A student used the apparatus shown to determine the molar heat of combustion of ethanol - HSC - SSCE Chemistry - Question 4 - 2006 - Paper 1

The temperature change of the water is calculated by subtracting the initial temperature from the final temperature:

$\Delta T = 45.5°C - 25.0°C = 20.5°C$

The heat gained by the water can be calculated using the formula:

$q = mc\Delta T$

Where:

• $m = 300 \text{ g}$ (mass of water)
• $c = 4.18 \text{ J g}^{-1} \text{°C}^{-1}$ (specific heat capacity of water)
• $\Delta T = 20.5 \text{°C}$

Thus,

$q = 300 \text{ g} \times 4.18 \text{ J g}^{-1} \text{°C}^{-1} \times 20.5°C = 25,683 \text{ J}$ or $25.683 \text{ kJ}$

To find the moles of ethanol burned, divide the mass of ethanol by its molar mass ($C_2H_5OH$ has a molar mass of approximately 46.07 g/mol):

$\text{Moles of ethanol} = \frac{1.15 \text{ g}}{46.07 \text{ g mol}^{-1}} \approx 0.02495 \text{ mol}$

Finally, the molar heat of combustion can be found by dividing the heat gained by the water by the moles of ethanol burned:

$\text{Molar heat of combustion} = \frac{25.683 \text{ kJ}}{0.02495 \text{ mol}} \approx 1029.8 \text{ kJ mol}^{-1}$

Thus, rounding gives us: 1030 kJ mol⁻¹.