1.30 g of glucose (M = 180 g mol^{-1}) underwent complete combustion - VCE - SSCE Chemistry - Question 13 - 2012 - Paper 1

The energy released during the combustion of glucose can be calculated using the standard enthalpy change of combustion, which is approximately -2800 kJ/mol. Therefore, the total energy released (Q) is:

$Q = n \times \Delta H_c$

Substituting the values: Q = 0.00722 \, \text{mol} \times (-2800 \, \text{kJ mol^{-1}}) = -20.18 \, \text{kJ}

This is the energy used to heat the water.

Using the formula for heat transfer:

$Q = m c \Delta T$

where

• $Q$ is the heat energy (20.18 kJ),
• $m$ is the mass of water,
• $c$ is the specific heat capacity of water (4.18 kJ/(kg·°C)), and
• $\Delta T$ is the temperature change (24.3 °C).

We can rearrange this to find the mass of the water: $m = \frac{Q}{c \Delta T}$

Substituting the values (converting kJ to J): $m = \frac{20180 \, \text{J}}{4180 \, \text{J/(kg·°C)} \times 24.3 \, \text{°C}} = 2.00 \times 10^{2} \, \text{g}$