Answer
$Q_{in}=20.90\text{ kJ/kg}$
Work Step by Step
We take the water as the system, which is a closed system, for which the energy balance on the system $E_{\text {in }}-E_{\text {out }}=\Delta E_{\text {system }}$ with $W=0$ can be written as
$$
Q_{\text {in }}=\Delta U
$$ or $$
\begin{aligned}
Q_{\text {in }} & =m c \Delta T \\
& =(2 \mathrm{~kg})\left(4.18 \mathrm{~kJ} / \mathrm{kg} \cdot{ }^{\circ} \mathrm{C}\right)\left(2.5^{\circ} \mathrm{C}\right) \\
& =20.90 \mathrm{~kJ}(\text { per gram of fuel })
\end{aligned}
$$ Therefore, heat transfer per kg of the fuel would be $20,900
\text{ kJ/kg}$ fuel. Disregarding the slight energy stored in the gases of the combustion chamber, this value corresponds to the heating value of the fuel.
