Worked example 1
Calculate the total energy required to heat 0.50 kg of water from 20 °C to 100 °C and then completely boil it. \ (Given: Specific heat capacity of water = 4200 J kg⁻¹ K⁻¹, Specific latent heat of vaporisation of water = 2.26 × 10⁶ J kg⁻¹).
Show solution outline
- Energy to heat the water (Q₁): \ We use Q = mcΔT. \ Mass (m) = 0.50 kg \ Specific heat capacity (c) = 4200 J kg⁻¹ K⁻¹ \ Temperature change (ΔT) = 100 °C - 20 °C = 80 K (Note: A change of 80 °C is the same as 80 K). \ Q₁ = 0.50 kg × 4200 J kg⁻¹ K⁻¹ × 80 K = 168,000 J \ \ 2. Energy to boil the water (Q₂): \ We use Q = mL. \ Mass (m) = 0.50 kg \ Specific latent heat of vaporisation (Lᵥ) = 2.26 × 10⁶ J kg⁻¹ \ Q₂ = 0.50 kg × 2.26 × 10⁶ J kg⁻¹ = 1,130,000 J \ \ 3. Total energy (Q_total): \ Q_total = Q₁ + Q₂ \ Q_total = 168,000 J + 1,130,000 J = 1,298,000 J \ Therefore, the total energy required is approximately 1.30 × 10⁶ J (to 3 significant figures).