The Heat Conduction Formula, expressed as \( Q = rac{kA(T_1 – T_2)t}{d} \), describes the rate at which thermal energy transfers through a solid material due to a temperature gradient. This formula is a core concept in Class 11 Physics (NCERT Chapter 11 — Thermal Properties of Matter) and appears regularly in CBSE board exams, JEE Main, and NEET. In this article, you will find the complete derivation, a comprehensive formula sheet, three progressively difficult solved examples, CBSE exam tips for 2025-26, common mistakes, and JEE/NEET application patterns.
Key Heat Conduction Formulas at a Glance
Quick reference for the most important heat conduction formulas used in CBSE and competitive exams.
- Heat conducted: \( Q = \frac{kA(T_1 – T_2)t}{d} \)
- Rate of heat flow: \( \frac{Q}{t} = \frac{kA \Delta T}{d} \)
- Thermal resistance: \( R = \frac{d}{kA} \)
- Fourier’s Law (differential form): \( \frac{dQ}{dt} = -kA \frac{dT}{dx} \)
- Thermal conductivity: \( k = \frac{Qd}{A(T_1 – T_2)t} \)
- Series combination of slabs: \( R_{total} = R_1 + R_2 + R_3 \)
- Parallel combination of slabs: \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \)
What is the Heat Conduction Formula?
The Heat Conduction Formula quantifies the amount of heat energy transferred through a solid body when a temperature difference exists across it. Heat always flows from a region of higher temperature to a region of lower temperature. This spontaneous flow continues until thermal equilibrium is reached.
Conduction is one of the three modes of heat transfer. The other two are convection and radiation. In conduction, energy transfers through molecular vibrations and collisions. There is no bulk movement of the material itself.
The Heat Conduction Formula is derived from Fourier’s Law of Heat Conduction, introduced by the French mathematician Jean-Baptiste Joseph Fourier in 1822. According to this law, the rate of heat transfer through a material is proportional to the temperature gradient and the cross-sectional area, and inversely proportional to the thickness of the material.
In the NCERT Class 11 Physics textbook, this concept is covered in Chapter 11 — Thermal Properties of Matter. CBSE board exams frequently test numerical problems based on this formula. JEE Main and NEET also include questions on thermal conductivity, thermal resistance, and composite slabs.
Heat Conduction Formula — Expression and Variables
The standard form of the Heat Conduction Formula is:
\[ Q = \frac{kA(T_1 – T_2)t}{d} \]
Here, Q represents the total heat transferred in Joules, k is the thermal conductivity of the material, A is the cross-sectional area, \( T_1 – T_2 \) is the temperature difference across the slab, t is the time duration, and d is the thickness of the slab.
The rate of heat flow (heat current) is often more useful in steady-state problems:
\[ H = \frac{Q}{t} = \frac{kA(T_1 – T_2)}{d} \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| Q | Total heat transferred | Joule (J) |
| k | Thermal conductivity | W m¹ K¹ (or W/mK) |
| A | Cross-sectional area of the slab | Square metre (m²) |
| T₁ | Temperature of the hotter face | Kelvin (K) or °C |
| T₂ | Temperature of the cooler face | Kelvin (K) or °C |
| t | Time of heat transfer | Second (s) |
| d | Thickness of the slab | Metre (m) |
| H | Rate of heat flow (heat current) | Watt (W) |
| R | Thermal resistance | K/W |
Derivation of the Heat Conduction Formula
Fourier’s Law states that the heat flux (heat flow per unit area per unit time) is proportional to the negative temperature gradient. Consider a uniform slab of thickness d and cross-sectional area A.
Step 1: The temperature difference across the slab is \( \Delta T = T_1 – T_2 \), where \( T_1 > T_2 \).
Step 2: The temperature gradient is \( \frac{\Delta T}{d} = \frac{T_1 – T_2}{d} \).
Step 3: By Fourier’s Law, the rate of heat flow is proportional to the area and the temperature gradient:
\[ H = \frac{dQ}{dt} = kA \frac{(T_1 – T_2)}{d} \]
Step 4: Multiplying both sides by time t gives the total heat transferred:
\[ Q = \frac{kA(T_1 – T_2)t}{d} \]
The negative sign in the differential form indicates that heat flows in the direction of decreasing temperature. For practical calculations, we use the magnitude, which gives the formula above.
Complete Thermal Physics Formula Sheet
The following table covers all major formulas related to heat conduction and thermal properties. This is an essential reference for CBSE Class 11 and competitive exam preparation.
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Heat Conduction (Total) | \( Q = \frac{kA(T_1-T_2)t}{d} \) | k=thermal conductivity, A=area, d=thickness, t=time | Joule (J) | Class 11, Ch 11 |
| Rate of Heat Flow | \( H = \frac{kA \Delta T}{d} \) | H=heat current, ΔT=temperature difference | Watt (W) | Class 11, Ch 11 |
| Thermal Resistance | \( R = \frac{d}{kA} \) | d=thickness, k=thermal conductivity, A=area | K/W | Class 11, Ch 11 |
| Heat Flow via Thermal Resistance | \( H = \frac{\Delta T}{R} \) | H=heat current, ΔT=temp. difference, R=thermal resistance | Watt (W) | Class 11, Ch 11 |
| Series Slabs (Thermal Resistance) | \( R_{total} = R_1 + R_2 + \cdots \) | R₁, R₂=individual resistances | K/W | Class 11, Ch 11 |
| Parallel Slabs (Thermal Resistance) | \( \frac{1}{R_{total}} = \frac{1}{R_1} + \frac{1}{R_2} \) | R₁, R₂=individual resistances | K/W | Class 11, Ch 11 |
| Linear Thermal Expansion | \( \Delta L = L_0 \alpha \Delta T \) | L₀=original length, α=coefficient, ΔT=temp. change | Metre (m) | Class 11, Ch 11 |
| Newton’s Law of Cooling | \( \frac{dT}{dt} = -b(T – T_0) \) | T=body temp., T₀=surrounding temp., b=cooling constant | °C/s | Class 11, Ch 11 |
| Stefan-Boltzmann Law | \( P = \sigma A T^4 \) | σ=5.67×10&sup8; W m² K⁴, T=absolute temperature | Watt (W) | Class 11, Ch 11 |
| Wien’s Displacement Law | \( \lambda_{max} T = b \) | λₐₐₓ=peak wavelength, b=2.898×10⁻³ m·K | m·K | Class 11, Ch 11 |
| Specific Heat Capacity | \( Q = mc\Delta T \) | m=mass, c=specific heat, ΔT=temperature change | Joule (J) | Class 11, Ch 11 |
Heat Conduction Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A steel slab has a thickness of 0.02 m and a cross-sectional area of 0.5 m². The temperature of one face is 100°C and the other face is 20°C. The thermal conductivity of steel is 50 W/mK. Calculate the total heat conducted through the slab in 60 seconds.
Given:
- k = 50 W/mK
- A = 0.5 m²
- T₁ = 100°C, T₂ = 20°C, so ΔT = 80°C = 80 K
- d = 0.02 m
- t = 60 s
Step 1: Write the Heat Conduction Formula: \( Q = \frac{kA(T_1 – T_2)t}{d} \)
Step 2: Substitute all values:
\[ Q = \frac{50 \times 0.5 \times 80 \times 60}{0.02} \]
Step 3: Calculate the numerator: \( 50 \times 0.5 = 25 \); \( 25 \times 80 = 2000 \); \( 2000 \times 60 = 120000 \)
Step 4: Divide by thickness: \( Q = \frac{120000}{0.02} = 6{,}000{,}000 \) J
Answer
Total heat conducted = 6 × 10&sup6; J (6 MJ)
Example 2 (Class 11-12 Level)
Problem: Two slabs, one of copper (thickness 0.03 m, k = 385 W/mK) and one of wood (thickness 0.05 m, k = 0.17 W/mK), are placed in series. The outer face of the copper slab is at 80°C and the outer face of the wood slab is at 20°C. The cross-sectional area of both slabs is 1 m². Find the steady-state rate of heat flow through the composite slab.
Given:
- Copper: k₁ = 385 W/mK, d₁ = 0.03 m
- Wood: k₂ = 0.17 W/mK, d₂ = 0.05 m
- A = 1 m², T₁ = 80°C, T₂ = 20°C
Step 1: Calculate thermal resistance of copper:
\[ R_1 = \frac{d_1}{k_1 A} = \frac{0.03}{385 \times 1} = 7.79 \times 10^{-5} \text{ K/W} \]
Step 2: Calculate thermal resistance of wood:
\[ R_2 = \frac{d_2}{k_2 A} = \frac{0.05}{0.17 \times 1} = 0.294 \text{ K/W} \]
Step 3: Total thermal resistance in series:
\[ R_{total} = R_1 + R_2 = 0.0000779 + 0.294 \approx 0.2941 \text{ K/W} \]
Step 4: Rate of heat flow:
\[ H = \frac{\Delta T}{R_{total}} = \frac{80 – 20}{0.2941} = \frac{60}{0.2941} \approx 204.0 \text{ W} \]
Answer
Rate of heat flow through the composite slab ≈ 204 W. Note that the wood slab dominates the total thermal resistance because its thermal conductivity is much lower than copper’s.
Example 3 (JEE/NEET Level)
Problem: Three identical slabs each of thickness d, area A, and thermal conductivities k, 2k, and 3k respectively are connected in parallel between two reservoirs at temperatures T₁ and T₂ (T₁ > T₂). Find the effective thermal conductivity of the combination.
Given:
- Slab 1: thermal conductivity = k, thickness = d, area = A
- Slab 2: thermal conductivity = 2k, thickness = d, area = A
- Slab 3: thermal conductivity = 3k, thickness = d, area = A
- Temperature difference: ΔT = T₁ − T₂
Step 1: For slabs in parallel, the total heat flow rate is the sum of individual heat flow rates:
\[ H_{total} = H_1 + H_2 + H_3 \]
Step 2: Write each heat flow rate using the conduction formula:
\[ H_{total} = \frac{kA\Delta T}{d} + \frac{2kA\Delta T}{d} + \frac{3kA\Delta T}{d} = \frac{6kA\Delta T}{d} \]
Step 3: The equivalent slab has total area 3A (three slabs in parallel) and the same thickness d. The effective thermal conductivity kᵉᵉᵉ satisfies:
\[ H_{total} = \frac{k_{eff}(3A)\Delta T}{d} \]
Step 4: Equate the two expressions:
\[ \frac{k_{eff}(3A)\Delta T}{d} = \frac{6kA\Delta T}{d} \]
Step 5: Solve for kᵉᵉᵉ:
\[ k_{eff} = \frac{6k}{3} = 2k \]
Answer
The effective thermal conductivity of the parallel combination is kᵉᵉᵉ = 2k. This is the arithmetic mean of k, 2k, and 3k, which is a standard result for identical-dimension slabs in parallel.
CBSE Exam Tips 2025-26
- Identify the mode of heat transfer first. CBSE questions often ask you to state whether conduction, convection, or radiation is involved before applying the formula. Award-carrying marks are lost when students skip this step.
- Convert units before substituting. Thickness must be in metres, area in m², and temperature difference in Kelvin or °C (the difference is numerically the same). We recommend writing units alongside every substitution step.
- Use the thermal resistance analogy for composite slabs. Treat the problem like an electrical circuit. Series slabs add resistances; parallel slabs add conductances. This approach saves time and reduces errors in 2025-26 board exams.
- State Fourier’s Law in 3-mark theory questions. Write the formula, define each variable, and state the conditions (steady state, uniform cross-section). This earns full marks in definition-type questions.
- Memorise common thermal conductivity values. Silver (~420 W/mK), Copper (~385 W/mK), Aluminium (~205 W/mK), Glass (~1 W/mK), and Wood (~0.17 W/mK) are frequently used in CBSE numericals.
- Check significant figures. CBSE 2025-26 marking schemes expect answers rounded to 2-3 significant figures. Always state the unit in the final answer.
Common Mistakes to Avoid
- Confusing thickness with length: In the formula, d is the thickness of the slab in the direction of heat flow, not the length of the object. Always draw a diagram to identify the correct dimension.
- Using inconsistent temperature units: The temperature difference ΔT = T₁ − T₂ is numerically equal in Kelvin and Celsius. However, if the problem gives absolute temperatures in Kelvin, subtract correctly. Never mix units within a single calculation.
- Forgetting the area term in composite slab problems: When slabs have different areas, you cannot simply add thermal conductivities. Always calculate individual thermal resistances using \( R = d/(kA) \) and then combine them.
- Applying the formula to non-steady-state situations: The standard Heat Conduction Formula assumes steady-state conditions (temperature at each point does not change with time). For transient heat conduction, a more complex differential equation is needed.
- Mixing up series and parallel slab rules: In series, the same heat current flows through all slabs and resistances add. In parallel, the same temperature difference applies across all slabs and conductances add. Swapping these rules is a very common JEE and CBSE error.
JEE/NEET Application of Heat Conduction Formula
In our experience, JEE aspirants encounter the Heat Conduction Formula in at least 1-2 questions per year, either directly or as part of a multi-concept problem involving calorimetry or thermodynamics. NEET questions tend to focus on conceptual understanding and straightforward numericals.
Pattern 1: Composite Slab Problems (JEE Main)
JEE Main frequently tests composite slabs in series. The key insight is that in steady state, the heat current H is the same through all slabs. You set up \( H = \frac{\Delta T_1}{R_1} = \frac{\Delta T_2}{R_2} \) and solve for the interface temperature. This tests both the Heat Conduction Formula and the thermal resistance concept simultaneously.
Pattern 2: Effective Thermal Conductivity (JEE Advanced)
JEE Advanced problems often ask for the “effective” or “equivalent” thermal conductivity of a combination of slabs. For slabs in series of equal area:
\[ k_{eff} = \frac{d_1 + d_2}{\frac{d_1}{k_1} + \frac{d_2}{k_2}} \]
For slabs in parallel of equal thickness:
\[ k_{eff} = \frac{k_1 A_1 + k_2 A_2}{A_1 + A_2} \]
Memorising these results and their derivations can save valuable time in the JEE Advanced paper.
Pattern 3: NEET Conceptual Questions
NEET questions on heat conduction typically ask which material is a better conductor, what happens to heat flow when thickness doubles, or how thermal conductivity relates to electrical conductivity (Wiedemann-Franz Law). Our experts suggest practising at least 15 NEET previous year questions on this topic. Focus on the physical interpretation of each variable in the Heat Conduction Formula rather than rote memorisation.
For further reading, refer to the NCERT Class 11 Physics Chapter 11 on the official NCERT website.
FAQs on Heat Conduction Formula
Explore more related formulas on ncertbooks.net to strengthen your understanding of thermal physics. Read our detailed article on the Electric Current Formula to understand the analogy between heat flow and electric current. Study the Current Density Formula alongside thermal conductivity to appreciate the Wiedemann-Franz Law. You can also visit our comprehensive Physics Formulas hub for a complete list of all Class 11 and Class 12 physics formulas. For motion-related topics, our article on the Terminal Velocity Formula provides another excellent example of steady-state physics analysis.