The Heat of vaporisation Formula gives the amount of heat energy required to convert a unit mass of liquid into vapour at constant temperature and pressure. Expressed as \( Q = mL_v \), this formula is a core concept in Class 11 Physics and Chemistry (NCERT Chapter 11 — Thermal Properties of Matter). It also appears regularly in JEE Main, JEE Advanced, and NEET thermodynamics questions. This article covers the formula, its derivation, a complete formula sheet, three solved examples, CBSE exam tips, common mistakes, and JEE/NEET applications.
Key Heat of Vaporisation Formulas at a Glance
Quick reference for the most important formulas related to heat of vaporisation.
- Basic formula: \( Q = mL_v \)
- Latent heat of vaporisation: \( L_v = \dfrac{Q}{m} \)
- Mass from heat: \( m = \dfrac{Q}{L_v} \)
- Water vaporisation value: \( L_v(\text{water}) = 2.26 \times 10^6 \, \text{J/kg} \)
- Clausius–Clapeyron equation: \( \dfrac{dP}{dT} = \dfrac{L_v}{T \Delta v} \)
- Trouton’s rule: \( \dfrac{L_v}{T_b} \approx 88 \, \text{J mol}^{-1}\text{K}^{-1} \)
- Enthalpy form: \( \Delta H_{vap} = Q \) at constant pressure
What is the Heat of Vaporisation Formula?
The Heat of vaporisation Formula describes the relationship between the heat energy supplied to a liquid and the mass of liquid that converts to vapour. When a liquid reaches its boiling point, it does not immediately become a gas. It needs additional energy to break the intermolecular forces holding its molecules together. This energy is called the latent heat of vaporisation.
The term “latent” means hidden. The temperature of the substance does not change during this phase transition. All the energy goes into breaking molecular bonds. NCERT Class 11 Physics, Chapter 11 (Thermal Properties of Matter) defines this quantity and provides standard values for common substances.
The latent heat of vaporisation of water is one of the highest known values for any substance. This property makes water an exceptional coolant and explains many natural phenomena, from sweating to cloud formation. Understanding this formula is essential for scoring well in CBSE board exams and for cracking thermodynamics problems in JEE and NEET.
Heat of Vaporisation Formula — Expression and Variables
The fundamental expression for the heat of vaporisation is:
\[ Q = m L_v \]
where Q is the total heat energy absorbed or released, m is the mass of the substance, and \ (L_v\) is the specific latent heat of vaporisation.
Rearranging the formula gives:
\[ L_v = \frac{Q}{m} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \ (Q\) | Heat energy absorbed or released | Joule (J) |
| \ (m\) | Mass of the liquid | Kilogram (kg) |
| \ (L_v\) | Specific latent heat of vaporisation | J/kg |
| \ (\Delta H_{vap}\) | Enthalpy of vaporisation (molar) | J/mol or kJ/mol |
| \ (T_b\) | Boiling point temperature | Kelvin (K) |
Derivation of the Heat of Vaporisation Formula
The derivation begins with the definition of latent heat. When a substance undergoes a phase change, the heat supplied does not raise the temperature. Instead, it changes the phase of the substance.
Step 1: Define heat supplied as \ (Q\) and mass as \ (m\).
Step 2: Experiments show that \ (Q\) is directly proportional to \ (m\) at a fixed temperature and pressure. So \ (Q \propto m\).
Step 3: Introduce the proportionality constant \ (L_v\), called the specific latent heat of vaporisation. This gives \ (Q = m L_v\).
Step 4: Rearranging, \ (L_v = Q/m\). This is the heat required per unit mass for the phase change from liquid to vapour.
The SI unit of \ (L_v\) is J/kg. For water, \ (L_v = 2.26 \times 10^6 \) J/kg at 100°C and standard atmospheric pressure.
Complete Thermal Physics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat of Vaporisation | \ (Q = mL_v\) | Q = heat, m = mass, L_v = latent heat | J | Class 11, Ch 11 |
| Specific Latent Heat | \ (L_v = Q/m\) | Q = heat, m = mass | J/kg | Class 11, Ch 11 |
| Heat of Fusion | \ (Q = mL_f\) | L_f = latent heat of fusion | J | Class 11, Ch 11 |
| Sensible Heat (Temperature Change) | \ (Q = mc\Delta T\) | c = specific heat, ΔT = temp change | J | Class 11, Ch 11 |
| Clausius–Clapeyron Equation | \ (\dfrac{dP}{dT} = \dfrac{L_v}{T\Delta v}\) | P = pressure, T = temp, Δv = specific volume change | Pa/K | Class 11, Ch 12 |
| Trouton’s Rule | \ (\dfrac{\Delta H_{vap}}{T_b} \approx 88\) | ΔH_vap = molar enthalpy, T_b = boiling point | J mol−1 K−1 | Class 11 Chemistry, Ch 6 |
| Newton’s Law of Cooling | \ (\dfrac{dQ}{dt} = -hA(T – T_0)\) | h = heat transfer coeff, A = area, T_0 = ambient temp | W | Class 11, Ch 11 |
| Stefan–Boltzmann Law | \ (P = \sigma A T^4\) | σ = 5.67×10−8 W m−2 K−4, A = area | W | Class 11, Ch 11 |
| Thermal Conductivity | \ (Q/t = kA(T_1 – T_2)/d\) | k = conductivity, A = area, d = thickness | W | Class 11, Ch 11 |
| Ideal Gas Law | \ (PV = nRT\) | P = pressure, V = volume, n = moles, R = 8.314 J/mol K | Pa·m³ | Class 11, Ch 13 |
Heat of Vaporisation Formula — Solved Examples
Example 1 (Class 9–10 Level)
Problem: How much heat is required to completely vaporise 2 kg of water at 100°C? (Given: \ (L_v\) of water = \ (2.26 \times 10^6\) J/kg)
Given: m = 2 kg, \ (L_v = 2.26 \times 10^6\) J/kg
Step 1: Write the heat of vaporisation formula: \ (Q = mL_v\)
Step 2: Substitute the given values: \ (Q = 2 \times 2.26 \times 10^6\)
Step 3: Calculate: \ (Q = 4.52 \times 10^6\) J
Answer
Heat required = \ (4.52 \times 10^6\) J = 4.52 MJ
Example 2 (Class 11–12 Level)
Problem: A 500 W heater is used to vaporise water at 100°C. How long will it take to vaporise 300 g of water? (Given: \ (L_v\) of water = \ (2.26 \times 10^6\) J/kg)
Given: Power P = 500 W, m = 300 g = 0.3 kg, \ (L_v = 2.26 \times 10^6\) J/kg
Step 1: Calculate the total heat required using \ (Q = mL_v\):
\ (Q = 0.3 \times 2.26 \times 10^6 = 6.78 \times 10^5\) J
Step 2: Use the power formula \ (P = Q/t\), so \ (t = Q/P\):
\ (t = \dfrac{6.78 \times 10^5}{500}\)
Step 3: Calculate time:
\ (t = 1356\) seconds \ (\approx 22.6\) minutes
Answer
Time required = 1356 s ≈ 22.6 minutes
Example 3 (JEE/NEET Level)
Problem: A calorimeter contains 200 g of water at 50°C. Steam at 100°C is passed into it until the temperature of the mixture reaches 80°C. Find the mass of steam condensed. (Given: specific heat of water \ (c = 4200\) J/kg°C, \ (L_v = 2.26 \times 10^6\) J/kg)
Given: \ (m_w\) = 200 g = 0.2 kg, \ (T_{\text{initial}} = 50\)°C, \ (T_{\text{final}} = 80\)°C, steam temperature = 100°C, c = 4200 J/kg°C, \ (L_v = 2.26 \times 10^6\) J/kg. Let mass of steam condensed = \ (m_s\).
Step 1: Heat released by steam has two parts. First, steam condenses at 100°C: \ (Q_1 = m_s L_v\). Second, condensed water cools from 100°C to 80°C: \ (Q_2 = m_s c (100 – 80) = m_s \times 4200 \times 20\).
Step 2: Total heat released by steam: \ (Q_{\text{released}} = m_s(2.26 \times 10^6 + 84000) = m_s \times 2.344 \times 10^6\) J
Step 3: Heat absorbed by calorimeter water (ignoring calorimeter heat capacity): \ (Q_{\text{absorbed}} = m_w c \Delta T = 0.2 \times 4200 \times (80 – 50) = 0.2 \times 4200 \times 30 = 25200\) J
Step 4: Apply heat balance (heat lost = heat gained): \ (m_s \times 2.344 \times 10^6 = 25200\)
Step 5: Solve for \ (m_s\): \ (m_s = \dfrac{25200}{2.344 \times 10^6} \approx 0.01075\) kg = 10.75 g
Answer
Mass of steam condensed ≈ 10.75 g
CBSE Exam Tips 2025-26
- Memorise standard values: Always remember \ (L_v\) of water = \ (2.26 \times 10^6\) J/kg and \ (L_f\) of ice = \ (3.34 \times 10^5\) J/kg. CBSE 2025-26 papers frequently use these values directly.
- Unit conversion is critical: Convert grams to kilograms before substituting into \ (Q = mL_v\). A mass left in grams will give an answer that is 1000 times too large.
- Show all steps clearly: CBSE awards step marks. Write the formula first, then substitute values, then calculate. Do not skip steps in 3-mark or 5-mark questions.
- Distinguish latent and sensible heat: Use \ (Q = mL_v\) for phase changes (no temperature change). Use \ (Q = mc\Delta T\) when temperature changes. Mixing these two up is the most common error in board exams.
- We recommend practising calorimetry problems that combine both formulas. These mixed problems appear almost every year in CBSE Class 11 term exams.
- Check sign conventions: Heat is absorbed (\ (Q > 0\)) during vaporisation. Heat is released (\ (Q < 0\)) during condensation. State this in your answer to earn full marks.
Common Mistakes to Avoid
- Using grams instead of kilograms: The SI unit of mass is kg. The standard value of \ (L_v\) for water is given in J/kg. If you substitute mass in grams, your answer will be wrong by a factor of 1000. Always convert first.
- Confusing \ (L_v\) with \ (L_f\): Latent heat of vaporisation (\ (L_v\)) applies to liquid-to-gas transitions. Latent heat of fusion (\ (L_f\)) applies to solid-to-liquid transitions. These are completely different values and must not be interchanged.
- Ignoring the cooling step in steam problems: When steam condenses and then cools to a lower temperature, students often forget to include the heat released during cooling. Both the condensation heat (\ (mL_v\)) and the cooling heat (\ (mc\Delta T\)) must be included in the heat balance.
- Applying the formula during temperature change: The formula \ (Q = mL_v\) applies only at the boiling point, when phase change occurs at constant temperature. If the liquid is still heating up, use \ (Q = mc\Delta T\) instead.
- Forgetting that condensation is the reverse process: The same amount of heat \ (Q = mL_v\) is released when vapour condenses into liquid. Many students think vaporisation and condensation involve different amounts of heat.
JEE/NEET Application of the Heat of Vaporisation Formula
In our experience, JEE aspirants encounter the heat of vaporisation formula most often in calorimetry problems. These problems combine phase changes with temperature changes. Mastering the heat balance equation is essential for scoring in this section.
Application Pattern 1: Steam Condensation Calorimetry
JEE problems frequently involve mixing steam with cold water in a calorimeter. You must account for three separate heat exchanges: steam condensing at 100°C, condensed water cooling to the final temperature, and cold water warming to the final temperature. Set up the equation: heat released = heat absorbed. This pattern appears in JEE Main almost every year.
Application Pattern 2: Clausius–Clapeyron Equation
JEE Advanced questions sometimes ask how boiling point changes with pressure. The Clausius–Clapeyron equation \ (\dfrac{dP}{dT} = \dfrac{L_v}{T \Delta v}\) connects the heat of vaporisation to the pressure–temperature relationship of a liquid–vapour system. Understanding this equation requires a solid grasp of \ (L_v\).
Application Pattern 3: NEET Physiology Context
NEET Biology and Physics both use the concept of heat of vaporisation. The high \ (L_v\) of water (2.26 MJ/kg) explains why sweating is an effective cooling mechanism. Evaporation of just 1 gram of sweat removes 2260 J of heat from the body. NEET questions often ask students to calculate heat lost through sweating using \ (Q = mL_v\).
Our experts suggest practising at least 20 calorimetry problems from previous JEE and NEET papers. Pay special attention to problems that combine \ (Q = mL_v\) and \ (Q = mc\Delta T\) in a single heat balance equation.
FAQs on Heat of Vaporisation Formula
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