define and use specific latent heat and distinguish between specific latent heat of fusion and specific latent heat of vaporisation

Specific Latent Heat

1. Definition (AO1)

The specific latent heat \(L\) of a substance is the amount of heat energy required to change the phase of 1 kg of that substance at constant pressure and constant temperature. It is a material property, expressed in J kg⁻¹.

2. Fundamental Equation (AO2)

For any phase change the heat transferred, \(Q\), is given by

• Q – heat energy (J)
• m – mass of the substance (kg)
• L – specific latent heat (J kg⁻¹)

3. Types of Specific Latent Heat (AO1)

4. How \(Lf\) and \(Lv\) Differ (AO1)

• Fusion – \(L_f\) (ΔH_fus)

  • Occurs at the melting point of the substance.
  • Only the ordered lattice of a solid is broken; most nearest‑neighbour contacts remain.
  • Consequently the energy required is relatively modest.

• Vapourisation – \(L_v\) (ΔH_vap)

  • Occurs at the boiling point (or wherever the vapour pressure equals the external pressure).
  • Nearly all intermolecular attractions are overcome, producing widely separated gas molecules.
  • Therefore L_v > L_f for the same material.

5. Typical Values at 1 atm (J kg⁻¹)

6. Checklist for Using \(Q = mL\) (AO2)

1. Identify the phase change. Is it fusion or vapourisation? Choose \(Lf\) or \(Lv\) accordingly.
2. Confirm constant pressure. The Cambridge syllabus assumes the change occurs at the ambient pressure (normally 1 atm).
3. Read the correct specific latent‑heat value. Use the data sheet or a table; check that the units are J kg⁻¹.
4. Convert mass to kilograms. (e.g. 250 g → 0.250 kg.)
5. Substitute into \(Q = mL\). Perform the multiplication, keeping the appropriate number of significant figures.
6. Assign the sign of \(Q\). Positive for heat absorbed (melting, boiling); negative for heat released (freezing, condensation).

7. Worked Example – Melting Ice (Fusion)

Problem: Calculate the heat required to melt 250 g of ice at 0 °C.

1. Phase change: solid → liquid → use \(L_f\) (ΔH_fus).
2. Given: \(L_f(\text{ice}) = 3.34 × 10⁵ \text{J kg}^{-1}\).
3. Mass: \(m = 250 \text{g} = 0.250 \text{kg}\).
4. Apply the formula:
   
   \(Q = mL_f = 0.250 \text{kg} × 3.34 × 10⁵ \text{J kg}^{-1}= 8.35 × 10⁴ \text{J}\)

5. Result: Q = +8.35 × 10⁴ J** (heat absorbed).

8. Worked Example – Boiling Water (Vapourisation)

Problem: How much energy is needed to convert 1.5 kg of water at 100 °C into steam at 100 °C?

1. Phase change: liquid → gas → use \(L_v\) (ΔH_vap).
2. Given: \(L_v(\text{water}) = 2.26 × 10⁶ \text{J kg}^{-1}\).
3. Mass: \(m = 1.5 \text{kg}\).
4. Apply the formula:
   
   \(Q = mL_v = 1.5 \text{kg} × 2.26 × 10⁶ \text{J kg}^{-1}= 3.39 × 10⁶ \text{J}\)

5. Result: Q = +3.39 × 10⁶ J** (heat absorbed).

9. Relationship to the Wider Syllabus (AO3)

• Specific latent heat is a type of energy change covered in the “Work, Energy & Power” block (topic 5). It illustrates the principle that energy can be transferred without a temperature change.
• The equation \(Q = mL\) is analogous to the work‑energy theorem \(W = \Delta E_{\text{kin}}\); both link a material property (latent heat or mass) to an energy transfer.
• Understanding latent heat underpins thermodynamic questions about phase diagrams, heating‑cooling cycles, and calorimetry experiments (practical skills in the syllabus).

10. Key Points to Remember (AO1)

• Latent heat is absorbed or released at constant temperature and constant pressure.
• Specific latent heats are denoted L_f (fusion) and L_v (vapourisation); the corresponding enthalpy changes are ΔH_fus and ΔH_vap.
• For a given substance, L_v > L_f because vapourisation breaks virtually all intermolecular forces.
• The same formula, Q = mL, applies to melting/freezing and boiling/condensation – only the value of \(L\) changes.
• Use the checklist in section 6 to avoid common mistakes (wrong phase, wrong units, sign of \(Q\)).