Latent heat and multi-stage thermal calculations

What you will learn

ΔQ = mL; fusion/vaporisation distinctions; combined heating and phase changes.

Specification mapping: C5.2

Prerequisites: C5.1, C4.4.2

Physical principles

The latent-energy equation is ΔQ = mL, where m is the mass changing phase and L is specific latent heat in J kg⁻¹. Rearrange to L = ΔQ/m or m = ΔQ/L. The temperature does not appear because the energy changes the state at the transition condition, rather than producing a temperature rise.

Melting and vaporisation absorb energy; freezing and condensation release the corresponding energy magnitudes under the reverse conditions. Use a sign convention if tracking the sample’s energy change, or report a positive released-energy magnitude with its direction. Do not insert specific heat capacity into a latent calculation.

A multi-stage problem may require warming a solid, melting, warming a liquid, boiling and warming vapour. Compute each applicable mcΔT or mL term separately and add. Material properties and transition temperatures must be supplied or sourced for the specified conditions; do not treat illustrative water data as universal constants. If input power is known and losses negligible, total time is total energy divided by power.

Scientific diagram and interpretation

For vaporisation, electrical energy is compared with mass lost while steadily boiling. Steam, hot liquid and electricity need controlled laboratory procedures. Fusion uses a distinct melting method described in the lesson.

Equations and relationships

ΔQ = mL; L in J kg⁻¹

Worked examples and applications

Melt a known mass

Using supplied Lf = 3.34 × 10⁵ J kg⁻¹, melting 0.200 kg at its melting point requires ΔQ = 0.200×3.34×10⁵ = 66800 J = 66.8 kJ.

Mass condensed

If 226 kJ is released during condensation and supplied Lv = 2.26 × 10⁶ J kg⁻¹, m = 226000/(2.26×10⁶) = 0.100 kg.

Heat then boil

For 0.100 kg water from 20 to 100 °C, use supplied c = 4200 J kg⁻¹ K⁻¹ and Lv = 2.26 × 10⁶ J kg⁻¹. Sensible energy = 0.100×4200×80 = 33600 J; vaporisation = 226000 J. Total = 259600 J ≈ 260 kJ. At 500 W with negligible losses, t = 259600/500 = 519.2 s ≈ 520 s.

Common misconceptions

For multi-stage heating, mL covers only the phase change, not warming before or after it.

Examination technique

Lay out a stage table with starting state, ending state, equation and energy.

For an explanation, connect the physical cause to its effect. For a calculation, state the model, rearrange before substitution, retain intermediate digits, and finish with units. For a data question, refer to the actual readings and distinguish a trend from proof of a cause.

Check your understanding

Five questions. Choose an option, then Check Answer. Read the feedback and use Try Again to reset. Use Tab between controls, arrow keys within radio groups, and Enter or Space on buttons. These local practice answers are not saved.