1. Heat absorbed by the calorimeter

In a bomb calorimeter, the heat released by combustion at constant volume is absorbed by the calorimeter:

$q_{\text{cal}} = C_{\text{cal}}\Delta T$

Given:

$C_{\text{cal}} = 5\ \text{kJ K}^{-1}, \qquad \Delta T = 5\ \text{K}$

So,

$q_{\text{cal}} = 5 \times 5 = 25\ \text{kJ}$

Thus, the combustion of the given sample of octane releases 25 kJ of heat.

2. Moles of }n\text{-octane burned}

Octane is $\mathrm{C_8H_{18}}$.

Molar mass:

$M = 8(12) + 18(1) = 96 + 18 = 114\ \text{g mol}^{-1}$

Given mass:

$m = 1.14\ \text{g}$

Number of moles:

$n = \frac{1.14}{114} = 0.01\ \text{mol}$

3. Heat of combustion per mole

If $0.01$ mol releases $25$ kJ, then $1$ mol releases:

$\Delta U_{\text{comb}} = \frac{25}{0.01} = 2500\ \text{kJ mol}^{-1}$

Since the question asks for the magnitude, we report:

$\boxed{2500\ \text{kJ mol}^{-1}}$

(Actual heat of combustion would be negative, but magnitude is positive.)

4. Comparison with stored answer

Stored correct answer = 2500

Our derived answer = 2500

So they agree.