac{{10}^{14.3}\text{P}}{{10}^{9.9}\text{P}}=\exp \left[\frac{430\text{kJ}/\text{mol}}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{570°\text{C}}-\frac{1}{T_2}\right)\right]\\ {10}^{4.4}=\exp \left[\frac{430\text{kJ}/\text{mol}}{8.314\text{J}/\text{mol}\cdot \text{K}}\left(\frac{1}{843\text{K}}-\frac{1}{T_2}\right)\right]\\ {10}^{4.4}=\exp \left(51,720\left(\frac{1}{843\text{K}}-\frac{1}{T_2}\right)\right)\\ \ln {10}^{4.4}=51,720\left(\frac{1}{843\text{K}}-\frac{1}{T_2}\right)\end{array}$

 $\begin{array}{c}{T}_2=\frac{1}{0.0011862-0.0001959}\\ =1009.7\text{K}\\ =736.7°\text{C}\end{array}$

Thus, the temperature of the soda-lime glass is $736.7°\text{C}$.