The heat of combustion, \(Q\) is given by:
\(Q=m c \Delta \theta\)
Where \(m\) - mass of water in calorimeter
\(c-\) specific heat capacity of water
\(\Delta \theta\) - temp. rise of water
So,
\(\begin{aligned} Q &=100 \times 4.2 \times(312.3-298) \\ &=6.006 kJ \end{aligned}\)
\(=6.006 kJ\)
No of mole of ethanol
$$
\begin{aligned}
&=\frac{0.92 g }{[(12 \times 2)+(1 \times 5)+16+1]}=\frac{0.92}{46} \\
&=0.02 mol
\end{aligned}
$$
If \(0.02 mol of _{2} H _{5} OH \rightarrow 6.006 kJ\)
\(1 mol\) of \(C _{2} H _{5} OH \rightarrow x\)
\(x=\frac{6.006 kJ \times 1}{0.02}=300.3 kJ\)
\(\therefore\) The heat of combustion is \(-300.3 kJmol ^{-1}\), the reaction being exothermic