The Heat Formula \ ( Q = mc\Delta T \) is one of the most fundamental expressions in thermal physics, used to calculate the amount of heat energy absorbed or released by a substance when its temperature changes. Covered in NCERT Class 10 (Chapter 11) and revisited in Class 11 Physics (Chapter 11 — Thermal Properties of Matter), this formula is essential for CBSE board exams and appears regularly in JEE Main and NEET. This article covers the complete heat formula, its derivation, a full formula sheet, three solved examples at progressive difficulty, CBSE exam tips for 2025-26, and JEE/NEET application strategies.
Key Heat Formulas at a Glance
Quick reference for the most important heat and thermal energy formulas.
- Heat absorbed or released: \( Q = mc\Delta T \)
- Temperature change: \( \Delta T = T_f – T_i \)
- Heat of fusion (melting): \( Q = mL_f \)
- Heat of vaporisation: \( Q = mL_v \)
- Newton’s Law of Cooling: \( \frac{dQ}{dt} = -k(T – T_0) \)
- Thermal conductivity: \( Q = \frac{kA(T_1 – T_2)t}{d} \)
- Stefan–Boltzmann Law: \( P = \sigma A T^4 \)
What is the Heat Formula?
The Heat Formula expresses the relationship between heat energy, mass, specific heat capacity, and temperature change. Heat is a form of energy that flows from a body at a higher temperature to a body at a lower temperature. It is not the same as temperature. Temperature measures the average kinetic energy of molecules. Heat measures the total thermal energy transferred.
In NCERT Class 10 Science (Chapter 11 — Work and Energy) and Class 11 Physics (Chapter 11 — Thermal Properties of Matter), heat energy is defined as the energy exchanged between a system and its surroundings due to a temperature difference. The three modes of heat transfer are conduction, convection, and radiation.
The primary heat formula is written as:
Q = mcΔT
Here, Q is the heat energy in Joules, m is the mass of the substance in kilograms, c is the specific heat capacity in J kg−1 K−1, and ΔT is the change in temperature in Kelvin or Celsius. This formula applies when there is no change of state. When a substance melts or boils, the latent heat formulas apply instead.
Heat Formula — Expression and Variables
\[ Q = mc\Delta T \]
Where \( \Delta T = T_f – T_i \), with \( T_f \) being the final temperature and \( T_i \) the initial temperature.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | Heat energy absorbed or released | Joule (J) |
| m | Mass of the substance | Kilogram (kg) |
| c | Specific heat capacity | J kg−1 K−1 |
| ΔT | Change in temperature | Kelvin (K) or Celsius (°C) |
| Tf | Final temperature | K or °C |
| Ti | Initial temperature | K or °C |
Derivation of the Heat Formula
The heat formula is derived from the definition of specific heat capacity. Specific heat capacity (c) of a substance is defined as the amount of heat required to raise the temperature of 1 kg of that substance by 1 K.
Step 1: By definition, heat required to raise 1 kg by 1 K equals c (in J kg−1 K−1).
Step 2: For a mass m kg, heat required to raise temperature by 1 K is \( m \times c \).
Step 3: For a temperature change of \( \Delta T \) K, the total heat is \( Q = mc\Delta T \).
If Q is positive, the substance absorbs heat. If Q is negative, the substance releases heat. This sign convention is critical in calorimetry problems.
Specific Heat Capacity of Common Substances
| Substance | Specific Heat Capacity (J kg−1 K−1) |
|---|---|
| Water | 4186 |
| Ice | 2090 |
| Steam | 2010 |
| Aluminium | 900 |
| Iron | 450 |
| Copper | 385 |
| Lead | 128 |
| Air (at constant pressure) | 1005 |
Complete Thermal Physics Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat Energy (sensible heat) | \( Q = mc\Delta T \) | m = mass, c = specific heat, ΔT = temp. change | Joule (J) | Class 11, Ch 11 |
| Latent Heat of Fusion | \( Q = mL_f \) | m = mass, Lf = latent heat of fusion | J | Class 11, Ch 11 |
| Latent Heat of Vaporisation | \( Q = mL_v \) | m = mass, Lv = latent heat of vaporisation | J | Class 11, Ch 11 |
| Thermal Conductivity (Fourier’s Law) | \( Q = \frac{kA(T_1 – T_2)t}{d} \) | k = thermal conductivity, A = area, d = thickness, t = time | J | Class 11, Ch 11 |
| Newton’s Law of Cooling | \( \frac{dQ}{dt} = -k(T – T_0) \) | k = cooling constant, T = body temp., T0 = ambient temp. | W (J s−1) | Class 11, Ch 11 |
| Stefan–Boltzmann Law | \( P = \sigma A T^4 \) | σ = 5.67 × 10−8 W m−2 K−4, A = area, T = absolute temp. | Watt (W) | Class 11, Ch 11 |
| Wien’s Displacement Law | \( \lambda_{max} T = b \) | λmax = peak wavelength, b = 2.898 × 10−3 m K | m K | Class 11, Ch 11 |
| Heat Capacity | \( C = mc \) | m = mass, c = specific heat capacity | J K−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 refer to the two substances | J | Class 11, Ch 11 |
| Ideal Gas Internal Energy Change | \( \Delta U = nC_v \Delta T \) | n = moles, Cv = molar heat capacity at constant volume | J | Class 11, Ch 12 |
| First Law of Thermodynamics | \( \Delta U = Q – W \) | ΔU = internal energy change, Q = heat added, W = work done | J | Class 11, Ch 12 |
Heat Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: How much heat 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, Ti = 25°C
- Final temperature, Tf = 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 (or 50 K)
Step 2: Apply the heat formula.
\( Q = mc\Delta T \)
Step 3: Substitute the values.
\( Q = 2 \times 4200 \times 50 \)
Step 4: Calculate.
\( Q = 420{,}000 \) J = 420 kJ
Answer
Heat required = 420,000 J (420 kJ)
Example 2 (Class 11-12 Level — Calorimetry)
Problem: A 0.5 kg iron block at 200°C is dropped into 2 kg of water at 20°C in a calorimeter. Find the final equilibrium temperature. (ciron = 450 J kg−1 K−1, cwater = 4200 J kg−1 K−1. Ignore heat loss to the calorimeter.)
Given:
- Mass of iron, m1 = 0.5 kg; c1 = 450 J kg−1 K−1; T1 = 200°C
- Mass of water, m2 = 2 kg; c2 = 4200 J kg−1 K−1; T2 = 20°C
Step 1: Apply the principle of calorimetry. Heat lost by iron = Heat gained by water.
\( m_1 c_1 (T_1 – T_f) = m_2 c_2 (T_f – T_2) \)
Step 2: Substitute values.
\( 0.5 \times 450 \times (200 – T_f) = 2 \times 4200 \times (T_f – 20) \)
Step 3: Simplify both sides.
\( 225(200 – T_f) = 8400(T_f – 20) \)
\( 45000 – 225T_f = 8400T_f – 168000 \)
Step 4: Solve for Tf.
\( 45000 + 168000 = 8400T_f + 225T_f \)
\( 213000 = 8625 T_f \)
\( T_f = \frac{213000}{8625} \approx 24.7 \)°C
Answer
Final equilibrium temperature ≈ 24.7°C
Example 3 (JEE/NEET Level — Phase Change + Sensible Heat)
Problem: Calculate the total heat required to convert 500 g of ice at −10°C to steam at 120°C. (cice = 2100 J kg−1 K−1, Lf = 3.36 × 105 J kg−1, cwater = 4200 J kg−1 K−1, Lv = 2.26 × 106 J kg−1, csteam = 2010 J kg−1 K−1)
Given: m = 0.5 kg, with five distinct stages of heating.
Step 1: Heat ice from −10°C to 0°C (no phase change).
\( Q_1 = mc_{ice}\Delta T = 0.5 \times 2100 \times 10 = 10{,}500 \) J
Step 2: Melt ice at 0°C (phase change — latent heat of fusion).
\( Q_2 = mL_f = 0.5 \times 3.36 \times 10^5 = 1{,}68{,}000 \) J
Step 3: Heat water from 0°C to 100°C.
\( Q_3 = mc_{water}\Delta T = 0.5 \times 4200 \times 100 = 2{,}10{,}000 \) J
Step 4: Vaporise water at 100°C (latent heat of vaporisation).
\( Q_4 = mL_v = 0.5 \times 2.26 \times 10^6 = 11{,}30{,}000 \) J
Step 5: Heat steam from 100°C to 120°C.
\( Q_5 = mc_{steam}\Delta T = 0.5 \times 2010 \times 20 = 20{,}100 \) J
Step 6: Total heat = Q1 + Q2 + Q3 + Q4 + Q5
\( Q_{total} = 10500 + 168000 + 210000 + 1130000 + 20100 \)
\( Q_{total} = 15{,}38{,}600 \) J ≈ 1538.6 kJ
Answer
Total heat required = 1,538,600 J ≈ 1538.6 kJ
CBSE Exam Tips 2025-26
- Always write the formula first. In CBSE board exams, examiners award marks for writing \( Q = mc\Delta T \) before substituting values. Never skip this step.
- Check units carefully. Mass must be in kilograms, not grams. Convert grams to kilograms by dividing by 1000 before substituting. This is the single most common error in board exams.
- Note the sign of ΔT. If the substance cools down, ΔT is negative and Q is negative. This means heat is released. CBSE questions sometimes ask whether heat is absorbed or released — the sign tells you directly.
- Distinguish sensible heat from latent heat. The formula \( Q = mc\Delta T \) applies only when there is no change of state. For melting or boiling, use \( Q = mL \). Many students incorrectly apply the wrong formula during phase changes.
- Memorise specific heat of water. The value 4200 J kg−1 K−1 (or 4186 J kg−1 K−1 precisely) appears in nearly every CBSE heat problem. We recommend writing it at the top of your answer sheet during the exam for quick reference.
- Calorimetry questions are high-yield. In 2025-26 board exams, 3-mark and 5-mark questions frequently involve the principle of calorimetry. Practise setting up \( m_1 c_1 \Delta T_1 = m_2 c_2 \Delta T_2 \) quickly and solving for the unknown variable.
Common Mistakes to Avoid
Our experts have identified the most frequent errors students make when applying the heat formula. Avoiding these will directly improve your score.
-
Mistake 1: Using grams instead of kilograms.
The SI unit of mass is the kilogram. If the problem gives mass in grams, convert it first. For example, 500 g = 0.5 kg. Substituting 500 directly gives an answer 1000 times too large. -
Mistake 2: Applying \( Q = mc\Delta T \) during a phase change.
During melting or boiling, temperature does not change (ΔT = 0). The heat formula gives Q = 0, which is wrong. Use \( Q = mL_f \) or \( Q = mL_v \) during phase changes. -
Mistake 3: Confusing heat capacity with specific heat capacity.
Heat capacity C = mc (unit: J K−1) refers to the whole object. Specific heat capacity c (unit: J kg−1 K−1) is a property per unit mass. These are different quantities with different units. -
Mistake 4: Ignoring the sign convention in calorimetry.
Heat lost by the hot body must equal heat gained by the cold body. Always write: Heat lost = Heat gained. Do not add both as positive on the same side of the equation. -
Mistake 5: Mixing up Celsius and Kelvin for ΔT.
A change of 1°C equals a change of 1 K. So ΔT is numerically the same in Celsius and Kelvin. However, absolute temperature (used in Stefan–Boltzmann Law) must always be in Kelvin. Students sometimes substitute Celsius values in radiation formulas — this is a critical error.
JEE/NEET Application of the Heat Formula
In our experience, JEE aspirants encounter the heat formula in at least 2–3 questions per paper, often combined with thermodynamics or kinetic theory. NEET questions focus more on straightforward calorimetry and phase-change calculations. Here are the key application patterns to master.
Pattern 1: Multi-Stage Heating Problems (JEE Main & NEET)
These problems involve taking a substance through multiple phases. You must calculate Q for each stage separately and add them. The ice-to-steam problem (Example 3 above) is the classic template. JEE Main frequently asks for the heat required for one specific stage, so read the question carefully.
Pattern 2: Calorimetry and Mixing Problems (JEE Main)
Two or more substances at different temperatures are mixed. You must find the equilibrium temperature using \( \sum m_i c_i \Delta T_i = 0 \). JEE problems sometimes include a solid melting inside water, requiring you to combine \( Q = mc\Delta T \) with \( Q = mL \). Always check whether the final state involves a phase change before assuming a single formula applies.
Pattern 3: Heat Transfer via Conduction (JEE Advanced)
JEE Advanced tests Fourier’s Law of conduction: \( Q = \frac{kA(T_1 – T_2)t}{d} \). Problems often involve composite slabs in series or parallel. The thermal resistance analogy (similar to electrical resistance) is the key tool. In series, \( R_{total} = R_1 + R_2 \). In parallel, \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \).
Pattern 4: Newton’s Law of Cooling (NEET & JEE Main)
NEET frequently asks numerical problems based on Newton’s Law of Cooling. The simplified form is: \( \frac{T_1 – T_2}{t} = k\left(\frac{T_1 + T_2}{2} – T_0\right) \). This is the average temperature form used for short cooling intervals. Practise identifying T0 (ambient temperature) correctly from the problem statement.
FAQs on the Heat Formula
Explore More Physics Formulas
Now that you have mastered the Heat Formula, strengthen your physics preparation with these related topics. Understanding thermal energy connects directly to electrical energy and wave behaviour.
- Explore the Electric Current Formula to see how energy transfer applies in circuits.
- Study the Current Density Formula for deeper understanding of charge flow and energy.
- Review the Terminal Velocity Formula to connect energy concepts with mechanics.
- Visit our complete Physics Formulas hub for the full list of NCERT-aligned formula guides.
- For official NCERT textbook content, refer to the NCERT official website.