For the complete combustion of ethanol, $\mathrm{\Delta U}=\mathrm{q}$ the amount of heat produced as measured in bomb calorimeter is $1364.47 KJ mo{l}^{-1 }$at 25⁰C. Assuming ideality, the enthalpy of combustion, ${\mathrm{\Delta H}}_{\mathrm{C}}$  for the reaction will be $[\mathrm{R}=8.314{\mathrm{JK}}^{-1}{\mathrm{mol}}^{-1}]$

${\text{ΔU\hspace{0.17em}=\hspace{0.17em}-1364.47\hspace{0.17em}KJmol}}^{-1}$ (Amount of heat produced)
Enthalpy of combustion is 
${\text{ΔH\hspace{0.17em}=\hspace{0.17em}ΔU\hspace{0.17em}+\hspace{0.17em}Δn}}_{\text{g}}\text{RT}$
According to equation ${\text{Δn}}_{\text{g}}\text{=\hspace{0.17em}2-3\hspace{0.17em}=\hspace{0.17em}-1}$
$\left({\text{C}}_{\text{2}}{\text{H}}_{\text{5}}{\text{OH}}_{\left(l\right)}+{30}_{2_{\left(g\right)}}\to {\text{2\hspace{0.17em}CO}}_{{\text{2}}_{\left(\text{g}\right)}}{\text{+\hspace{0.17em}3H}}_{\text{2}}{\text{O}}_{\left(l\right)}\right)$
$\mathrm{T}={25}^{\circ }\mathrm{C}=298 \mathrm{K}$ thus ;
$\text{ΔH\hspace{0.17em}=\hspace{0.17em}-1364\hspace{0.17em}.\hspace{0.17em}47\hspace{0.17em}+\hspace{0.17em}}\left[\frac{\left(\text{-1}\right)\text{\hspace{0.17em}×\hspace{0.17em}8\hspace{0.17em}.\hspace{0.17em}314\hspace{0.17em}×\hspace{0.17em}298}}{\text{1000}}\right]$
${\text{ΔH\hspace{0.17em}=\hspace{0.17em}-1364\hspace{0.17em}.\hspace{0.17em}47\hspace{0.17em}-\hspace{0.17em}2.477\hspace{0.17em}=\hspace{0.17em}-\hspace{0.17em}1366.947\hspace{0.17em}KJ\hspace{0.17em}mol}}^{\text{-1}}$