Latent Heat Numericals: Melting Does Not Always Raise the Temperature

For a specified phase change at its transition conditions, the transferred energy magnitude is Q = mL. This differs from mcΔT for changing temperature within one phase; a problem containing both stages needs both contributions.

List the stages rather than forcing one formula

A heating description can involve warming a solid, melting it, warming the resulting liquid and perhaps another phase change. Each stage has its own energy relation and material quantity. The simple Q = mL calculation here applies only to the stated change of phase.

Specific latent heat L has units J/kg, while specific heat capacity c has units J/(kg·K). The different units reflect different questions. During an ideal equilibrium transition of a pure substance at fixed pressure, added energy can change the phase without increasing temperature until the transition is complete.

Worked melting calculation

A school exercise supplies latent heat of fusion L = 334,000 J/kg. Find the energy required to melt 0.20 kg of a sample already at its melting temperature. The magnitude is Q = 0.20 × 334,000 = 66,800 J, or 66.8 kJ.

No warming term is included because the starting state is explicitly at the transition temperature. If 33,400 J is available under the same ideal assumptions, the mass melted is Q/L = 33,400/334,000 = 0.10 kg. A 0.20 kg starting sample would therefore not all melt from that energy alone.

Choose the sign convention separately

Melting and vaporisation absorb energy from the surroundings. Freezing and condensation release energy. When the question asks for a heat-transfer magnitude, report a positive amount and say whether it is absorbed or released. If a signed energy balance is requested, define positive heat into the chosen system.

The supplied L may depend on substance and stated conditions. Do not use a fusion value for vaporisation or a value for one material for another. An ideal supplied constant is sufficient for the paper exercise without implying laboratory-grade precision.

Common mistake

Convert grams to kilograms if L is per kilogram. Do not multiply by an invented temperature change during the phase-change stage. These examples do not describe instructions for handling boiling liquids, steam or extreme temperatures; they are calculations using specified hypothetical states.

Phase badal raha hai ya temperature? Formula uske baad chuno.

Three practice questions with worked answers

1. For L = 200,000 J/kg, find energy to change the phase of 0.30 kg.

   Q = mL = 0.30 × 200,000 = 60,000 J.

2. A phase change uses 40,000 J with L = 250,000 J/kg. Find the mass changed.

   m = Q/L = 40,000/250,000 = 0.16 kg.

3. A 50 g sample freezes with L = 300,000 J/kg. Find the energy magnitude released.

   50 g = 0.050 kg. The released amount is 0.050 × 300,000 = 15,000 J.

Related study guides

• Heat and Temperature: Specific Heat Numericals with a Simple Method
• Thermodynamics for JEE and NEET: Formulas, Sign Conventions and Solved Problems

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References and authorship

Original examples and practice questions, prepared with AI assistance and editorial arithmetic review. Related definitions and methods: OpenStax: Phase Change and Latent Heat. No textbook or exam-board affiliation is claimed.