The Heat Flux Formula gives the rate of thermal energy transfer per unit area across a surface, expressed as \( q = Q / (A \cdot t) \) or equivalently \( q = k \cdot \Delta T / d \). It is a fundamental concept covered in Class 11 Physics (Thermal Properties of Matter) and is equally important for JEE Main, JEE Advanced, and NEET examinations. This article covers the complete derivation, variables, a ready-to-use formula sheet, three progressive solved examples, CBSE exam tips for 2025-26, and common mistakes students make.
Key Heat Flux Formulas at a Glance
Quick reference for the most important heat flux and heat transfer formulas.
- Basic heat flux: \( q = \dfrac{Q}{A \cdot t} \)
- Conduction (Fourier’s Law): \( q = -k \dfrac{dT}{dx} \)
- Convection (Newton’s Law of Cooling): \( q = h (T_s – T_\infty) \)
- Radiation (Stefan–Boltzmann): \( q = \varepsilon \sigma T^4 \)
- Heat transfer rate: \( Q = m c \Delta T \)
- Thermal resistance: \( R = d / (k A) \)
- Overall heat transfer: \( Q = U A \Delta T \)
What is Heat Flux Formula?
The Heat Flux Formula quantifies the amount of thermal energy that flows through a unit area of a surface per unit time. In simple terms, heat flux tells us how “intense” the heat transfer is at any given surface. It is denoted by the symbol q (or sometimes φ) and its SI unit is W/m² (Watts per square metre).
Heat flux is covered in NCERT Class 11 Physics, Chapter 11 — Thermal Properties of Matter. The chapter discusses conduction, convection, and radiation as the three mechanisms of heat transfer. The heat flux formula applies to all three modes, though its mathematical form changes slightly for each mode.
Physically, heat flux is a vector quantity. It has both magnitude and direction. The direction is always from the region of higher temperature to the region of lower temperature. This is consistent with the Second Law of Thermodynamics. Engineers use heat flux extensively in designing heat exchangers, building insulation systems, and cooling systems for electronic devices. For CBSE students, understanding this formula is essential for scoring well in the Thermal Properties of Matter chapter. For JEE and NEET aspirants, heat flux problems often appear in combination with Fourier’s Law and Newton’s Law of Cooling.
Heat Flux Formula — Expression and Variables
The general Heat Flux Formula is written as:
\[ q = \frac{Q}{A \cdot t} \]
For conduction, Fourier’s Law gives the heat flux as:
\[ q = -k \frac{dT}{dx} \]
For a slab of uniform thickness, this simplifies to:
\[ q = k \frac{(T_1 – T_2)}{d} \]
For convection, Newton’s Law of Cooling gives:
\[ q = h (T_s – T_\infty) \]
For radiation, the Stefan–Boltzmann Law gives:
\[ q = \varepsilon \sigma T^4 \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| q | Heat flux (thermal energy per unit area per unit time) | W/m² |
| Q | Total heat energy transferred | Joule (J) |
| A | Cross-sectional area of the surface | m² |
| t | Time of heat transfer | Second (s) |
| k | Thermal conductivity of the material | W/(m·K) |
| T1, T2 | Temperatures at the two faces of the slab | Kelvin (K) or °C |
| d | Thickness of the slab (distance between faces) | Metre (m) |
| dT/dx | Temperature gradient along direction x | K/m |
| h | Convective heat transfer coefficient | W/(m²·K) |
| Ts | Surface temperature | Kelvin (K) |
| T∞ | Ambient (fluid) temperature far from surface | Kelvin (K) |
| ϵ | Emissivity of the surface (dimensionless, 0 to 1) | — |
| σ | Stefan–Boltzmann constant | 5.67 × 10²&sup8; W/(m²·K&sup4;) |
| T | Absolute temperature of the radiating body | Kelvin (K) |
Derivation of the Basic Heat Flux Formula
The derivation begins with the definition of heat transfer rate (thermal power). The rate of heat transfer is \( \dot{Q} = Q/t \), measured in Watts. Heat flux is simply this rate divided by the area through which heat flows:
Step 1: Define heat transfer rate: \( \dot{Q} = Q / t \)
Step 2: Divide by area A to get flux: \( q = \dot{Q} / A = Q / (A \cdot t) \)
Step 3 (Conduction): Fourier observed experimentally that \( \dot{Q} \propto A \cdot \Delta T / d \). Introducing the proportionality constant k (thermal conductivity): \( \dot{Q} = k A (T_1 – T_2) / d \). Dividing both sides by A gives: \( q = k (T_1 – T_2) / d \).
This derivation is consistent with the NCERT Class 11 treatment of conduction in Chapter 11.
Complete Heat Transfer Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Basic Heat Flux | \( q = Q / (A \cdot t) \) | Q = heat energy, A = area, t = time | W/m² | Class 11, Ch 11 |
| Fourier’s Law (Conduction) | \( q = k (T_1 – T_2) / d \) | k = thermal conductivity, d = thickness | W/m² | Class 11, Ch 11 |
| Newton’s Law of Cooling (Convection) | \( q = h (T_s – T_\infty) \) | h = convective coefficient, Ts = surface temp | W/m² | Class 11, Ch 11 |
| Stefan–Boltzmann Law (Radiation) | \( q = \varepsilon \sigma T^4 \) | ϵ = emissivity, σ = 5.67×10²&sup8; W/m²K&sup4; | W/m² | Class 11, Ch 11 |
| Total Heat Transferred | \( Q = m c \Delta T \) | m = mass, c = specific heat, ΔT = temp change | Joule (J) | Class 11, Ch 11 |
| Thermal Resistance | \( R_{th} = d / (k A) \) | d = thickness, k = conductivity, A = area | K/W | Class 11, Ch 11 |
| Heat Transfer Rate (Conduction) | \( \dot{Q} = k A (T_1 – T_2) / d \) | k = conductivity, A = area, d = thickness | W (Watt) | Class 11, Ch 11 |
| Net Radiation Heat Flux | \( q_{net} = \varepsilon \sigma (T_s^4 – T_{surr}^4) \) | Ts = surface temp, Tsurr = surrounding temp | W/m² | Class 11, Ch 11 |
| Wien’s Displacement Law | \( \lambda_{max} T = b \) | b = 2.898×10²³ m·K, T = temperature | m·K | Class 11, Ch 11 |
| Overall Heat Transfer Coefficient | \( Q = U A \Delta T \) | U = overall coefficient, A = area, ΔT = temp diff | W | Class 11, Ch 11 |
Heat Flux Formula — Solved Examples
Example 1 (Class 9–10 Level): Basic Heat Flux Calculation
Problem: A metal plate has a surface area of 0.5 m². A total of 1800 J of heat energy passes through it in 60 seconds. Calculate the heat flux through the plate.
Given:
- Total heat energy, Q = 1800 J
- Area, A = 0.5 m²
- Time, t = 60 s
Step 1: Write the basic heat flux formula:
\( q = \dfrac{Q}{A \cdot t} \)
Step 2: Substitute the given values:
\( q = \dfrac{1800}{0.5 \times 60} \)
Step 3: Calculate the denominator: \( 0.5 \times 60 = 30 \)
Step 4: Divide: \( q = \dfrac{1800}{30} = 60 \) W/m²
Answer
The heat flux through the plate is 60 W/m².
Example 2 (Class 11–12 Level): Conduction Using Fourier’s Law
Problem: A brick wall of thickness 0.20 m and thermal conductivity 0.72 W/(m·K) separates the inside of a building (maintained at 25°C) from the outside (at 5°C). Calculate the heat flux through the wall.
Given:
- Thermal conductivity, k = 0.72 W/(m·K)
- Temperature on warm side, T1 = 25°C
- Temperature on cold side, T2 = 5°C
- Thickness, d = 0.20 m
Step 1: Write Fourier’s Law for a uniform slab:
\( q = k \dfrac{(T_1 – T_2)}{d} \)
Step 2: Calculate the temperature difference:
\( \Delta T = T_1 – T_2 = 25 – 5 = 20 \) K
Step 3: Substitute all values:
\( q = 0.72 \times \dfrac{20}{0.20} \)
Step 4: Simplify: \( q = 0.72 \times 100 = 72 \) W/m²
Step 5: Interpret the result. The heat flux is 72 W/m², meaning 72 Joules of thermal energy escape through every square metre of the wall every second.
Answer
The heat flux through the brick wall is 72 W/m².
Example 3 (JEE/NEET Level): Combined Conduction and Comparison
Problem: Two slabs A and B are placed in contact. Slab A has thermal conductivity kA = 2 W/(m·K) and thickness dA = 0.04 m. Slab B has thermal conductivity kB = 0.5 W/(m·K) and thickness dB = 0.02 m. The outer face of A is at 100°C and the outer face of B is at 20°C. Assuming steady-state heat transfer, find (i) the heat flux through the composite slab and (ii) the temperature at the interface.
Given:
- kA = 2 W/(m·K), dA = 0.04 m
- kB = 0.5 W/(m·K), dB = 0.02 m
- Thot = 100°C, Tcold = 20°C
Step 1: In steady state, the heat flux through both slabs is equal. Use the total thermal resistance concept.
Thermal resistance per unit area of slab A: \( R_A = d_A / k_A = 0.04 / 2 = 0.02 \) m²·K/W
Thermal resistance per unit area of slab B: \( R_B = d_B / k_B = 0.02 / 0.5 = 0.04 \) m²·K/W
Step 2: Total resistance per unit area:
\( R_{total} = R_A + R_B = 0.02 + 0.04 = 0.06 \) m²·K/W
Step 3: Calculate heat flux:
\( q = \dfrac{T_{hot} – T_{cold}}{R_{total}} = \dfrac{100 – 20}{0.06} = \dfrac{80}{0.06} \approx 1333 \) W/m²
Step 4: Find the interface temperature Ti. The heat flux through slab A equals the overall heat flux:
\( q = \dfrac{T_{hot} – T_i}{R_A} \)
\( 1333 = \dfrac{100 – T_i}{0.02} \)
\( 100 – T_i = 1333 \times 0.02 = 26.67 \)
\( T_i = 100 – 26.67 \approx 73.3 \)°C
Answer
(i) Heat flux through the composite slab ≈ 1333 W/m².
(ii) Interface temperature ≈ 73.3°C.
CBSE Exam Tips 2025-26
- Know all three modes: CBSE questions often ask you to identify whether conduction, convection, or radiation is involved. Write the correct formula for each mode before substituting values.
- Units are critical: We recommend always converting temperature to Kelvin when using the Stefan–Boltzmann radiation formula. For Fourier’s Law, °C and K give the same temperature difference, so either works.
- Direction of heat flow: Heat always flows from higher to lower temperature. In Fourier’s Law, the negative sign in \( q = -k (dT/dx) \) accounts for this. In CBSE numericals, use the magnitude \( q = k \Delta T / d \) and state the direction separately.
- Thermal conductivity values: Memorise approximate values for common materials — metals (copper ~400, iron ~80), glass (~1), brick (~0.7), wood (~0.1–0.2). These are often given in problems but knowing them helps with estimation questions.
- Composite slabs: For two or more slabs in series, add thermal resistances. This is a favourite 3-mark question in CBSE Class 11 exams. Practice the formula \( R_{total} = R_1 + R_2 \) thoroughly.
- Distinguish heat flux from heat transfer rate: Heat flux (q) is per unit area in W/m². Heat transfer rate (\( \dot{Q} \)) is total power in Watts. CBSE sometimes asks for one when students calculate the other. Read the question carefully.
Common Mistakes to Avoid
- Mistake 1 — Confusing Q with q: Many students use Q (total heat in Joules) and q (heat flux in W/m²) interchangeably. Remember, \( q = Q / (A \cdot t) \). Always check whether the question asks for total heat or heat flux.
- Mistake 2 — Forgetting area in radiation problems: The Stefan–Boltzmann Law \( P = \varepsilon \sigma A T^4 \) gives total radiated power. The heat flux is \( q = \varepsilon \sigma T^4 \) (without A). Students often multiply by area twice, giving a wrong answer.
- Mistake 3 — Using Celsius instead of Kelvin in radiation: The Stefan–Boltzmann Law requires absolute temperature in Kelvin. Using °C here gives a drastically wrong answer. Always add 273 (or 273.15) to convert.
- Mistake 4 — Ignoring the negative sign in Fourier’s Law: The full form is \( q = -k (dT/dx) \). The negative sign shows heat flows opposite to the temperature gradient. In numerical problems, use magnitudes and state direction. Ignoring this in theory questions loses marks.
- Mistake 5 — Adding conductivities instead of resistances for composite slabs: For slabs in series, thermal resistances add up, not conductivities. The correct approach is \( R_{total} = d_1/(k_1 A) + d_2/(k_2 A) \), not \( k_{eff} = k_1 + k_2 \).
JEE/NEET Application of Heat Flux Formula
In our experience, JEE aspirants encounter the Heat Flux Formula in several recurring question patterns. Understanding these patterns saves time during the exam.
Pattern 1: Composite Slab Problems (JEE Main)
JEE Main frequently tests composite slabs with two or three layers. The key principle is that heat flux is equal through all layers in steady state. Students must set up the thermal resistance network and use \( q = \Delta T_{total} / R_{total} \) to find the heat flux, then back-calculate interface temperatures. This pattern appeared in JEE Main 2022 and 2023 papers.
Pattern 2: Radiation and Emissivity (NEET)
NEET questions on radiation often involve comparing heat loss from two bodies at different temperatures or with different emissivities. The net radiation heat flux formula \( q_{net} = \varepsilon \sigma (T_s^4 – T_{surr}^4) \) is essential. NEET 2023 had a direct question on which body loses more heat based on emissivity and temperature values.
Pattern 3: Thermal Conductivity Comparison (JEE Advanced)
JEE Advanced problems often present a scenario where two rods of different materials are connected end-to-end or in parallel, and ask for the effective thermal conductivity or the heat flux ratio. For rods in series, the effective conductivity is given by \( k_{eff} = (d_1 + d_2) / (d_1/k_1 + d_2/k_2) \). For rods in parallel, \( k_{eff} = (k_1 A_1 + k_2 A_2) / (A_1 + A_2) \). Our experts suggest drawing a circuit analogy (thermal circuit) to avoid confusion between series and parallel configurations.
In our experience, students who master the thermal resistance analogy (treating heat flux like current, temperature difference like voltage, and thermal resistance like electrical resistance) solve these problems significantly faster than those who rely on memorised formulas alone.
FAQs on Heat Flux Formula
For more related formulas, explore our detailed articles on the Electric Current Formula and the Current Density Formula, which follow a similar flux-based approach. You can also visit our Complete Physics Formulas Hub for a full list of Class 11 and Class 12 formula sheets. For the official NCERT syllabus reference, visit the NCERT official website. Additionally, our article on the de Broglie Wavelength Formula covers another key Class 11 topic that frequently appears alongside thermal physics in JEE papers.