The Heat Of Fusion Formula gives the amount of heat energy required to convert a solid substance into a liquid at its melting point, without any change in temperature. Expressed as \( Q = mL_f \), this formula is a core concept in thermodynamics covered in NCERT Class 11 Physics (Chapter 11 — Thermal Properties of Matter). It is equally important for NEET and JEE Main aspirants who encounter latent heat problems regularly. This article covers the formula expression, derivation, a complete physics formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Heat Of Fusion Formulas at a Glance
Quick reference for the most important heat of fusion and related latent heat formulas.
- Heat of fusion: \( Q = mL_f \)
- Heat of vaporisation: \( Q = mL_v \)
- Specific heat capacity: \( Q = mc\Delta T \)
- Total heat (heating + melting): \( Q = mc\Delta T + mL_f \)
- Latent heat of fusion of water: \( L_f = 3.34 \times 10^5 \text{ J/kg} \)
- Enthalpy of fusion: \( \Delta H_{fus} = L_f \times M \) (molar form)
- Heat released on freezing: \( Q_{released} = mL_f \) (same magnitude)
What is the Heat Of Fusion Formula?
The Heat Of Fusion Formula quantifies the thermal energy absorbed by a solid substance as it melts into a liquid at its melting point. During this phase transition, the temperature of the substance does not change. All the absorbed energy goes into breaking the intermolecular bonds that hold the solid lattice together.
This concept is introduced in NCERT Class 11 Physics, Chapter 11 — Thermal Properties of Matter. The heat involved in a phase change is called latent heat, meaning “hidden heat,” because it does not raise the thermometer reading. The specific latent heat of fusion, denoted \( L_f \), is the heat required per unit mass of the substance.
For water, \( L_f = 3.34 \times 10^5 \text{ J/kg} \) (or 80 cal/g). This value is widely used in both CBSE board problems and competitive exam questions. The reverse process — liquid freezing into solid — releases exactly the same amount of heat. Understanding both directions is essential for solving calorimetry problems in JEE and NEET.
Heat Of Fusion Formula — Expression and Variables
The standard expression for the heat of fusion is:
\[ Q = m L_f \]
where Q is the heat absorbed or released during the phase transition from solid to liquid (or liquid to solid).
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( Q \) | Heat energy absorbed or released | Joule (J) |
| \( m \) | Mass of the substance | Kilogram (kg) |
| \( L_f \) | Specific latent heat of fusion | J/kg |
| \( \Delta H_{fus} \) | Molar enthalpy of fusion | J/mol |
| \( T_m \) | Melting point temperature | Kelvin (K) or °C |
Derivation of the Heat Of Fusion Formula
The derivation starts from the definition of latent heat. When a substance undergoes a phase change, the heat exchanged is proportional to the mass of the substance. The constant of proportionality is the specific latent heat \( L_f \).
Step 1: Consider a solid of mass \( m \) at its melting point \( T_m \).
Step 2: Supply heat \( dQ \) to a small element of mass \( dm \). The heat required is \( dQ = L_f \, dm \).
Step 3: Integrate over the total mass:
\[ Q = \int_0^m L_f \, dm = L_f \cdot m \]
Step 4: Since \( L_f \) is constant for a given substance at a fixed pressure, we get the final formula \( Q = mL_f \). This derivation confirms that the heat of fusion depends only on mass and the material property \( L_f \), not on the rate of heating.
Complete Thermal Physics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat of Fusion | \( Q = mL_f \) | m = mass, L₁ = specific latent heat of fusion | J | Class 11, Ch 11 |
| Heat of Vaporisation | \( Q = mL_v \) | m = mass, L₂ = specific latent heat of vaporisation | J | Class 11, Ch 11 |
| Specific Heat Capacity | \( Q = mc\Delta T \) | m = mass, c = specific heat, ΔT = temperature change | J | Class 11, Ch 11 |
| Calorimetry (heat gained = heat lost) | \( m_1 c_1 \Delta T_1 = m_2 c_2 \Delta T_2 \) | m = mass, c = specific heat, ΔT = temperature change | J | Class 11, Ch 11 |
| Total Heat (heating + melting) | \( Q = mc\Delta T + mL_f \) | m = mass, c = specific heat, ΔT = rise to melting point | J | Class 11, Ch 11 |
| Molar Enthalpy of Fusion | \( \Delta H_{fus} = L_f \times M \) | M = molar mass in kg/mol | J/mol | Class 11, Ch 11 |
| Newton’s Law of Cooling | \( \frac{dT}{dt} = -k(T – T_0) \) | T = temperature, T₀ = ambient temp, k = cooling constant | K/s | Class 11, Ch 11 |
| Stefan–Boltzmann Law | \( P = \sigma A T^4 \) | σ = 5.67×10⁻&sup8; W/m²K&sup4;, A = area, T = temperature | W | Class 11, Ch 11 |
| Wien’s Displacement Law | \( \lambda_{max} T = b \) | b = 2.898×10⁻³ m·K | m·K | Class 11, Ch 11 |
| Thermal Conductivity | \( H = \frac{KA(T_1 – T_2)}{L} \) | K = thermal conductivity, A = area, L = length | W | Class 11, Ch 11 |
Heat Of Fusion Formula — Solved Examples
Example 1 (Class 9-10 Level): Melting Ice
Problem: How much heat is required to completely melt 500 g of ice at 0°C? (Latent heat of fusion of water = 3.34 × 10&sup5; J/kg)
Given:
- Mass, \( m = 500 \text{ g} = 0.5 \text{ kg} \)
- Latent heat of fusion, \( L_f = 3.34 \times 10^5 \text{ J/kg} \)
Step 1: Write the Heat Of Fusion Formula: \( Q = mL_f \)
Step 2: Substitute the values:
\[ Q = 0.5 \times 3.34 \times 10^5 \]
Step 3: Calculate:
\[ Q = 1.67 \times 10^5 \text{ J} = 167{,}000 \text{ J} \]
Answer
Heat required = 1.67 × 10&sup5; J (167 kJ)
Example 2 (Class 11-12 Level): Heating Ice to Steam — Total Heat Calculation
Problem: Calculate the total heat required to convert 200 g of ice at −10°C to water at 20°C. Given: specific heat of ice = 2100 J/kg·K, latent heat of fusion of water = 3.34 × 10&sup5; J/kg, specific heat of water = 4200 J/kg·K.
Given:
- Mass, \( m = 200 \text{ g} = 0.2 \text{ kg} \)
- Initial temperature of ice, \( T_i = -10^\circ\text{C} \)
- Final temperature of water, \( T_f = 20^\circ\text{C} \)
- Specific heat of ice, \( c_{ice} = 2100 \text{ J/kg·K} \)
- Latent heat of fusion, \( L_f = 3.34 \times 10^5 \text{ J/kg} \)
- Specific heat of water, \( c_w = 4200 \text{ J/kg·K} \)
Step 1: Heat to warm ice from −10°C to 0°C:
\[ Q_1 = mc_{ice}\Delta T = 0.2 \times 2100 \times 10 = 4200 \text{ J} \]
Step 2: Heat to melt ice at 0°C (Heat of Fusion Formula):
\[ Q_2 = mL_f = 0.2 \times 3.34 \times 10^5 = 66{,}800 \text{ J} \]
Step 3: Heat to warm water from 0°C to 20°C:
\[ Q_3 = mc_w\Delta T = 0.2 \times 4200 \times 20 = 16{,}800 \text{ J} \]
Step 4: Total heat:
\[ Q_{total} = Q_1 + Q_2 + Q_3 = 4200 + 66800 + 16800 = 87{,}800 \text{ J} \]
Answer
Total heat required = 87,800 J = 87.8 kJ
Example 3 (JEE/NEET Level): Calorimetry with Phase Change
Problem: 50 g of ice at 0°C is mixed with 200 g of water at 60°C in a thermally insulated calorimeter. Find the final temperature of the mixture. (L₁ = 3.34 × 10&sup5; J/kg; c₂ = 4200 J/kg·K)
Given:
- Mass of ice, \( m_i = 50 \text{ g} = 0.05 \text{ kg} \)
- Temperature of ice, \( T_i = 0^\circ\text{C} \)
- Mass of water, \( m_w = 200 \text{ g} = 0.2 \text{ kg} \)
- Temperature of water, \( T_w = 60^\circ\text{C} \)
Step 1: Find heat available from warm water cooling to 0°C:
\[ Q_{available} = m_w c_w T_w = 0.2 \times 4200 \times 60 = 50{,}400 \text{ J} \]
Step 2: Find heat needed to melt all the ice:
\[ Q_{melt} = m_i L_f = 0.05 \times 3.34 \times 10^5 = 16{,}700 \text{ J} \]
Step 3: Since \( Q_{available} (50{,}400) > Q_{melt} (16{,}700) \), all ice melts. Remaining heat available after melting:
\[ Q_{rem} = 50{,}400 – 16{,}700 = 33{,}700 \text{ J} \]
Step 4: This remaining heat raises the temperature of the total water (0.05 + 0.2 = 0.25 kg) from 0°C:
\[ Q_{rem} = (m_i + m_w) c_w \Delta T \]
\[ 33700 = 0.25 \times 4200 \times \Delta T \]
\[ \Delta T = \frac{33700}{1050} \approx 32.1^\circ\text{C} \]
Step 5: Final temperature = \( 0 + 32.1 = 32.1^\circ\text{C} \)
Answer
Final temperature of the mixture = 32.1°C
CBSE Exam Tips 2025-26
- Memorise standard values: We recommend learning \( L_f(\text{water}) = 3.34 \times 10^5 \text{ J/kg} \) and \( L_v(\text{water}) = 22.6 \times 10^5 \text{ J/kg} \) by heart. These appear in almost every calorimetry problem.
- Unit conversion is critical: Always convert mass from grams to kilograms before substituting into \( Q = mL_f \). A missed conversion is the most common reason for wrong answers.
- Phase change means no temperature change: In CBSE 2025-26 papers, questions often test whether students know that temperature stays constant during melting. State this explicitly in your answer for full marks.
- Multi-step heating problems: Break the problem into stages — heating to melting point, melting, then heating the liquid. Calculate \( Q \) for each stage separately and add them up.
- Direction matters: Heat is absorbed during melting and released during freezing. Use the correct sign convention when applying the principle of calorimetry.
- NCERT examples first: Our experts suggest solving all NCERT in-text and exercise problems from Chapter 11 before attempting previous year papers. The CBSE board directly adapts several problems from NCERT.
Common Mistakes to Avoid
| Common Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Using \( Q = mc\Delta T \) for phase changes | During melting, \( \Delta T = 0 \), so this formula gives zero heat, which is incorrect. | Use \( Q = mL_f \) for phase transitions. Use \( Q = mc\Delta T \) only when temperature changes. |
| Forgetting to convert grams to kilograms | The SI unit of \( L_f \) is J/kg. Using grams gives an answer 1000 times smaller. | Always convert: \( m(\text{kg}) = m(\text{g}) \div 1000 \) before substituting. |
| Confusing heat of fusion with heat of vaporisation | They are different phase transitions with very different values (\( L_v \gg L_f \) for water). | Fusion = solid to liquid. Vaporisation = liquid to gas. Use the correct \( L \) value. |
| Assuming temperature rises during melting | Phase changes are isothermal. The absorbed heat breaks bonds, not raises temperature. | State clearly: “Temperature remains constant at the melting point during the phase transition.” |
| Ignoring the sign of heat in calorimetry | In a calorimeter, heat gained by one body must equal heat lost by the other. Mixing up signs leads to wrong equilibrium temperatures. | Apply: heat gained = heat lost. Treat all \( Q \) values as positive magnitudes and balance them. |
JEE/NEET Application of Heat Of Fusion Formula
In our experience, JEE aspirants encounter the Heat Of Fusion Formula in three main problem patterns. Understanding these patterns saves significant time during the exam.
Pattern 1: Multi-Stage Calorimetry Problems
JEE Main frequently presents problems where a substance is heated from a sub-zero temperature to above its melting point. Students must calculate heat for each stage: warming the solid, melting it (using \( Q = mL_f \)), and then warming the resulting liquid. These problems test systematic thinking. Always draw a temperature-time graph mentally to identify each stage before calculating.
Pattern 2: Equilibrium Temperature with Phase Change
NEET often presents a mixture of ice and warm water in an insulated container. The key step is to first check whether all the ice melts or not. Compare the heat available from the warm water with the heat needed to melt all the ice. This comparison determines the approach. If all ice melts, find the final temperature using the remaining heat. If not all ice melts, the final temperature is 0°C.
Pattern 3: Enthalpy and Thermodynamic Linkage (JEE Advanced)
JEE Advanced sometimes connects the Heat Of Fusion Formula to the Clausius-Clapeyron equation and entropy of fusion. The entropy change during fusion is given by:
\[ \Delta S_{fus} = \frac{\Delta H_{fus}}{T_m} \]
where \( T_m \) is the melting point in Kelvin. Students should know that \( \Delta H_{fus} = mL_f \) for a given mass, and that fusion is always an endothermic process (\( \Delta H_{fus} > 0 \)). In our experience, JEE aspirants who connect thermodynamics and thermal physics concepts score significantly higher in Section B of JEE Advanced Paper 2.
For NEET, focus on the numerical application of \( Q = mL_f \) and calorimetry. NEET 2024 had two questions directly involving latent heat of fusion. Practising at least 30 calorimetry problems before your exam is strongly recommended.
FAQs on Heat Of Fusion Formula
Explore More Physics Formulas
Now that you have mastered the Heat Of Fusion Formula, strengthen your thermal physics preparation with related topics. Visit our complete Physics Formulas hub for a full list of NCERT-aligned formula articles. You may also find these articles useful:
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For official NCERT textbook content, refer to the NCERT official website to download the Class 11 Physics textbook and verify all standard values used in this article.