The Thermal Energy Formula gives the total internal energy of a system due to the random motion of its particles, expressed as \ ( Q = mc\Delta T \) for heat transfer calculations. This formula is a cornerstone of Class 10 and Class 11 Physics, appearing in the chapters on Heat and Thermodynamics. It is equally critical for JEE Main, JEE Advanced, and NEET aspirants who must solve calorimetry and thermodynamics problems under time pressure. This article covers the formula expression, derivation, a complete formula sheet, three progressive solved examples, CBSE exam tips, common mistakes, and JEE/NEET application patterns.
Key Thermal Energy Formulas at a Glance
Quick reference for the most important thermal energy and heat formulas.
- Heat absorbed or released: \( Q = mc\Delta T \)
- Thermal energy (kinetic theory): \( E_{thermal} = \frac{3}{2} nRT \)
- Heat for phase change (latent heat): \( Q = mL \)
- Newton's Law of Cooling: \( \frac{dT}{dt} = -k(T – T_0) \)
- Stefan–Boltzmann Law: \( P = \sigma A T^4 \)
- Thermal conductivity: \( Q = \frac{kA(T_1 – T_2)t}{d} \)
- First Law of Thermodynamics: \( \Delta U = Q – W \)
What is the Thermal Energy Formula?
The Thermal Energy Formula describes the energy stored within a substance due to the kinetic energy of its constituent particles. Every object above absolute zero temperature possesses thermal energy. When heat flows into or out of a body, this energy changes. The primary formula used to calculate this change is \( Q = mc\Delta T \), where Q is the heat energy transferred, m is the mass of the substance, c is the specific heat capacity, and \( \Delta T \) is the change in temperature.
In the NCERT curriculum, thermal energy concepts appear in Class 10 Science (Chapter 11: Heat) and Class 11 Physics (Chapter 11: Thermal Properties of Matter). At the molecular level, the kinetic theory of gases defines the thermal energy of an ideal gas as \( E_{thermal} = \frac{3}{2} nRT \), where n is the number of moles and R is the universal gas constant. Understanding both expressions is essential for a complete grasp of the topic.
Thermal energy is distinct from heat. Heat is the energy in transit, while thermal energy is the energy stored inside the system. This distinction is frequently tested in CBSE board exams and JEE.
Thermal Energy Formula — Expression and Variables
The two most important expressions for thermal energy are given below.
1. Heat Transfer Formula (Calorimetry):
\[ Q = mc\Delta T \]
2. Thermal Energy of an Ideal Gas (Kinetic Theory):
\[ E_{thermal} = \frac{3}{2} nRT \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | Heat energy transferred | Joule (J) |
| m | Mass of the substance | Kilogram (kg) |
| c | Specific heat capacity | J kg¹ K¹ |
| \( \Delta T \) | Change in temperature \( (T_f – T_i) \) | Kelvin (K) or °C |
| \( E_{thermal} \) | Thermal energy of ideal gas | Joule (J) |
| n | Number of moles of gas | mol |
| R | Universal gas constant | 8.314 J mol¹ K¹ |
| T | Absolute temperature | Kelvin (K) |
Derivation of Q = mcΔT
The derivation begins with the experimental observation that the heat absorbed by a body is directly proportional to its mass and to the rise in temperature. Mathematically:
\[ Q \propto m \quad \text{and} \quad Q \propto \Delta T \]
Combining both proportionalities:
\[ Q \propto m \Delta T \]
Introducing the constant of proportionality c, called the specific heat capacity of the substance, we get:
\[ Q = mc\Delta T \]
The specific heat capacity c depends on the nature of the material. For water, \( c = 4200 \) J kg¹ K¹. This high value explains why water is used as a coolant in engines and radiators.
Derivation of Thermal Energy for an Ideal Gas
From the kinetic theory of gases, the average kinetic energy per molecule along one degree of freedom is \( \frac{1}{2}k_BT \). A monatomic ideal gas has 3 translational degrees of freedom. For n moles containing \( N = nN_A \) molecules:
\[ E_{thermal} = N \times \frac{3}{2}k_BT = nN_A \times \frac{3}{2}k_BT = \frac{3}{2}nRT \]
Here \( R = N_A k_B = 8.314 \) J mol¹ K¹ is the universal gas constant.
Complete Thermal Energy Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat Transfer (Calorimetry) | \( Q = mc\Delta T \) | m = mass, c = specific heat, ΔT = temp change | Joule (J) | Class 11, Ch 11 |
| Thermal Energy of Ideal Gas | \( E = \frac{3}{2}nRT \) | n = moles, R = gas constant, T = temperature | Joule (J) | Class 11, Ch 13 |
| Latent Heat Formula | \( Q = mL \) | m = mass, L = latent heat | Joule (J) | Class 11, Ch 11 |
| First Law of Thermodynamics | \( \Delta U = Q – W \) | ΔU = internal energy change, Q = heat, W = work | Joule (J) | Class 11, Ch 12 |
| Thermal Conductivity | \( Q = \frac{kA(T_1-T_2)t}{d} \) | k = conductivity, A = area, d = thickness, t = time | Joule (J) | Class 11, Ch 11 |
| Newton's Law of Cooling | \( \frac{dT}{dt} = -k(T – T_0) \) | T = object temp, T⊂0; = surrounding temp, k = constant | K/s | Class 11, Ch 11 |
| Stefan–Boltzmann Law | \( P = \sigma A T^4 \) | σ = 5.67×10&sup8; W m² K&sup4;, A = area, T = temp | Watt (W) | Class 11, Ch 11 |
| Wien's Displacement Law | \( \lambda_{max} T = b \) | λmax = peak wavelength, b = 2.898×10³ m·K | m·K | Class 11, Ch 11 |
| Equipartition of Energy | \( E = \frac{f}{2}k_BT \) | f = degrees of freedom, k⊂B; = Boltzmann constant | Joule (J) | Class 11, Ch 13 |
| Internal Energy Change (Ideal Gas) | \( \Delta U = nC_v\Delta T \) | C⊂v; = molar heat capacity at constant volume | Joule (J) | Class 11, Ch 12 |
Thermal Energy Formula — Solved Examples
Example 1 (Class 9–10 Level): Heat Absorbed by Water
Problem: A 2 kg sample of water is heated from 25°C to 75°C. Calculate the heat energy absorbed. (Specific heat capacity of water = 4200 J kg¹ K¹)
Given: m = 2 kg, \( T_i \) = 25°C, \( T_f \) = 75°C, c = 4200 J kg¹ K¹
Step 1: Find the change in temperature.
\( \Delta T = T_f – T_i = 75 – 25 = 50 \) °C = 50 K
Step 2: Apply the Thermal Energy Formula \( Q = mc\Delta T \).
\( Q = 2 \times 4200 \times 50 \)
Step 3: Calculate.
\( Q = 420000 \) J = 420 kJ
Answer
Heat absorbed = 420,000 J (420 kJ)
Example 2 (Class 11–12 Level): Calorimetry — Mixing Two Liquids
Problem: 500 g of water at 80°C is mixed with 300 g of water at 20°C in a thermally insulated container. Find the final equilibrium temperature. (Specific heat capacity of water = 4200 J kg¹ K¹)
Given: \( m_1 \) = 0.5 kg, \( T_1 \) = 80°C; \( m_2 \) = 0.3 kg, \( T_2 \) = 20°C
Step 1: Apply the principle of calorimetry. Heat lost by hot water = Heat gained by cold water.
\( m_1 c (T_1 – T_f) = m_2 c (T_f – T_2) \)
Step 2: Since c is the same for both, cancel it.
\( 0.5(80 – T_f) = 0.3(T_f – 20) \)
Step 3: Expand and solve.
\( 40 – 0.5T_f = 0.3T_f – 6 \)
\( 46 = 0.8T_f \)
\( T_f = \frac{46}{0.8} = 57.5 \)°C
Answer
Final equilibrium temperature = 57.5°C
Example 3 (JEE/NEET Level): Thermal Energy of an Ideal Gas and Internal Energy Change
Problem: Two moles of a monatomic ideal gas are heated at constant volume from 300 K to 500 K. Calculate (a) the increase in thermal energy of the gas and (b) the heat supplied to the gas. (R = 8.314 J mol¹ K¹)
Given: n = 2 mol, \( T_1 \) = 300 K, \( T_2 \) = 500 K, monatomic gas (degrees of freedom f = 3)
Step 1: For a monatomic ideal gas, molar heat capacity at constant volume is \( C_v = \frac{3}{2}R \).
Step 2: Calculate the change in internal (thermal) energy using \( \Delta U = nC_v\Delta T \).
\( \Delta T = 500 – 300 = 200 \) K
\( \Delta U = 2 \times \frac{3}{2} \times 8.314 \times 200 \)
\( \Delta U = 2 \times 1.5 \times 8.314 \times 200 = 4988.4 \) J
Step 3: Since the process is at constant volume, no work is done (W = 0). By the First Law of Thermodynamics \( \Delta U = Q – W \):
\( Q = \Delta U + W = 4988.4 + 0 = 4988.4 \) J
Answer
(a) Increase in thermal energy = 4988.4 J ≈ 4988 J
(b) Heat supplied = 4988.4 J (equal to ΔU since W = 0 at constant volume)
CBSE Exam Tips 2025-26
- Always convert temperature to Kelvin when using the ideal gas thermal energy formula \( E = \frac{3}{2}nRT \). Using Celsius will give a completely wrong answer.
- Memorise specific heat values: Water = 4200 J kg¹ K¹, Ice = 2100 J kg¹ K¹, Aluminium = 900 J kg¹ K¹. These appear directly in CBSE numericals.
- Distinguish between Q and ΔU. In the 2025-26 CBSE syllabus, questions on the First Law frequently ask students to identify which quantity is thermal energy and which is heat. We recommend practising at least 10 such distinction questions.
- For calorimetry problems, always write “Heat lost = Heat gained” as the first line. This earns a step mark even if the final answer has an arithmetic error.
- Check units before substituting. Mass must be in kg, temperature change in K (or °C for ΔT only), and specific heat in J kg¹ K¹. A unit mismatch is the most common reason for losing marks.
- For latent heat questions, remember that \( Q = mL \) has no ΔT term. Temperature does not change during a phase transition. Our experts suggest drawing a heating curve to visualise this clearly.
Common Mistakes to Avoid
- Mistake 1: Confusing thermal energy with temperature. Temperature is a measure of average kinetic energy per particle. Thermal energy is the total internal energy of all particles. A large cold lake has more thermal energy than a small hot cup of tea.
- Mistake 2: Using Celsius instead of Kelvin for the gas formula. The expression \( E = \frac{3}{2}nRT \) requires absolute temperature T in Kelvin. Always add 273 (or 273.15) to convert from Celsius.
- Mistake 3: Applying Q = mcΔT during a phase change. When a substance is melting or boiling, temperature stays constant. Use \( Q = mL \) instead. Using the wrong formula here is a very common board exam error.
- Mistake 4: Ignoring the sign of Q. When a body loses heat, Q is negative. When it gains heat, Q is positive. In calorimetry, the sign convention ensures heat lost + heat gained = 0 for an isolated system.
- Mistake 5: Using the wrong value of degrees of freedom. For JEE problems, monatomic gases have f = 3, diatomic gases have f = 5 (at room temperature), and polyatomic gases have f = 6. Using f = 3 for a diatomic gas like O⊂2; is a very frequent error in competitive exams.
JEE/NEET Application of Thermal Energy Formula
In our experience, JEE aspirants encounter the Thermal Energy Formula in three major contexts: calorimetry problems, thermodynamic process calculations, and kinetic theory questions. Each pattern requires a slightly different approach.
Pattern 1: Calorimetry and Heat Exchange
JEE Main frequently tests problems where two or more substances at different temperatures are mixed. The principle is conservation of energy: total heat lost = total heat gained. Students must apply \( Q = mc\Delta T \) for each substance and solve the resulting equation. Phase changes add complexity, requiring the use of \( Q = mL \) at transition points.
Pattern 2: Internal Energy and Thermodynamic Processes
JEE Advanced questions often combine the thermal energy formula with the First Law \( \Delta U = Q – W \). For an isothermal process, \( \Delta U = 0 \), so Q = W. For an adiabatic process, Q = 0, so \( \Delta U = -W \). Knowing when thermal energy changes and when it does not is key to scoring in this section. NEET also tests these concepts in the context of biological systems and metabolic heat.
Pattern 3: Equipartition Theorem and Degrees of Freedom
JEE Advanced problems on kinetic theory ask students to calculate the total thermal energy using \( E = \frac{f}{2}nRT \), where f is the number of degrees of freedom. For a diatomic gas, \( E = \frac{5}{2}nRT \). The ratio of specific heats \( \gamma = C_p/C_v = 1 + \frac{2}{f} \) is derived directly from this. Our experts suggest memorising the values of \( \gamma \) for monatomic (1.67), diatomic (1.40), and polyatomic (1.33) gases, as they appear in both JEE and NEET.
FAQs on Thermal Energy Formula
Explore More Physics Formulas
Now that you have mastered the Thermal Energy Formula, strengthen your Physics preparation with these related topics. Study the Buoyancy Formula to understand pressure and fluid mechanics. Explore Bernoulli's Equation Formula for fluid flow and energy conservation in fluids. For optics, the Brewster's Law Formula covers polarisation of light. You can also browse our complete Physics Formulas hub for a full list of NCERT-aligned formula articles for Class 6–12 and competitive exams. For the official NCERT textbook content, refer to the NCERT official website.