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

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.

Worked example

A student burns 2.30 g of ethanol (M = 46.0 g mol^-1) in a spirit burner under a copper can holding 200 mL of water. The water temperature rises from 22.0°C to 65.0°C.

Step 1. Mass of water: 200 mL × 1.00 g mL^-1 = 200 g.

Step 2. ΔT = 65.0 - 22.0 = 43.0°C.

Step 3. q = mcΔT = 200 × 4.18 × 43.0 = 35,948 J = 35.9 kJ (absorbed by water, released by ethanol).

Step 4. n(C2H5OH) = 2.30 / 46.0 = 0.0500 mol.

Step 5. ΔH_c (experimental) = -35.9 / 0.0500 = -718 kJ mol^-1.

The literature value is about -1367 kJ mol^-1. The experimental value is roughly half. The deficit reflects heat losses to the air, heat absorbed by the copper can, and incomplete combustion (a yellow flame produces soot). For a SAC discussion answer, name all three.

Common traps

Forgetting the negative sign on ΔH. Combustion is exothermic. ΔH must be negative even though the magnitude is reported as a positive q.

Using mass of fuel for m in q = mcΔT. m is the mass of water (the thing being heated), not the mass of fuel.

Mixing units. Specific heat capacity is in J g^-1 K^-1, so m must be in grams. If you use kg, divide c by 1000 first, or you will be out by 1000.

Using mL as if it were grams without checking density. For pure water near room temperature this is fine (1 mL ≈ 1 g). For salt solutions or other solvents, density matters.

Confusing q with ΔH. q is the heat for that specific experiment (a number of joules). ΔH is the heat per mole of reaction. Always divide by n.

Calibrating a bomb calorimeter with q = mcΔT for water only. The bomb has many components. Use CF = E / ΔT from an electrical calibration instead.

In one sentence

Calorimetry measures the heat released by a reaction by tracking the temperature change of water (using q = mcΔT in solution calorimetry, or a calibration factor in bomb calorimetry) and dividing by the moles of fuel to give the molar enthalpy of combustion as a negative ΔH.