
A polystyrene cup can act as a calorimeter to find enthalpy changes in a chemical reaction

Equation for calculating energy transferred in a calorimeter
Specific heat capacity calculationsIn a calorimetry experiment 2.50 g of methane is burnt in excess oxygen. 30% of the energy released during the combustion is absorbed by 500 g of water, the temperature of which rises from 25 °C to 68 °CThe specific heat capacity of water is 4.18 J g-1°C−1 What is the total energy released per gram of methane burnt?
Answer
Step 1: q = m x c x ΔT
m (of water) = 500 g
c (of water) = 4.18 J g-1 °C-1
ΔT (of water) = 68 oC - 25 oC
= 43 oC
Step 2: q = 500 x 4.18 x 43
= 89 870 J
Step 3: This is only 30% of the total energy released by methane
Total energy x 0.3 = 89 870 J
Total energy = 299 567 J
Step 4: This is released by 2.50 g of methane
Energy released by 1.00 g of methane = 299 567 ÷ 2.50
= 119 827 J = 120 000 J
= 120 kJ g-1
When new bonds are formed the amount of energy released is equal to the amount of energy absorbed when the same bonds are broken.For example:O2 (g) → 2O (g) E (O=O) = +498 kJ mol-12O (g) → O2 (g) E (O=O) = -498 kJ mol-1 Aqueous solutions of acid, alkalis and salts are assumed to be largely water so you can just use the m and c values of water when calculating the energy transferred. To calculate any changes in enthalpy per mole of a reactant or product the following relationship can be used:
When there is a rise in temperature, the value for ΔH becomes negative suggesting that the reaction is exothermic and when the temperature falls, the value for ΔH becomes positive suggesting that the reaction is endothermic.
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