Physics·Revision Notes

Latent Heat — Revision Notes

NEET UG
Updated 13 Sep 2026

⚡ 30-Second Revision

  • Latent Heat ($L$) — Heat absorbed/released during phase change at constant temperature.
  • Formula — Q=mLQ = mL
  • Latent Heat of Fusion ($L_f$) — Solid ⇌\rightleftharpoons Liquid. For water: Lf=3.34×105 J/kgL_f = 3.34 \times 10^5\,\text{J/kg} (80 cal/g80\,\text{cal/g}).
  • Latent Heat of Vaporization ($L_v$) — Liquid ⇌\rightleftharpoons Gas. For water: Lv=2.26×106 J/kgL_v = 2.26 \times 10^6\,\text{J/kg} (540 cal/g540\,\text{cal/g}).
  • Key Principle — Energy changes potential energy (intermolecular forces), not kinetic energy (temperature).
  • Heating Curve — Flat plateaus indicate phase changes (Q=mLQ=mL), sloped sections indicate temperature changes (Q=mcDeltaTQ=mcDelta T).

2-Minute Revision

Latent heat is the energy involved in changing the phase of a substance (solid, liquid, gas) without changing its temperature. This 'hidden' heat is used to alter the potential energy of molecules by overcoming or establishing intermolecular forces.

The two main types are latent heat of fusion (LfL_f) for solid-liquid transitions and latent heat of vaporization (LvL_v) for liquid-gas transitions. The amount of heat (QQ) involved is given by Q=mLQ = mL, where mm is mass and LL is the specific latent heat.

For water, LfL_f is 3.34×105 J/kg3.34 \times 10^5\,\text{J/kg} and LvL_v is 2.26×106 J/kg2.26 \times 10^6\,\text{J/kg}. Note that LvL_v is significantly higher than LfL_f. In multi-stage problems, combine latent heat calculations with specific heat calculations (Q=mcDeltaTQ=mcDelta T) for temperature changes.

Heating curves graphically represent this, with flat plateaus indicating phase changes. Remember that heat is absorbed during melting and vaporization, and released during freezing and condensation.

5-Minute Revision

Latent heat is a critical concept in thermodynamics, describing the heat energy absorbed or released during a phase transition (like melting, boiling, freezing, or condensation) at a constant temperature.

Unlike specific heat, which causes a temperature change, latent heat is entirely dedicated to changing the potential energy of the molecules by altering their intermolecular forces and arrangement. For example, when ice melts at 0∘C0^\circ\text{C}, the absorbed latent heat of fusion breaks the rigid bonds, allowing water molecules to move freely as a liquid, without increasing their average kinetic energy (and thus temperature).

The fundamental formula is Q=mLQ = mL, where QQ is the heat, mm is the mass, and LL is the specific latent heat. There are two primary types:

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  1. Latent Heat of Fusion ($L_f$) — For solid ⇌\rightleftharpoons liquid transitions. For water, Lf=3.34×105 J/kgL_f = 3.34 \times 10^5\,\text{J/kg} (80 cal/g80\,\text{cal/g}). This heat is absorbed during melting and released during freezing.
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  3. Latent Heat of Vaporization ($L_v$) — For liquid ⇌\rightleftharpoons gas transitions. For water, Lv=2.26×106 J/kgL_v = 2.26 \times 10^6\,\text{J/kg} (540 cal/g540\,\text{cal/g}). This heat is absorbed during vaporization (boiling/evaporation) and released during condensation. Note that LvL_v is significantly larger than LfL_f because more energy is needed to completely separate molecules into a gas.

Example: To convert 10 g10\,\text{g} of ice at 0∘C0^\circ\text{C} to water at 50∘C50^\circ\text{C}:

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  1. Melt ice — Q1=mLf=(0.01 kg)×(3.34×105 J/kg)=3340 JQ_1 = mL_f = (0.01\,\text{kg}) \times (3.34 \times 10^5\,\text{J/kg}) = 3340\,\text{J}.
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  3. Heat water — Q2=mcwaterΔT=(0.01 kg)×(4186 J/kg∘C)×(50∘C−0∘C)=2093 JQ_2 = mc_{\text{water}}\Delta T = (0.01\,\text{kg}) \times (4186\,\text{J/kg}^\circ\text{C}) \times (50^\circ\text{C} - 0^\circ\text{C}) = 2093\,\text{J}.
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  5. Total Heat — Qtotal=Q1+Q2=3340 J+2093 J=5433 JQ_{\text{total}} = Q_1 + Q_2 = 3340\,\text{J} + 2093\,\text{J} = 5433\,\text{J}.

Real-world applications include sweating for cooling, refrigeration, and the severity of steam burns due to the release of LvL_v. Heating curves visually represent these processes, with flat sections indicating phase changes at constant temperature.

Prelims Revision Notes

Latent heat is the energy associated with phase changes at constant temperature. It's crucial for NEET as it often combines with specific heat in multi-step problems.

Key Definitions & Formulas:

  • Latent Heat ($L$) — Heat per unit mass for a phase change. Q=mLQ = mL.
  • Latent Heat of Fusion ($L_f$) — Solid ⇌\rightleftharpoons Liquid. Absorbed during melting, released during freezing. For water, Lf=3.34×105 J/kgL_f = 3.34 \times 10^5\,\text{J/kg} (80 cal/g80\,\text{cal/g}). This energy breaks/forms bonds in the solid lattice.
  • Latent Heat of Vaporization ($L_v$) — Liquid ⇌\rightleftharpoons Gas. Absorbed during vaporization (boiling/evaporation), released during condensation. For water, Lv=2.26×106 J/kgL_v = 2.26 \times 10^6\,\text{J/kg} (540 cal/g540\,\text{cal/g}). This energy overcomes intermolecular forces completely.
  • Specific Heat ($c$) — Heat per unit mass for 1∘C1^\circ\text{C} temperature change. Q=mcDeltaTQ = mcDelta T.

Important Points for NEET:

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  1. Constant Temperature — During a phase change, temperature remains constant. The supplied heat changes potential energy, not kinetic energy.
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  3. $L_v > L_f$ — Vaporization requires more energy than fusion because molecules need to completely separate.
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  5. Multi-Stage Problems — Most problems involve multiple steps: heating a phase (Q=mcDeltaTQ=mcDelta T), then phase change (Q=mLQ=mL), then heating the new phase, etc. Sum all QQ values.

* Example: Ice at −10∘C-10^\circ\text{C} to steam at 110∘C110^\circ\text{C} involves 5 steps: Qice+QLf+Qwater+QLv+QsteamQ_{\text{ice}} + Q_{L_f} + Q_{\text{water}} + Q_{L_v} + Q_{\text{steam}}.

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  1. Calorimetry — For mixing problems (e.g., ice and water), apply 'Heat Lost = Heat Gained'. Identify all possible phase changes and temperature changes for each component.
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  3. Heating Curves — Understand that flat plateaus represent phase changes (latent heat), and sloped sections represent temperature changes (specific heat). The slope is inversely proportional to specific heat capacity.
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  5. Units — Be consistent. Convert grams to kg, calories to joules, etc., as needed. 1 cal=4.186 J1\,\text{cal} = 4.186\,\text{J}.
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  7. Real-world Applications — Know why steam burns are severe (release of LvL_v), and how sweating cools the body (absorption of LvL_v).

Vyyuha Quick Recall

L.A.T.E.N.T. H.E.A.T. Liquid to gas, Absorbs heat. Temperature Equal, No change at all. Transforms state, Hidden energy's call. Energy for bonds, Alters molecular space. Two types: Fusion and Vaporization's embrace.