The rate constant is doubled when temperature increases from $27°\mathrm{C}$ to $37°\mathrm{C}$. Activation energy in $\mathrm{kJ}$ is

We are given that:

$$\begin{gathered} \text{ When }{\mathrm{T}}_1=27+273=300\mathrm{K} \\ \text{ Let }{\mathrm{k}}_1=\mathrm{k} \\ \text{ When }{\mathrm{T}}_2=37+273=310\mathrm{K} \\ {\mathrm{k}}_2=2\mathrm{k} \\ \text{ Substituting these values the equation: } \\ \log \left(\frac{{\mathrm{k}}_2}{{\mathrm{k}}_1}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303}\times \left(\frac{{\mathrm{T}}_2-{\mathrm{T}}_1}{{\mathrm{T}}_1{\mathrm{T}}_2}\right) \\ \text{ We will get: } \\ \log \left(\frac{2\mathrm{k}}{\mathrm{k}}\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{310-300}{300\times 310}\right) \\ \log \left(2\right)=\frac{{\mathrm{E}}_{\mathrm{a}}}{2.303\times 8.314}\left(\frac{10}{300\times 310}\right) \\ {\mathrm{E}}_{\mathrm{a}}=53598.6{\mathrm{Jmol}}^{-1} \\ {\mathrm{E}}_{\mathrm{a}}=53.6{\mathrm{kJmol}}^{-1} \end{gathered}$$

Hence, the activation energy of the reaction is $53.6 \mathrm{KJ}$ which is equal to $54 \mathrm{KJ}$.