Reaction is always exothermic since it involves a combustion reaction.
Used to measure heat content of foods and fuels.

Calibration Calculations

Electrical energy is supplied for known time and the temperature change is recorded.
E = V x I x t
E = energy (J, joules)
V = voltage (V, volts)
I = current (A, amps)
t = time (s, seconds)
Calibration factor, C.F. = E ÷ ΔT
ΔT = T_{f} - T_{i} where T_{f} is the final temperature reached during calibration
and T_{i} is the initial temperature.

ΔH Calculations

Energy released = C.F. x ΔT(reaction)
ΔT(reaction) = final temperature of reaction mixture - initial temperature reaction mixture
ΔH = Energy released ÷ moles of substance
ΔH is negative since energy is released in the combustion reaction.

Solution Calorimeter

Reaction may be exothermic or endothermic.
Used to measure heat of solution, heat of neutralisation or other heats of reaction in solution.

Calibration Calculations

Electrical energy is supplied for known time and the temperature change is recorded.
E = V x I x t E = energy (J, joules) V = voltage (V, volts)
I = current (A, amps)
t = time (s, seconds)
Calibration factor, C.F. = E ÷ ΔT
ΔT = T_{f} - T_{i} where T_{f} is the final temperature reached during calibration
and T_{i} is the initial temperature.

ΔH Calculations

Energy released = C.F. x ΔT(reaction)
ΔT(reaction) = highest temperature of reaction mixture - lowest temperature reaction mixture ΔH = Energy released ÷ moles of substance
ΔH is positive for an endothermic reaction (heat absorbed, temperature decreases)
ΔH is negative for an exothermic reaction (heat released, temperature increases)

Example

In an experiment to determine the energy content of naphthlane, C_{10}H_{8}, a bomb calorimeter was used.

Initially a constant current of 1.80 A was passed through the electric heater for 75.00 seconds. The potential difference was 4.95 V. The temperature increased from 18.25^{o}C to 18.32^{o}C.

A 1.19 g sample of naphthalene was then burnt in the bomb calorimeter and the temperature increased from 18.32^{o}C to 23.74^{o}C.

Calculate the heat of combustion for naphthalene.

Calibration Factor (C.F.) Calculations

E = energy supplied by the electric heater in Joules, J
V = potential difference (voltage) in Volts, V = 4.95 V
I = current in Amps (A) = 1.80 A
t = time in seconds = 75.00 sec

Energy released = C.F. x (T_{f reaction} - T_{i reaction})
C.F. = 9546.43 J^{o}C^{-1} T_{f reaction} = 23.74^{o}C
T_{i reaction} = 18.32^{o}C

Energy released = 9546.43 x (23.74 - 18.32) = 9546.43 x 5.42
= 51741.65 J (per 1.19 g naphthalene)

Heat of Combustion of Naphthalene Calculations

moles naphthalene = mass ÷ molar mass
mass = 1.19 g
molar mass = (10 x 12.01) + (8 x 1.008) = 120.1 + 8.064 = 128.164 g mol^{-1} moles (n) = 1.19 ÷ 128.164 = 0.00928 mol (9.28 x 10^{-3} mol)

Energy released per mole = Energy released ÷ moles
Energy released = 51741.65 J
n = 0.00928 mol
Energy released per mole = 51741.65 ÷ 0.00928
= 5575608.8 J mol^{-1} = 5575.6 kJ mol^{-1} = 5.58 x 10^{3} kJ mol^{-1}

ΔH = - 5.58 x 10^{3} kJ mol^{-1}

ΔH is negative since the reaction is exothermic, heat is released.