Question

Question was provided via image upload.

Explanation:

Step1: Calculate water mass

Since density of water is $1\ \text{g/mL}$, mass $m = 500\ \text{mL} \times 1\ \text{g/mL} = 500\ \text{g}$

Step2: Find temperature change

$\Delta T = T_{\text{final}} - T_{\text{initial}} = 41.0^\circ\text{C} - 22.1^\circ\text{C} = 18.9^\circ\text{C}$

Step3: Calculate heat absorbed by water

Use $q = mc\Delta T$, $c=4.184\ \text{J/g}^\circ\text{C}$:
$q = 500\ \text{g} \times 4.184\ \text{J/g}^\circ\text{C} \times 18.9^\circ\text{C} = 39550.8\ \text{J} = 39.5508\ \text{kJ}$

Step4: Moles of pentane burned

Molar mass of $\text{C}_5\text{H}_{12} = (5\times12.01)+(12\times1.008) = 72.15\ \text{g/mol}$
$n = \frac{0.89\ \text{g}}{72.15\ \text{g/mol}} \approx 0.01234\ \text{mol}$

Step5: Calculate molar enthalpy of combustion

Heat released by pentane = $-q$ (combustion is exothermic):
$\Delta H = -\frac{39.5508\ \text{kJ}}{0.01234\ \text{mol}} \approx -3200\ \text{kJ/mol}$

Answer:

$\boldsymbol{-3200}$ kJ/mol (rounded to 2 significant figures)