Calorimetry (OCR A Level Chemistry A): Revision Note

Specific heat capacity calculations

In 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 °C.

The specific heat capacity of water is 4.18 J g^-1 K⁻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 = 120 000 J g^-1 (to 3 s.f.) or 120 kJ g^-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:

ΔH = $\frac{q}{n}\mathrm{or}\frac{m\times c\times \mathrm{\Delta }T}{n}$

When there is a rise in temperature, the value for ΔH becomes negative suggesting that the reaction is exothermic

• This means that your value should be negative for an exothermic reaction, e.g. combustion

When the temperature falls, the value for ΔH becomes positive suggesting that the reaction is endothermic

• This means that your value should be positive for an endothermic reaction