The Heat Gain Formula calculates the amount of thermal energy absorbed by a substance or a building element when it is exposed to a temperature difference or solar radiation, and it is expressed as \( Q = mc\Delta T \) for sensible heat gain in materials. This formula is a core concept in NCERT Class 11 Physics (Chapter 11 — Thermal Properties of Matter) and is equally important for CBSE Board examinations and competitive exams like 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 to avoid, and JEE/NEET application patterns.
Key Heat Gain Formulas at a Glance
Quick reference for the most important heat gain and heat transfer formulas.
- Sensible Heat Gain: \( Q = mc\Delta T \)
- Latent Heat Gain: \( Q = mL \)
- Heat Gain by Conduction: \( Q = \frac{kA\Delta T \cdot t}{d} \)
- Heat Gain by Radiation (Stefan-Boltzmann): \( Q = \sigma A T^4 t \)
- Solar Heat Gain: \( Q_{solar} = SHGC \times A \times E \)
- Heat Gain Rate (Power): \( P = \frac{Q}{t} \)
- Newton’s Law of Cooling: \( \frac{dQ}{dt} = hA(T_s – T_{\infty}) \)
What is the Heat Gain Formula?
The Heat Gain Formula describes the quantity of thermal energy that a body, substance, or enclosed space absorbs from its surroundings due to a temperature difference, solar radiation, or an internal heat source. Heat gain occurs whenever thermal energy flows from a region of higher temperature to a region of lower temperature. This is consistent with the Second Law of Thermodynamics.
In NCERT Class 11 Physics, Chapter 11 (Thermal Properties of Matter), heat gain is introduced through the concept of specific heat capacity. The standard expression is \( Q = mc\Delta T \), where the heat gained depends on the mass of the substance, its specific heat capacity, and the change in temperature. For phase changes, the Heat Gain Formula changes to \( Q = mL \), where \( L \) is the latent heat.
In building physics and engineering applications, heat gain also includes solar heat gain through glass windows and conductive heat gain through walls and roofs. The Heat Gain Formula is therefore used in thermodynamics, calorimetry, HVAC engineering, and environmental physics. It is a foundational concept tested in CBSE Class 10, Class 11, Class 12, JEE Main, JEE Advanced, and NEET examinations.
Heat Gain Formula — Expression and Variables
The primary Heat Gain Formula for a substance undergoing a temperature change (sensible heat) is:
\[ Q = mc\Delta T \]
Where \( \Delta T = T_{final} – T_{initial} \). For heat gain through a wall or slab by conduction over time \( t \):
\[ Q = \frac{kA\Delta T \cdot t}{d} \]
For solar heat gain through a glazing element:
\[ Q_{solar} = SHGC \times A \times E \]
| Symbol | Quantity | SI Unit |
|---|---|---|
| \( Q \) | Heat gained by the substance or body | Joule (J) |
| \( m \) | Mass of the substance | Kilogram (kg) |
| \( c \) | Specific heat capacity of the substance | J kg⁻¹ K⁻¹ |
| \( \Delta T \) | Change in temperature (T₁ − T₀) | Kelvin (K) or °C |
| \( L \) | Latent heat (for phase change) | J kg⁻¹ |
| \( k \) | Thermal conductivity of the material | W m⁻¹ K⁻¹ |
| \( A \) | Cross-sectional area of the surface | m² |
| \( d \) | Thickness of the material (slab/wall) | Metre (m) |
| \( t \) | Time duration of heat transfer | Second (s) |
| \( SHGC \) | Solar Heat Gain Coefficient (dimensionless) | No unit (0 to 1) |
| \( E \) | Incident solar irradiance on the surface | W m⁻² |
| \( \sigma \) | Stefan-Boltzmann constant | W m⁻² K⁻⁴ |
Derivation of the Heat Gain Formula
The Heat Gain Formula \( Q = mc\Delta T \) is derived from the definition of specific heat capacity. Specific heat capacity \( c \) is defined as the amount of heat required to raise the temperature of 1 kg of a substance by 1 K.
Step 1: By definition, \( c = \frac{Q}{m \Delta T} \).
Step 2: Rearranging for \( Q \): \( Q = mc\Delta T \).
Step 3: If \( \Delta T > 0 \), the substance gains heat. If \( \Delta T < 0 \), the substance loses heat.
Step 4: For conductive heat gain through a slab, Fourier’s Law gives the rate as \( \frac{dQ}{dt} = \frac{kA\Delta T}{d} \). Integrating over time \( t \) yields \( Q = \frac{kA\Delta T \cdot t}{d} \).
This derivation is directly referenced in NCERT Class 11 Physics, Chapter 11, under the section on specific heat capacity and calorimetry.
Complete Heat Transfer Formula Sheet
| Formula Name | Expression | Variables | SI Units | NCERT Chapter |
|---|---|---|---|---|
| Sensible Heat Gain | \( Q = mc\Delta T \) | m = mass, c = specific heat, ΔT = temp change | J | Class 11, Ch 11 |
| Latent Heat Gain | \( Q = mL \) | m = mass, L = latent heat | J | Class 11, Ch 11 |
| Conductive Heat Gain (Slab) | \( Q = \frac{kA\Delta T \cdot t}{d} \) | k = conductivity, A = area, d = thickness, t = time | J | Class 11, Ch 11 |
| Rate of Heat Conduction | \( H = \frac{kA\Delta T}{d} \) | H = heat flow rate (power) | W (J/s) | Class 11, Ch 11 |
| Stefan-Boltzmann Law (Radiation) | \( P = \sigma A T^4 \) | σ = 5.67 × 10⁻&sup8; W m⁻² K⁻⁴, T = absolute temp | W | Class 11, Ch 11 |
| Newton’s Law of Cooling | \( \frac{dQ}{dt} = hA(T_s – T_{\infty}) \) | h = convection coefficient, Tₛ = surface temp, T∞ = fluid temp | W | Class 11, Ch 11 |
| Solar Heat Gain | \( Q_{solar} = SHGC \times A \times E \) | SHGC = solar heat gain coefficient, E = irradiance | J or W | Applied Physics |
| Wien’s Displacement Law | \( \lambda_{max} T = b \) | b = 2.898 × 10⁻³ m·K, λ = peak wavelength | m | Class 11, Ch 11 |
| Thermal Resistance | \( R = \frac{d}{kA} \) | d = thickness, k = conductivity, A = area | K/W | Class 11, Ch 11 |
| Heat Gain Rate (Power) | \( P = \frac{Q}{t} \) | Q = total heat, t = time | W | Class 11, Ch 6 |
Heat Gain Formula — Solved Examples
Example 1 (Class 9-10 Level)
Problem: A 2 kg block of iron is heated from 25°C to 75°C. Calculate the heat gained by the iron block. (Specific heat capacity of iron = 450 J kg⁻¹ K⁻¹)
Given: m = 2 kg, T₁ = 25°C, T₂ = 75°C, c = 450 J kg⁻¹ K⁻¹
Step 1: Find the change in temperature.
\( \Delta T = T_2 – T_1 = 75 – 25 = 50 \) °C = 50 K
Step 2: Apply the Heat Gain Formula.
\( Q = mc\Delta T \)
Step 3: Substitute values.
\( Q = 2 \times 450 \times 50 \)
Step 4: Calculate.
\( Q = 45{,}000 \) J = 45 kJ
Answer
The heat gained by the iron block is 45,000 J (45 kJ).
Example 2 (Class 11-12 Level)
Problem: A brick wall has a thickness of 0.20 m, a cross-sectional area of 10 m², and a thermal conductivity of 0.72 W m⁻¹ K⁻¹. The temperature on the outer surface is 40°C and on the inner surface is 25°C. Calculate the total heat gained by the room through this wall in 6 hours.
Given: d = 0.20 m, A = 10 m², k = 0.72 W m⁻¹ K⁻¹, ΔT = 40 − 25 = 15 K, t = 6 h = 6 × 3600 = 21,600 s
Step 1: Write the conductive heat gain formula.
\( Q = \frac{kA\Delta T \cdot t}{d} \)
Step 2: Substitute all values.
\( Q = \frac{0.72 \times 10 \times 15 \times 21{,}600}{0.20} \)
Step 3: Calculate the numerator.
\( 0.72 \times 10 = 7.2 \)
\( 7.2 \times 15 = 108 \)
\( 108 \times 21{,}600 = 2{,}332{,}800 \) J
Step 4: Divide by thickness.
\( Q = \frac{2{,}332{,}800}{0.20} = 11{,}664{,}000 \) J
Step 5: Convert to kJ.
\( Q = 11{,}664 \) kJ ≈ 11.66 MJ
Answer
The total heat gained through the wall in 6 hours is approximately 11.66 MJ.
Example 3 (JEE/NEET Level)
Problem: A calorimeter of heat capacity 50 J/K contains 200 g of water at 20°C. Steam at 100°C is passed into it until the final temperature reaches 40°C. Find the mass of steam condensed. (Specific heat of water = 4200 J kg⁻¹ K⁻¹; Latent heat of steam = 2.26 × 10&sup6; J kg⁻¹)
Given: Heat capacity of calorimeter C₁ = 50 J/K, m₁ = 200 g = 0.2 kg (water), T₁ = 20°C, T₂ = 40°C, Tₛₜₛ₁₀ = 100°C, c₂ = 4200 J kg⁻¹ K⁻¹, L = 2.26 × 10&sup6; J kg⁻¹
Step 1: Calculate total heat gained by water and calorimeter.
\( Q_{gained} = m_1 c_2 \Delta T + C_1 \Delta T \)
\( \Delta T = 40 – 20 = 20 \) K
\( Q_{gained} = (0.2 \times 4200 \times 20) + (50 \times 20) \)
\( Q_{gained} = 16{,}800 + 1{,}000 = 17{,}800 \) J
Step 2: Heat lost by steam has two parts. First, steam condenses at 100°C (latent heat). Second, condensed water cools from 100°C to 40°C.
Let mass of steam condensed = \( m_s \) kg.
\( Q_{lost} = m_s L + m_s c_2 (100 – 40) \)
\( Q_{lost} = m_s \times 2.26 \times 10^6 + m_s \times 4200 \times 60 \)
\( Q_{lost} = m_s (2{,}260{,}000 + 252{,}000) = m_s \times 2{,}512{,}000 \)
Step 3: Apply heat balance (heat gained = heat lost).
\( m_s \times 2{,}512{,}000 = 17{,}800 \)
\( m_s = \frac{17{,}800}{2{,}512{,}000} \approx 7.08 \times 10^{-3} \) kg
Step 4: Convert to grams.
\( m_s \approx 7.08 \) g
Answer
The mass of steam condensed is approximately 7.08 g.
CBSE Exam Tips 2025-26
- Always convert units before substituting. Temperature must be in Kelvin for radiation problems. For \( Q = mc\Delta T \), both °C and K give the same ΔT, so either unit works for that formula only.
- State the formula first. In CBSE board exams, writing the formula before substituting always earns method marks. Never skip this step.
- Distinguish sensible and latent heat gain. If a phase change is involved (melting or boiling), use \( Q = mL \). If only temperature changes, use \( Q = mc\Delta T \). Mixing these two is a very common error.
- Heat balance principle is frequently tested. In calorimetry problems, always write: Heat gained by cold body = Heat lost by hot body. This principle forms the basis of most 5-mark problems in CBSE 2025-26.
- We recommend memorising standard specific heat values. Water (4200 J kg⁻¹ K⁻¹), copper (385 J kg⁻¹ K⁻¹), iron (450 J kg⁻¹ K⁻¹), and aluminium (900 J kg⁻¹ K⁻¹) are the most frequently used values in CBSE questions.
- For conduction problems, check thickness units. Thickness must always be in metres. Students frequently leave it in centimetres, which leads to a factor-of-100 error.
Common Mistakes to Avoid
- Mistake 1 — Using °C instead of K in radiation formulas. Stefan-Boltzmann Law \( P = \sigma A T^4 \) requires temperature in Kelvin (K), not Celsius. Always add 273 (or 273.15) to convert. Using °C here gives a completely wrong answer.
- Mistake 2 — Confusing heat gain with temperature. Heat gain \( Q \) is energy (in Joules). Temperature is not energy. A large mass can gain a lot of heat but show a small temperature change if its specific heat is high.
- Mistake 3 — Ignoring the calorimeter’s heat capacity. In calorimetry problems, students often forget to include the heat absorbed by the calorimeter itself. Always include the term \( C_{cal} imes \Delta T \) in the heat balance equation.
- Mistake 4 — Wrong sign convention for ΔT. Heat gain means \( \Delta T = T_{final} – T_{initial} > 0 \). If \( \Delta T \) is negative, the body is losing heat, not gaining it. Always check the direction of heat flow.
- Mistake 5 — Applying \( Q = mc\Delta T \) during a phase change. During melting or boiling, temperature stays constant. The correct formula is \( Q = mL \). Using \( Q = mc\Delta T \) with \( \Delta T = 0 \) will give \( Q = 0 \), which is physically incorrect.
JEE/NEET Application of Heat Gain Formula
The Heat Gain Formula appears in multiple forms across JEE Main, JEE Advanced, and NEET papers. In our experience, JEE aspirants encounter this formula in calorimetry, thermodynamics, and heat transfer problems almost every year. Understanding the different forms and knowing when to apply each one is critical for scoring well.
Pattern 1 — Calorimetry and Heat Balance
JEE and NEET frequently test the heat balance principle. A hot object is mixed with a cold liquid in a calorimeter. Students must equate heat gained by the cold substance to heat lost by the hot substance. The key is to include all components — liquid, calorimeter, and any phase changes. This pattern accounts for roughly 2-3 questions per year across JEE Main papers.
Pattern 2 — Conduction Through Composite Slabs
JEE Advanced tests heat gain through composite slabs (two or more materials in series or parallel). The concept of thermal resistance \( R = rac{d}{kA} \) is used. For slabs in series, total resistance \( R_{total} = R_1 + R_2 \). The heat flow rate is then \( H = rac{\Delta T}{R_{total}} \). Our experts suggest practising at least 10 composite slab problems before JEE Advanced.
Pattern 3 — Radiation and Newton’s Law of Cooling
NEET frequently asks about Newton’s Law of Cooling and blackbody radiation. The Stefan-Boltzmann Law \( P = \sigma A T^4 \) and the net radiation heat gain formula \( P_{net} = \sigma A (T^4 – T_0^4) \) are standard. In our experience, JEE aspirants often confuse emissive power with heat gain — remember that \( P_{net} \) represents the net heat gained or lost, not the total emission.
For the latest NCERT-aligned problems and official syllabus references, visit the NCERT official website and the CBSE official website.
FAQs on Heat Gain Formula
Explore more related physics formulas on ncertbooks.net. Visit our complete Physics Formulas hub for a full list of NCERT-aligned formula articles. You may also find the Terminal Velocity Formula and the Electric Current Formula useful for your CBSE and competitive exam preparation. For wave-particle duality topics, refer to our article on the de Broglie Wavelength Formula.