EDEXCEL IGCSE CHEMISTRY: DOUBLE SCIENCE 复习笔记:3.1.3 Energetics Calculations

EDEXCEL IGCSE CHEMISTRY: DOUBLE SCIENCE 复习笔记:3.1.3 Energetics Calculations

Calculate Heat Energy Change


  • In order the calculate heat energy changes you need to know the mass of the substance being heated, the temperature change and the specific heat capacity of the substance
  • The  specific heat capacity (c) is the energy needed to raise the temperature of 1 g of a substance by 1 K
  • The specific heat capacity of water is 4.18 J g-1 K-1
  • The energy transferred as heat can be calculated by:



Equation for calculating energy transferred in a calorimeter


  • The temperature change in Kelvin is the same as the temperature change in degrees Celsius



Worked Example

Excess iron powder was added to 100.0 cmof 0.200 mol dm-3  copper(II)sulfate solution in a calorimeter. The reaction equation was as followsFe (s) + CuSO(aq)     FeSO(aq) + Cu (s)The maximum temperature rise was 7.5 oC. Determine the heat energy change of the reaction, in kJ




The solution is assumed to have the same density as water, so 100.0 cm has a mass of 100 g


q = m x c x ΔT


q = 100 g x 4.18 J g-1 K-1 x 7.5 K = – 3135 J = -3.13 kJ



Worked Example

1.023 g of propan-1-ol (M = 60.11 g mol-1) was burned in a spirit burner and used to heat 200 g of water in a copper calorimeter. The temperature of the water rose by 30 oC.Calculate the heat energy change for the combustion of propan-1-ol using this data.




Calculate q


q = m x c x ΔT


q = 200 g x 4.18 J g-1 K-1 x 30 K = – 25 080 J = -25 kJ



Calculate Molar Enthalpy Change


  • Molar enthalpy change is the heat energy change per mole of substance
  • The symbol is ΔH and it has the unit kJ per mole
  • If is found by first determining the heat energy change for the reaction, q, and then dividing by the number of moles, n, of the substance


molar enthalpy change  = heat change for the reaction ÷ number of moles


ΔH = q ÷ n 


Worked Example

The energy from 0.01 mol of propan-1-ol was used to heat up 250 g of water. The temperature of the water rose from 298 K to 310 K (the specific heat capacity of water is 4.18 J g-1 K-1.Calculate the molar enthalpy of combustion.




Step 1: q = m x c x ΔT


(of water) = 250 g


(of water) = 4.18 J g-1 K-1


Δ(of water) = 310 – 298 K


= 12 K


Step 2: q = 250 x 4.18 x 12


= 12 540 J


Step 3:  This is the energy released by 0.01 mol of propan-1-ol


Total energy    ΔH = q ÷ n = 12 540 J ÷ 0.01 mol = 1 254 000 J mol-1


Total energy = – 1254 kJ mol-1