The Heat Capacity Formula gives the relationship between the heat energy absorbed or released by a substance and the resulting change in its temperature, expressed as Q = CΔT. This fundamental thermodynamics concept is covered in NCERT Class 11 Physics (Chapter 11 — Thermal Properties of Matter) and is equally important for CBSE board exams and competitive exams like JEE Main, JEE Advanced, and NEET. This article covers the complete formula, specific heat capacity, molar heat capacity, a full formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET applications.
Key Heat Capacity Formulas at a Glance
Quick reference for the most important heat capacity formulas used in CBSE and competitive exams.
- Heat Capacity: \( C = \frac{Q}{\Delta T} \)
- Heat Absorbed/Released: \( Q = C \Delta T \)
- Specific Heat Capacity: \( Q = mc\Delta T \)
- Specific Heat from Heat Capacity: \( c = \frac{C}{m} \)
- Molar Heat Capacity: \( Q = nC_m \Delta T \)
- Molar Heat Capacity from Heat Capacity: \( C_m = \frac{C}{n} \)
- Latent Heat: \( Q = mL \)
What is the Heat Capacity Formula?
The Heat Capacity Formula defines how much heat energy a substance requires to change its temperature by one degree. Heat capacity (symbol C) is a thermodynamic property of a given sample of a substance. It depends on both the type of material and the amount (mass or moles) of that material present.
Formally, heat capacity is the ratio of the heat energy supplied to the temperature rise produced. A substance with a high heat capacity needs more energy to raise its temperature by the same amount compared to a substance with a low heat capacity.
This concept is introduced in NCERT Class 11 Physics, Chapter 11 — Thermal Properties of Matter. It also appears in NCERT Class 11 Chemistry (Chapter 6 — Thermodynamics) in the context of calorimetry. Understanding the Heat Capacity Formula is essential for scoring well in CBSE board exams and for solving numerical problems in JEE Main and NEET.
The SI unit of heat capacity is joule per kelvin (J K−1). The formula connects three measurable quantities: heat energy supplied, the mass or amount of substance, and the temperature change observed.
Heat Capacity Formula — Expression and Variables
The basic Heat Capacity Formula is written as:
\[ C = \frac{Q}{\Delta T} \]
Rearranging to find the heat absorbed or released:
\[ Q = C \Delta T \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( Q \) | Heat energy absorbed or released | Joule (J) |
| \( C \) | Heat capacity of the substance sample | Joule per Kelvin (J K−1) |
| \( \Delta T \) | Change in temperature \( (T_f – T_i) \) | Kelvin (K) or °C |
| \( T_f \) | Final temperature | Kelvin (K) |
| \( T_i \) | Initial temperature | Kelvin (K) |
Derivation of the Heat Capacity Formula
The derivation follows directly from the experimental observation that the heat energy required to change the temperature of a body is proportional to the temperature change.
Step 1: Experimentally, heat supplied \( Q \) is directly proportional to the temperature change \( \Delta T \).
\[ Q \propto \Delta T \]
Step 2: Introduce a proportionality constant \( C \), which depends on the nature and amount of the substance.
\[ Q = C \Delta T \]
Step 3: Rearranging gives the definition of heat capacity.
\[ C = \frac{Q}{\Delta T} \]
This constant \( C \) is the heat capacity of the specific sample. It is an extensive property — it depends on the size of the sample.
Specific Heat Capacity
Specific heat capacity (symbol c or s) is the heat capacity per unit mass. It is an intensive property, meaning it does not depend on the amount of substance. It only depends on the type of material.
\[ Q = mc\Delta T \]
Here, \( m \) is the mass in kilograms and \( c \) is the specific heat capacity in J kg−1 K−1. The relationship between heat capacity and specific heat capacity is:
\[ c = \frac{C}{m} \quad \text{or} \quad C = mc \]
The specific heat capacity of water is 4186 J kg−1 K−1 (approximately 4200 J kg−1 K−1 for calculations). This high value makes water an excellent coolant and heat reservoir in nature.
| Substance | Specific Heat Capacity (J kg−1 K−1) |
|---|---|
| Water | 4186 |
| Aluminium | 900 |
| Iron | 450 |
| Copper | 385 |
| Lead | 128 |
| Ice | 2090 |
| Steam | 2010 |
Molar Heat Capacity
Molar heat capacity (symbol \( C_m \)) is the heat capacity per mole of substance. It is used extensively in chemistry and in the kinetic theory of gases.
\[ Q = nC_m \Delta T \]
Here, \( n \) is the number of moles and \( C_m \) is the molar heat capacity in J mol−1 K−1. The relationship between molar heat capacity and heat capacity is:
\[ C_m = \frac{C}{n} \]
For ideal gases, two important molar heat capacities exist. \( C_V \) is the molar heat capacity at constant volume and \( C_P \) is the molar heat capacity at constant pressure. They are related by Mayer’s relation:
\[ C_P – C_V = R \]
where \( R = 8.314 \) J mol−1 K−1 is the universal gas constant. The ratio \( \gamma = C_P / C_V \) is called the adiabatic index and is crucial for JEE problems on adiabatic processes.
Difference Between Heat Capacity and Specific Heat Capacity
| Parameter | Heat Capacity (C) | Specific Heat Capacity (c) |
|---|---|---|
| Definition | Heat needed to raise temperature of a given sample by 1 K | Heat needed to raise temperature of 1 kg of substance by 1 K |
| Symbol | C | c or s |
| SI Unit | J K−1 | J kg−1 K−1 |
| Type of property | Extensive (depends on amount) | Intensive (independent of amount) |
| Formula | \( C = Q / \Delta T \) | \( c = Q / (m \Delta T) \) |
| Depends on mass? | Yes | No |
Complete Thermodynamics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat Capacity | \( C = Q / \Delta T \) | C = heat capacity, Q = heat, ΔT = temp change | J K−1 | Class 11, Ch 11 |
| Specific Heat Capacity | \( Q = mc\Delta T \) | m = mass, c = specific heat, ΔT = temp change | J kg−1 K−1 | Class 11, Ch 11 |
| Molar Heat Capacity | \( Q = nC_m \Delta T \) | n = moles, C_m = molar heat capacity | J mol−1 K−1 | Class 11, Ch 11 |
| Mayer’s Relation | \( C_P – C_V = R \) | C_P = heat capacity at const. pressure, C_V = at const. volume | J mol−1 K−1 | Class 11, Ch 12 |
| Adiabatic Index | \( \gamma = C_P / C_V \) | γ = ratio of heat capacities | Dimensionless | Class 11, Ch 12 |
| Latent Heat | \( Q = mL \) | L = specific latent heat | J kg−1 | Class 11, Ch 11 |
| Calorimetry (heat exchange) | \( m_1 c_1 \Delta T_1 = m_2 c_2 \Delta T_2 \) | Subscripts 1 and 2 denote two substances | J | Class 11, Ch 11 |
| Thermal Expansion (linear) | \( \Delta L = L_0 \alpha \Delta T \) | α = linear expansion coefficient, L_0 = original length | m | Class 11, Ch 11 |
| Newton’s Law of Cooling | \( \frac{dQ}{dt} = -k(T – T_0) \) | k = cooling constant, T_0 = ambient temperature | 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 |
Heat Capacity Formula — Solved Examples
Example 1 (Class 9-10 Level): Finding Heat Absorbed by Water
Problem: How much heat energy is required to raise the temperature of 2 kg of water from 25°C to 75°C? (Specific heat capacity of water = 4200 J kg−1 K−1)
Given:
- Mass, \( m = 2 \) kg
- Initial temperature, \( T_i = 25 \)°C
- Final temperature, \( T_f = 75 \)°C
- Specific heat capacity, \( c = 4200 \) J kg−1 K−1
Step 1: Calculate the temperature change.
\( \Delta T = T_f – T_i = 75 – 25 = 50 \)°C = 50 K
Step 2: Write the specific heat capacity formula.
\( Q = mc\Delta T \)
Step 3: Substitute the values.
\( Q = 2 \times 4200 \times 50 \)
Step 4: Calculate.
\( Q = 420{,}000 \) J \( = 4.2 \times 10^5 \) J
Answer
Heat energy required = \( 4.2 \times 10^5 \) J (420 kJ)
Example 2 (Class 11-12 Level): Calorimetry — Finding Final Temperature
Problem: A 500 g iron block at 200°C is dropped into 1 kg of water at 20°C. Find the final equilibrium temperature. (Specific heat of iron = 450 J kg−1 K−1, specific heat of water = 4200 J kg−1 K−1. Assume no heat loss to surroundings.)
Given:
- Mass of iron, \( m_1 = 0.5 \) kg; \( c_1 = 450 \) J kg−1 K−1; \( T_1 = 200 \)°C
- Mass of water, \( m_2 = 1 \) kg; \( c_2 = 4200 \) J kg−1 K−1; \( T_2 = 20 \)°C
- Final temperature = \( T \) (unknown)
Step 1: Apply the principle of calorimetry. Heat lost by iron = Heat gained by water.
\( m_1 c_1 (T_1 – T) = m_2 c_2 (T – T_2) \)
Step 2: Substitute values.
\( 0.5 \times 450 \times (200 – T) = 1 \times 4200 \times (T – 20) \)
Step 3: Simplify the left side and right side.
\( 225(200 – T) = 4200(T – 20) \)
\( 45000 – 225T = 4200T – 84000 \)
Step 4: Collect terms and solve.
\( 45000 + 84000 = 4200T + 225T \)
\( 129000 = 4425T \)
\( T = \frac{129000}{4425} \approx 29.15 \)°C
Answer
Final equilibrium temperature ≈ 29.15°C
Example 3 (JEE/NEET Level): Molar Heat Capacity and Mayer’s Relation
Problem: One mole of an ideal diatomic gas is heated at constant pressure. The temperature rises from 300 K to 500 K. Calculate (a) the heat supplied, (b) the change in internal energy, and (c) the work done by the gas. (For diatomic gas: \( C_V = \frac{5}{2}R \), \( R = 8.314 \) J mol−1 K−1)
Given:
- \( n = 1 \) mol
- \( T_i = 300 \) K, \( T_f = 500 \) K, so \( \Delta T = 200 \) K
- \( C_V = \frac{5}{2}R \)
Step 1: Find \( C_P \) using Mayer’s relation.
\( C_P = C_V + R = \frac{5}{2}R + R = \frac{7}{2}R \)
\( C_P = \frac{7}{2} \times 8.314 = 29.099 \) J mol−1 K−1
Step 2: Calculate heat supplied at constant pressure.
\( Q = nC_P \Delta T = 1 \times 29.099 \times 200 = 5819.8 \) J \( \approx 5820 \) J
Step 3: Calculate change in internal energy.
\( \Delta U = nC_V \Delta T = 1 \times \frac{5}{2} \times 8.314 \times 200 = 4157 \) J
Step 4: Calculate work done using the first law of thermodynamics.
\( W = Q – \Delta U = 5820 – 4157 = 1663 \) J
(This equals \( nR\Delta T = 1 \times 8.314 \times 200 = 1662.8 \) J — consistent check.)
Answer
(a) Heat supplied \( Q \approx 5820 \) J (b) Change in internal energy \( \Delta U \approx 4157 \) J (c) Work done \( W \approx 1663 \) J
CBSE Exam Tips 2025-26
- Write the formula first: In CBSE answer sheets, always state the formula \( Q = mc\Delta T \) before substituting values. This earns formula marks even if the final answer has a calculation error.
- Unit consistency: Always convert mass to kilograms and temperature difference to kelvin (or °C — the numerical value of ΔT is the same in both). Forgetting to convert grams to kilograms is the most common error in board exams.
- Distinguish heat capacity from specific heat capacity: CBSE 2025-26 frequently tests this distinction. Heat capacity depends on the sample size; specific heat capacity does not. We recommend memorising this difference with an example.
- Calorimetry problems: For heat exchange problems, clearly state the principle — “heat lost by hot body = heat gained by cold body” — before writing the equation. This is worth 1 mark in CBSE marking schemes.
- Standard values to memorise: Learn the specific heat of water (4200 J kg−1 K−1), ice (2100 J kg−1 K−1), and iron (450 J kg−1 K−1). These appear repeatedly in CBSE numerical problems.
- Sign convention: When a body loses heat, \( Q \) is negative and \( \Delta T \) is negative. Their ratio (heat capacity) remains positive. Always check the sign of your answer for physical reasonableness.
Common Mistakes to Avoid
| Mistake | Why It’s Wrong | Correct Approach |
|---|---|---|
| Confusing heat capacity (C) with specific heat capacity (c) | Heat capacity depends on mass; specific heat capacity does not. Using them interchangeably gives wrong answers. | Use \( C = mc \) to relate them. Always check whether mass is given or not. |
| Using mass in grams instead of kilograms | SI unit of specific heat is J kg−1 K−1. Using grams gives an answer 1000 times too large. | Always convert: 500 g = 0.5 kg before substituting. |
| Using temperature in °C when kelvin is required | For ΔT, °C and K give the same numerical value. But for absolute temperature (e.g., Stefan’s law), kelvin is mandatory. | For ΔT, either unit works. For T in other formulas, always use kelvin. |
| Forgetting to apply the calorimetry principle correctly | Students sometimes add heat energies instead of equating heat lost to heat gained. | Write \( m_1 c_1 (T_1 – T_f) = m_2 c_2 (T_f – T_2) \) and ensure both sides are positive. |
| Mixing up \( C_P \) and \( C_V \) for gases | Using \( C_V \) for a constant-pressure process (or vice versa) gives incorrect heat values. | Identify the process first. Use \( C_P \) for constant pressure and \( C_V \) for constant volume. |
JEE/NEET Application of Heat Capacity Formula
The Heat Capacity Formula appears in multiple contexts across JEE Main, JEE Advanced, and NEET. In our experience, JEE aspirants encounter this formula in at least 2-3 questions per paper, spanning thermodynamics, kinetic theory, and calorimetry.
Application Pattern 1: Calorimetry and Mixing Problems (JEE Main, NEET)
These problems involve two or more substances exchanging heat until equilibrium. The key equation is \( \sum m_i c_i \Delta T_i = 0 \). JEE often adds a twist by including a phase change (latent heat), requiring you to check whether the final temperature involves melting or boiling.
Application Pattern 2: Molar Heat Capacities and Thermodynamic Processes (JEE Advanced)
JEE Advanced frequently tests Mayer’s relation \( C_P – C_V = R \) and the adiabatic index \( \gamma = C_P/C_V \). For a monatomic ideal gas, \( C_V = \frac{3}{2}R \) and \( C_P = \frac{5}{2}R \), giving \( \gamma = 5/3 \). For a diatomic gas, \( C_V = \frac{5}{2}R \) and \( C_P = \frac{7}{2}R \), giving \( \gamma = 7/5 \). These values are non-negotiable for adiabatic process problems.
Application Pattern 3: Polytropic Processes (JEE Advanced)
For a polytropic process \( PV^n = \text{constant} \), the effective molar heat capacity is:
\[ C = C_V – \frac{R}{n-1} \]
Here \( n \) is the polytropic index (not the number of moles). This formula combines the Heat Capacity Formula with the first law of thermodynamics. Our experts suggest practising at least 10 polytropic problems from previous JEE Advanced papers to build fluency.
For NEET, the focus is primarily on \( Q = mc\Delta T \) and calorimetry. NEET 2024 and 2023 both featured one direct numerical on specific heat capacity. Knowing the standard values of specific heat for common substances (water, ice, metals) is sufficient for NEET preparation.
FAQs on Heat Capacity Formula
Explore More Physics Formulas
Now that you have mastered the Heat Capacity Formula, strengthen your thermodynamics and physics preparation with these related resources on ncertbooks.net:
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For the official NCERT syllabus and textbooks, visit ncert.nic.in.