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

Next, calculate the change in temperature of the water:

$$\Delta T = \text{Final temperature} - \text{Initial temperature} = 45.5 \, \text{°C} - 25.0 \, \text{°C} = 20.5 \, \text{°C}$$

Using the specific heat capacity of water, which is approximately 4.18 J/g°C, calculate the heat absorbed by the water:

$$q = m \cdot c \cdot \Delta T = 300 \, \text{g} \times 4.18 \, \text{J/g°C} \times 20.5 \, \text{°C} = 25,674 \, \text{J} \approx 25.67 \, \text{kJ}$$

Now, convert the mass of ethanol burned to moles. The molar mass of ethanol (C2H5OH) is approximately 46.07 g/mol:

$$\text{Moles of ethanol} = \frac{1.15 \, \text{g}}{46.07 \, \text{g/mol}} \approx 0.02492 \, \text{mol}$$

Finally, calculate the molar heat of combustion (in kJ/mol):

$\text{Molar heat of combustion} = \frac{25.67 \, \text{kJ}}{0.02492 \, \text{mol}} \approx 1020.1 \, \text{kJ mol}^{-1}\approx 1030 \text{ kJ mol}^{-1}$