What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?

$\left(\mathrm{R}=8.314{\mathrm{Jmol}}^{-1}{\mathrm{K}}^{-1}\right)$

$$\begin{gathered} \log \frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\mathrm{R}}\left(\frac{1}{{\mathrm{T}}_1}-\frac{1}{{\mathrm{T}}_2}\right) \\ {\mathrm{k}}_2=2{\mathrm{k}}_1, \\ {\mathrm{T}}_1=20+273=293\mathrm{K}\mathrm{or}{\mathrm{T}}_2=35+273=308\mathrm{K} \\ \mathrm{R}=8.314\mathrm{J}{\mathrm{mol}}^{-1}{\mathrm{K}}^{-1} \\ \log 2=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{1}{293}-\frac{1}{308}\right) \\ 0.3010=\frac{{\mathrm{E}}_{\mathrm{a}}}{19.147}\times \frac{15}{293\times 308} \\ {\mathrm{E}}_{\mathrm{a}}=34673{\mathrm{Jmol}}^{-1}\mathrm{or}34.7{\mathrm{kJmol}}^{-1} \end{gathered}$$