Verify that an increase in 8º C will saturate a common emitter
You know that \$\text I_c\$ = \$\text I_s e^{V_{BE}/Vt}\$

To get the transistor to saturate you have to double the collector current, so if you consider \$\frac{\text I_{c1}}{\text I_{c2}} = 2\$, then 

\$\Delta V_{BE}\$ = ln(2)Vt = 18mV, assuming Vt = 26mV (room temperature)

So a base voltage delta of 18mV will cause the collector current to double. If we assume the base voltage divider is 'stiff', then a \$\Delta V_{BE}\$ from 0.600 to 0.582 will cause the transistor to saturate.