The Heat Transfer Formula describes how thermal energy moves between objects or systems due to a temperature difference, expressed mathematically for three distinct modes: conduction — \ ( Q = \frac{kA\Delta T \cdot t}{d} \), convection, and radiation. This topic is a core part of NCERT Class 11 Physics (Chapter 11: Thermal Properties of Matter) and appears regularly in CBSE board exams. JEE Main and NEET aspirants must master all three modes and their formulas. This article covers every heat transfer formula, derivations, a complete formula sheet, solved examples, exam tips, and common mistakes to avoid.
Key Heat Transfer Formulas at a Glance
Quick reference for the most important heat transfer formulas used in CBSE and competitive exams.
- Conduction: \ ( Q = \frac{kA\Delta T \cdot t}{d} \)
- Newton’s Law of Cooling: \ ( \frac{dQ}{dt} = hA(T – T_0) \)
- Stefan’s Law (Radiation): \ ( P = \sigma A e T^4 \)
- Wien’s Displacement Law: \ ( \lambda_{max} T = b \)
- Thermal Resistance: \ ( R = \frac{d}{kA} \)
- Heat Current: \ ( H = \frac{Q}{t} = \frac{kA\Delta T}{d} \)
- Net Radiation: \ ( P_{net} = \sigma A e (T^4 – T_0^4) \)
What is the Heat Transfer Formula?
The Heat Transfer Formula is a set of mathematical expressions that quantify how thermal energy flows from a region of higher temperature to a region of lower temperature. Heat always moves in the direction that reduces the temperature difference between two bodies or systems. This process continues until thermal equilibrium is reached, meaning both systems attain the same temperature.
In NCERT Class 11 Physics, Chapter 11 (Thermal Properties of Matter) introduces all three modes of heat transfer. Each mode operates through a different physical mechanism and therefore has its own governing formula. Conduction occurs through direct molecular interaction in solids. Convection involves the bulk movement of fluid (liquid or gas). Radiation transfers energy via electromagnetic waves and requires no medium at all.
Understanding the Heat Transfer Formula is essential for scoring well in CBSE board exams. It is equally critical for JEE Main, JEE Advanced, and NEET, where numerical problems and conceptual questions on thermal properties appear almost every year. Mastering these formulas helps students tackle a wide range of problems, from simple plug-and-calculate exercises to complex multi-step derivations.
Heat Transfer Formula — Expression and Variables
1. Conduction
Conduction is the transfer of heat through a solid material without any bulk movement of the material itself. The governing formula is Fourier’s Law of Heat Conduction:
\[ Q = \frac{kA\Delta T \cdot t}{d} \]
Here, Q is the heat transferred, k is the thermal conductivity, A is the cross-sectional area, ΔT is the temperature difference, t is the time, and d is the thickness of the material.
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | Heat Transferred | Joule (J) |
| k | Thermal Conductivity | W m-1 K-1 |
| A | Cross-sectional Area | m2 |
| ΔT | Temperature Difference | Kelvin (K) or °C |
| t | Time | Second (s) |
| d | Thickness / Length | Metre (m) |
Derivation of the Conduction Formula
Consider a slab of material with cross-sectional area A and thickness d. One face is maintained at temperature T1 and the other at T2, where T1 > T2. The temperature gradient across the slab is ΔT/d. Fourier observed experimentally that the rate of heat flow (H) is directly proportional to the area A and the temperature gradient, and inversely proportional to the thickness. Introducing the proportionality constant k (thermal conductivity of the material), we get the heat current:
\[ H = \frac{Q}{t} = kA\frac{\Delta T}{d} \]
Multiplying both sides by time t gives the total heat transferred Q. This is Fourier’s Law of Heat Conduction.
2. Newton’s Law of Cooling (Convection)
Newton’s Law of Cooling describes the rate at which a body loses heat to its surroundings by convection:
\[ \frac{dQ}{dt} = hA(T – T_0) \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| dQ/dt | Rate of Heat Loss | Watt (W) |
| h | Convective Heat Transfer Coefficient | W m-2 K-1 |
| A | Surface Area | m2 |
| T | Temperature of Body | Kelvin (K) |
| T0 | Temperature of Surroundings | Kelvin (K) |
3. Stefan’s Law of Radiation
Radiation is heat transfer through electromagnetic waves. Stefan’s Law gives the power radiated by a body:
\[ P = \sigma A e T^4 \]
The net power radiated by a body at temperature T in surroundings at temperature T0 is:
\[ P_{net} = \sigma A e (T^4 – T_0^4) \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| P | Power Radiated | Watt (W) |
| σ | Stefan-Boltzmann Constant (5.67 × 10-8 W m-2 K-4) | W m-2 K-4 |
| A | Surface Area | m2 |
| e | Emissivity (0 to 1) | Dimensionless |
| T | Absolute Temperature | Kelvin (K) |
Complete Heat Transfer Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Fourier’s Law (Conduction) | \ ( Q = \frac{kA\Delta T \cdot t}{d} \) | k=thermal conductivity, A=area, ΔT=temp. diff., t=time, d=thickness | J | Class 11, Ch 11 |
| Heat Current | \ ( H = \frac{kA\Delta T}{d} \) | k=thermal conductivity, A=area, ΔT=temp. diff., d=thickness | W | Class 11, Ch 11 |
| Thermal Resistance | \ ( R = \frac{d}{kA} \) | d=thickness, k=thermal conductivity, A=area | K W-1 | Class 11, Ch 11 |
| Newton’s Law of Cooling | \ ( \frac{dQ}{dt} = hA(T – T_0) \) | h=convective coefficient, A=area, T=body temp., T0=surrounding temp. | W | Class 11, Ch 11 |
| Stefan’s Law | \ ( P = \sigma A e T^4 \) | σ=Stefan constant, A=area, e=emissivity, T=temperature | W | Class 11, Ch 11 |
| Net Radiation Power | \ ( P_{net} = \sigma A e (T^4 – T_0^4) \) | T=body temp., T0=surrounding temp. | W | Class 11, Ch 11 |
| Wien’s Displacement Law | \ ( \lambda_{max} T = b \) | λmax=peak wavelength, T=temperature, b=2.898×10-3 m·K | m | Class 11, Ch 11 |
| Thermal Conductivity (Series Slabs) | \ ( R_{total} = R_1 + R_2 \) | R1, R2=individual thermal resistances | K W-1 | Class 11, Ch 11 |
| Thermal Conductivity (Parallel Slabs) | \ ( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \) | R1, R2=individual thermal resistances | K W-1 | Class 11, Ch 11 |
| Specific Heat Capacity | \ ( Q = mc\Delta T \) | m=mass, c=specific heat, ΔT=temperature change | J | Class 11, Ch 11 |
Heat Transfer Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A metal rod of length 0.5 m and cross-sectional area 4 × 10-4 m2 has a thermal conductivity of 200 W m-1 K-1. One end is kept at 100°C and the other at 20°C. Calculate the heat transferred through the rod in 5 minutes.
Given: k = 200 W m-1 K-1, A = 4 × 10-4 m2, ΔT = 100 − 20 = 80°C = 80 K, d = 0.5 m, t = 5 × 60 = 300 s
Step 1: Write the conduction formula: \ ( Q = \frac{kA\Delta T \cdot t}{d} \)
Step 2: Substitute the values: \ ( Q = \frac{200 \times 4 \times 10^{-4} \times 80 \times 300}{0.5} \)
Step 3: Calculate numerator: \ ( 200 \times 4 \times 10^{-4} \times 80 \times 300 = 1920 \) J
Step 4: Divide by d: \ ( Q = \frac{1920}{0.5} = 3840 \) J
Answer
Heat Transferred Q = 3840 J
Example 2 (Class 11-12 Level)
Problem: A spherical black body of radius 0.1 m is at a temperature of 727°C. Calculate the power radiated by it. Also find the net power radiated if the surrounding temperature is 27°C. (Stefan-Boltzmann constant σ = 5.67 × 10-8 W m-2 K-4)
Given: r = 0.1 m, T = 727 + 273 = 1000 K, T0 = 27 + 273 = 300 K, e = 1 (black body), σ = 5.67 × 10-8 W m-2 K-4
Step 1: Calculate surface area: \ ( A = 4\pi r^2 = 4 \times 3.14 \times (0.1)^2 = 0.1256 \) m2
Step 2: Apply Stefan’s Law: \ ( P = \sigma A e T^4 = 5.67 \times 10^{-8} \times 0.1256 \times 1 \times (1000)^4 \)
Step 3: Calculate: \ ( P = 5.67 \times 10^{-8} \times 0.1256 \times 10^{12} = 5.67 \times 0.1256 \times 10^{4} \approx 7122 \) W
Step 4: Net power: \ ( P_{net} = \sigma A e (T^4 – T_0^4) = 5.67 \times 10^{-8} \times 0.1256 \times (10^{12} – 300^4) \)
Step 5: Calculate 3004 = 8.1 × 109, which is negligible compared to 1012. So \ ( P_{net} \approx 7122 – 46 \approx 7076 \) W
Answer
Power Radiated P ≈ 7122 W; Net Power Radiated Pnet ≈ 7076 W
Example 3 (JEE/NEET Level)
Problem: Two slabs of equal thickness d are placed in series. Slab 1 has thermal conductivity k and Slab 2 has thermal conductivity 2k. The left face of Slab 1 is at temperature T1 = 100°C and the right face of Slab 2 is at temperature T2 = 20°C. Find the temperature at the interface and the heat current per unit area.
Given: Slab 1: k1 = k, thickness = d; Slab 2: k2 = 2k, thickness = d; T1 = 100°C, T2 = 20°C
Step 1: In steady state, heat current through both slabs is equal. Let Ti be the interface temperature.
Step 2: Heat current through Slab 1: \ ( H = \frac{k \cdot A \cdot (100 – T_i)}{d} \)
Step 3: Heat current through Slab 2: \ ( H = \frac{2k \cdot A \cdot (T_i – 20)}{d} \)
Step 4: Equate the two expressions: \ ( k(100 – T_i) = 2k(T_i – 20) \)
Step 5: Solve: \ ( 100 – T_i = 2T_i – 40 \Rightarrow 140 = 3T_i \Rightarrow T_i = 46.67^\circ C \)
Step 6: Heat current per unit area: \ ( \frac{H}{A} = \frac{k(100 – 46.67)}{d} = \frac{53.33k}{d} \) W m-2
Answer
Interface Temperature Ti = 46.67°C; Heat Current per unit area = 53.33k/d W m-2
CBSE Exam Tips 2025-26
- We recommend memorising the value of the Stefan-Boltzmann constant (σ = 5.67 × 10-8 W m-2 K-4) and Wien’s constant (b = 2.898 × 10-3 m·K) as they appear directly in CBSE numericals.
- Always convert temperature from °C to Kelvin before using Stefan’s Law or Wien’s Law. Add 273 to the Celsius value.
- For conduction problems involving multiple slabs in series, use the thermal resistance analogy (similar to electrical resistance in series) to simplify calculations.
- In CBSE 2025-26 exams, questions on Newton’s Law of Cooling often ask students to plot temperature vs. time graphs. Practise interpreting these graphs.
- Remember that emissivity (e) for a perfect black body is 1 and for a perfect reflector is 0. The CBSE paper often specifies the type of body to guide your choice of e.
- Our experts suggest writing the formula first, then substituting values, and finally stating the SI unit of the answer. This approach earns full step marks.
Common Mistakes to Avoid
- Not converting to Kelvin: Stefan’s Law uses absolute temperature (T in Kelvin). Using Celsius directly gives a completely wrong answer. Always add 273 to °C.
- Confusing ΔT in conduction: In Fourier’s Law, ΔT is the temperature difference (Thot − Tcold). Students sometimes substitute individual temperatures instead of the difference.
- Ignoring emissivity: When the problem says “grey body” or gives an emissivity value, always include e in Stefan’s Law. Assuming e = 1 for non-black bodies is a common error.
- Using wrong units for d: The thickness d in the conduction formula must be in metres. Converting centimetres or millimetres to metres is frequently missed under exam pressure.
- Misapplying Newton’s Law of Cooling: This law is valid only when the temperature difference between the body and surroundings is small. Do not apply it to large temperature differences.
JEE/NEET Application of Heat Transfer Formula
In our experience, JEE aspirants encounter the Heat Transfer Formula in at least one to two questions per paper, spanning both numerical and conceptual formats. NEET also tests radiation and Newton’s Law of Cooling regularly. Here are the key application patterns:
Pattern 1: Composite Slab Problems (JEE Main)
JEE Main frequently presents problems with two or three slabs of different materials placed in series or parallel. The strategy is to use thermal resistance analogy. For slabs in series, total resistance is \ ( R_{total} = R_1 + R_2 + \ldots \). For slabs in parallel, \ ( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \). The interface temperature is found by equating heat currents, as demonstrated in Example 3 above.
Pattern 2: Black Body Radiation and Wien’s Law (JEE Advanced / NEET)
JEE Advanced tests the concept of peak emission wavelength shifting with temperature. Wien’s Displacement Law states \ ( \lambda_{max} T = 2.898 \times 10^{-3} \) m·K. A typical question gives the peak wavelength and asks for the temperature of a star or furnace. NEET uses this concept to explain why hotter stars appear blue and cooler stars appear red.
Pattern 3: Newton’s Law of Cooling Graphs (NEET)
NEET often presents a cooling curve (temperature vs. time) and asks students to identify the rate of cooling at a specific temperature. Since \ ( \frac{dT}{dt} \propto (T – T_0) \), the rate is highest at the start (when the temperature difference is maximum) and decreases exponentially. Recognising this graph shape is a high-yield skill for NEET.
Our experts recommend practising at least 20 numerical problems on each mode of heat transfer before the JEE/NEET exam. Focus especially on composite slab problems and Stefan’s Law calculations, as these carry the highest marks.
FAQs on Heat Transfer Formula
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