the use of solution calorimetry and bomb calorimetry to measure the energy released by chemical reactions, including the use of the specific heat capacity of water and q = mcΔT to calculate the energy released by combustion of fuels and the molar enthalpy of combustion

What this dot point is asking

VCAA wants you to know how a calorimeter measures the heat released by a reaction, to apply q = mcΔT in solution calorimetry (with water's specific heat capacity), to use a calibration factor in bomb calorimetry, and to convert the measured heat to a molar enthalpy value.

The answer

The diagram shows a simple solution calorimeter and the variables in $q = mc\Delta T$.

Two kinds of calorimetry

Solution calorimetry. A reaction (often combustion or neutralisation) takes place in or under a known mass of water. The heat released raises the water's temperature. The heat absorbed by the water is calculated with q = mcΔT.

Bomb calorimetry. A sample is burned in pure oxygen inside a sealed steel "bomb" surrounded by water. Because heat goes into multiple components (the steel bomb, the water jacket, the stirrer, the thermometer), the whole apparatus is calibrated with a known electrical input first. The calibration factor (CF, in J °C^-1) converts the measured ΔT directly into energy released.

The formula

q = mcΔT

• q = heat absorbed (or released), in joules
• m = mass of water (or solution), in grams
• c = specific heat capacity, in J g^-1 K^-1 (water's value is 4.18 J g^-1 K^-1)
• ΔT = temperature change, in °C or K (a degree change is the same on both scales)

For 250 g of water raised by 5.0°C:
q = 250 × 4.18 × 5.0 = 5225 J = 5.23 kJ

The water has absorbed 5.23 kJ. Energy is conserved, so 5.23 kJ has been released by the reaction (assuming no heat loss).

From heat to molar enthalpy

The molar enthalpy of combustion is the energy released per mole of fuel burned, reported as a negative ΔH.

Steps:

1. Calculate q (heat absorbed by water) using q = mcΔT, or use the calibration factor for a bomb.
2. Convert q from J to kJ (divide by 1000).
3. Find n (moles of fuel) from n = m / M.
4. Molar enthalpy of combustion = q / n, applied with a negative sign because combustion is exothermic.

ΔH_c = -q / n (in kJ mol^-1)

Calibration factor for a bomb calorimeter

A bomb calorimeter is calibrated electrically. A heater of known voltage V and current I runs for time t, supplying:

E = V × I × t (in joules)

This raises the calorimeter by ΔT. The calibration factor is:

CF = E / ΔT (in J °C^-1)

In a later combustion run, ΔT is measured and the energy released is:

q = CF × ΔT

Then divide by moles of fuel to get ΔH per mole, with a negative sign.

Heat losses and the experimental error

Solution calorimetry usually underestimates the true ΔH because:

• Heat escapes to the surrounding air.
• Heat is absorbed by the calorimeter walls, the thermometer, the stirrer (not just the water).
• Combustion may be incomplete (forming CO and soot, which release less energy than CO2).
• Evaporation of water near the flame removes additional energy.

Markers expect you to state which way the experimental error pushes the result and why. For combustion experiments in open air, the experimental ΔH is always smaller in magnitude (less negative) than the literature value.

Bomb calorimeters are far more accurate because they are insulated and calibrated.

Examples in context

Example 1. Loy Yang brown-coal heat content by bomb calorimetry. AGL Loy Yang power-station chemists determine the gross calorific value of Latrobe Valley brown coal in a bomb calorimeter. A pellet of dried coal weighing $1.000 \, \text{g}$ is burned in oxygen at $30 \, \text{atm}$. The bomb has calibration factor $\text{CF} = 9.85 \, \text{kJ/}^{\circ}\text{C}$ from an electrical run. The water bath rises by $2.51^{\circ}\text{C}$. Heat released $q = 9.85 \times 2.51 = 24.7 \, \text{kJ}$. Per gram, the calorific value is $24.7 \, \text{MJ/kg}$. Compared with black coal at $30 \, \text{MJ/kg}$, brown coal's lower value reflects its $\sim 60\%$ moisture and oxygen content. The result feeds into the station's fuel-feed-rate calculations: at $2200 \, \text{MW}$, the plant burns roughly $320$ tonnes per hour.

Example 2. Bushfire calorimetry at CSIRO Bushfire Behaviour Lab. Researchers at CSIRO's bushfire wind-tunnel facility in Canberra use cone calorimetry to measure heat-release rates of native eucalypt litter. A $50 \, \text{g}$ sample of dried $\text{Eucalyptus regnans}$ leaves is exposed to a $50 \, \text{kW/m}^2$ radiant flux. Combustion gases warm a water-cooled heat-flux meter; integrating the signal gives total heat $q = 1180 \, \text{kJ}$. Heat of combustion $= 1180 / 0.050 = 23.6 \, \text{MJ/kg}$, comparable to black coal. This high value, combined with the leaves' low ignition temperature ($230^{\circ}\text{C}$) and volatile oil content, explains why eucalypt forests burn so intensely in Victoria during the Forest Fire Danger Index 100 conditions of Black Saturday 2009.

Try this

Q1. State two advantages of a bomb calorimeter over a simple solution calorimeter for measuring the enthalpy of combustion. [2 marks]

• Cue. Constant volume; sealed (no heat loss to environment); high-pressure oxygen ensures complete combustion.

Q2. A $1.500 \, \text{g}$ sample of methanol is burned in a bomb calorimeter with $\text{CF} = 3.45 \, \text{kJ/}^{\circ}\text{C}$. The temperature rises by $9.92^{\circ}\text{C}$. (a) Calculate the heat released. (b) Determine $\Delta H_c$ of methanol in $\text{kJ/mol}$. [4 marks]

• Cue. (a) $q = 3.45 \times 9.92 = 34.2 \, \text{kJ}$. (b) $n = 1.500 / 32.0 = 0.0469$ mol; $\Delta H_c = -34.2 / 0.0469 = -729 \, \text{kJ/mol}$ (literature $-726$).

Q3. A student uses solution calorimetry to measure neutralisation enthalpy. They mix $50.0 \, \text{mL}$ of $1.00 \, \text{mol/L}$ HCl with $50.0 \, \text{mL}$ of $1.00 \, \text{mol/L}$ NaOH and observe a temperature rise from $20.0$ to $26.8^{\circ}\text{C}$. (a) Calculate $q$. (b) Calculate $\Delta H$ per mole of water formed. (c) State two assumptions made. [2+2+2 marks]

• Cue. (a) $q = 100 \times 4.18 \times 6.8 = 2842 \, \text{J}$. (b) $n = 0.0500$ mol; $\Delta H = -2842 / 0.0500 = -56.8 \, \text{kJ/mol}$. (c) Density and specific heat of mixture equals water; no heat loss to surroundings.