Rate of cooling of a body is proportional to the temperature difference between body and surroundings: dT/dt = -k(T - T_surroundings). Holds for small temperature differences.
-- NCERT Class 11 Physics, Ch. 10, p. 220Heat Transfer
8 MCQs4 revision cards9-step worked example
Source: NCERT Properties of Solids and LiquidsPYQ coverage: NEET 2021, 2026Official key: NTA-verifiedLast updated: 26 Sep 2026
Heat Transfer, explained for NEET
Heat moves from a hotter body to a cooler one through three mechanisms — conduction, convection, and radiation. NEET questions on this topic test whether you can distinguish the mechanism, recall the governing law, and apply the Stefan-Boltzmann formula for radiation problems.
Conduction is heat transfer through a medium without bulk motion of matter. Metals are good conductors; wood, air, and rubber are poor conductors (insulators). The rate of heat flow through a uniform rod depends on thermal conductivity, cross-sectional area, temperature difference, and length. NCERT Class 11 Physics Chapter 10 (page 211) states the law governing steady-state conduction.
Convection is heat transfer by actual movement of fluid. Natural convection arises from density differences (hot fluid rises); forced convection uses an external agent like a fan or pump. Sea breezes and land breezes are standard NCERT examples. Convection cannot occur in solids.
Radiation is heat transfer through electromagnetic waves — no medium required. Every body above 0 K radiates. The Stefan-Boltzmann law gives the total power radiated: P = σεAT⁴, where σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, ε is emissivity (1 for a black body), A is surface area, and T is absolute temperature in kelvin (NCERT Class 11 Physics Chapter 10, page 219). The net power radiated to surroundings at temperature T₀ is P_net = σεA(T⁴ − T₀⁴).
Watch-out: A common confusion is computing radiation power using Celsius instead of kelvin — the T⁴ dependence amplifies even a small offset. Always convert to kelvin first. Also note that emissivity ε = 1 for an ideal black body; real surfaces have ε < 1, and a good absorber is also a good emitter (Kirchhoff's law, NCERT Class 11 Physics Chapter 10, page 218).
Can you answer these Heat Transfer MCQs?
Select an option to see the explanation. Wrong answers show why your choice was tempting — and name the exact trap it exploits.
MCQ 1Easy RecallPractice
Which mode of heat transfer does NOT require a material medium?
Show answer and why every option is right or wrong
Answer: A. Radiation is the transfer of heat via electromagnetic waves, which can propagate through vacuum. NCERT Class 11 Physics Chapter 10, page 211 distinguishes the three modes explicitly.
Why B is wrong: B is wrong because convection requires bulk fluid motion, which is impossible without a material medium.
Why C is wrong: C is wrong because conduction requires a medium — heat travels via molecular collisions in the material.
Why D is wrong: D is wrong because both conduction and convection require a medium; the question asks for the one that does not.
MCQ 2Easy RecallPractice
According to Kirchhoff's law of thermal radiation, at thermal equilibrium a good absorber of radiation is also a:
Show answer and why every option is right or wrong
Answer: B. Kirchhoff's law states that at thermal equilibrium the ratio of emissive power to absorptive power is the same for all bodies, so a good absorber must be a good emitter. Kirchhoff's law is not stated in NCERT; NCERT Class 11 Physics Chapter 10, page 218 notes that black bodies absorb and emit radiant energy better than lighter-coloured bodies.
Why A is wrong: A is wrong because a good absorber absorbs most incident radiation rather than reflecting it — high absorptivity means low reflectivity.
Why C is wrong: C is wrong because good absorption implies the radiation is absorbed, not transmitted through the body.
Why D is wrong: D is wrong because it directly contradicts Kirchhoff's law — a body that absorbs well must emit well at the same temperature.
MCQ 3Easy RecallPractice
In which of the following is heat transferred primarily by convection?
Show answer and why every option is right or wrong
Answer: D. Sea breeze is driven by natural convection: land heats faster during the day, air above land rises, and cooler air from the sea moves in to replace it. NCERT Class 11 Physics Chapter 10, pages 211–212.
Why A is wrong: A is wrong because heat from the Sun travels through the vacuum of space via radiation, not convection.
Why B is wrong: B is wrong because the spoon is a solid — heat travels along it by conduction (molecular vibration transfer), not bulk fluid motion.
Why C is wrong: C is wrong because heat flow along a solid copper rod is conduction, not convection.
MCQ 4Direct ApplicationPractice
A black body at temperature T radiates power P. If its absolute temperature is doubled to 2T, the radiated power becomes:
Show answer and why every option is right or wrong
Answer: B. By the Stefan-Boltzmann law, P = σεAT⁴. When T → 2T, power → σεA(2T)⁴ = 16 σεAT⁴ = 16P. NCERT Class 11 Physics Chapter 10, page 219.
Why A is wrong: A is wrong because it treats power as proportional to T (linear), ignoring the T⁴ dependence in the Stefan-Boltzmann law.
Why C is wrong: C is wrong because it uses T³ dependence (2³ = 8); the correct exponent is 4, giving 2⁴ = 16.
Why D is wrong: D is wrong because it uses T² dependence (2² = 4), but the Stefan-Boltzmann law has T⁴, not T².
MCQ 5Direct ApplicationPractice
Two identical black bodies are at temperatures 300 K and 600 K. The ratio of the net rate of energy radiated by the hotter body to that of the cooler body is:
Show answer and why every option is right or wrong
Answer: C. Both bodies have the same σ, ε, and A. Power ratio = (600)⁴ / (300)⁴ = (600/300)⁴ = 2⁴ = 16. So the ratio is 16 : 1. NCERT Class 11 Physics Chapter 10, page 219.
Why A is wrong: A is wrong because it assumes power scales linearly with temperature (600/300 = 2), ignoring the fourth-power law.
Why B is wrong: B is wrong because it uses a square dependence (2² = 4); the Stefan-Boltzmann law requires T⁴.
Why D is wrong: D is wrong because it uses a cubic dependence (2³ = 8); the correct power is the fourth, giving 16.
MCQ 6Direct ApplicationPractice
A body with surface area 0.10 m² and emissivity 0.60 is at 500 K. Taking σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴, the power radiated by this body is closest to:
Show answer and why every option is right or wrong
Answer: D. P = σεAT⁴ = 5.67 × 10⁻⁸ × 0.60 × 0.10 × (500)⁴. Now (500)⁴ = 6.25 × 10¹⁰. So P = 5.67 × 10⁻⁸ × 0.060 × 6.25 × 10¹⁰ = 5.67 × 0.060 × 6.25 × 10² = 0.3402 × 625 = 212.6 ≈ 213 W. NCERT Class 11 Physics Chapter 10, page 219.
Why A is wrong: A is wrong — this combines both errors: ε = 1.0 and A = 1.0 m², giving roughly 10/0.6 ≈ 17 times the correct answer.
Why B is wrong: B is wrong — this value results from using ε = 1.0 (black body) instead of the given emissivity of 0.60.
Why C is wrong: C is wrong — this comes from using A = 1.0 m² instead of the given 0.10 m², inflating the answer by a factor of 10.
MCQ 7Concept TrapPractice
A thermos flask reduces heat loss by minimising all three modes of transfer. The silvered inner walls primarily reduce heat loss by:
Show answer and why every option is right or wrong
Answer: A. Silvered (shiny) surfaces have low emissivity and high reflectivity, so they reflect radiant heat back and emit very little — this targets radiation loss. The vacuum between walls handles conduction and convection. NCERT Class 11 Physics Chapter 10, page 218.
Why B is wrong: B is wrong because convection requires a fluid medium; the vacuum between walls eliminates convection, not the silvering.
Why C is wrong: C is wrong because conduction is reduced by the vacuum gap between the walls, not by the silvering itself.
Why D is wrong: D is wrong because both conduction and convection are addressed by the vacuum, not the silvered coating.
MCQ 8CalculationPractice
A solid sphere of radius R at temperature T is surrounded by an environment at temperature T₀. If the sphere's radius is halved (to R/2) while its temperature and emissivity remain unchanged, by what factor does the net rate of radiative heat loss change?
Show answer and why every option is right or wrong
Answer: C. Net radiative loss = σε A (T⁴ − T₀⁴). Surface area of a sphere A = 4πR². When R → R/2, A → 4π(R/2)² = πR² = (1/4) × 4πR². Temperature terms are unchanged, so the net loss scales by 1/4. The factor 4 in the area formula and the constant π are exact geometric/mathematical quantities and do not affect significant-figure counting.
Why A is wrong: A is wrong because it inverts the relationship, suggesting smaller radius increases radiative loss; in fact, smaller surface area means less radiation.
Why B is wrong: B is wrong because it assumes power scales linearly with radius (halving R halves P); but area depends on R², so halving R reduces area — and power — by a factor of 4, not 2.
Why D is wrong: D is wrong because it suggests power increases by 4 when radius is halved; the area (and hence power) decreases by 4, not increases.
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Heat Transfer: quick recall before you leave
How do you solve a Heat Transfer question? A worked example
- 1
Given
A furnace wall (treated as a black body, ε = 1) has a small observation hole of area A = 4.0 × 10⁻⁴ m². The furnace interior is at T = 1.50 × 10³ K. The surrounding temperature is T₀ = 300 K. σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
- 2
Required
Find the net rate of radiative heat loss through the hole.
- 3
Concept
The hole acts as a black-body radiator (Kirchhoff's law — a small hole in a large cavity is an ideal black body). The net power radiated to the surroundings follows from the Stefan-Boltzmann law.
- 4
Formula
P_net = σεA(T⁴ − T₀⁴)
- 5
Substitution
P_net = (5.67 × 10⁻⁸)(1)(4.0 × 10⁻⁴)[(1.50 × 10³)⁴ − (300)⁴]
- 6
Calculation
T⁴ = (1.50 × 10³)⁴ = (1.50)⁴ × 10¹² = 5.0625 × 10¹² K⁴
T₀⁴ = (300)⁴ = (3.00)⁴ × 10⁸ = 81.00 × 10⁸ = 8.100 × 10⁹ K⁴
T⁴ − T₀⁴ = 5.0625 × 10¹² − 0.0081 × 10¹² = 5.0544 × 10¹² K⁴
σ × ε × A = 5.67 × 10⁻⁸ × 1 × 4.0 × 10⁻⁴ = 2.268 × 10⁻¹¹ W K⁻⁴
P_net = 2.268 × 10⁻¹¹ × 5.0544 × 10¹² = 2.268 × 5.0544 × 10¹ = 114.6 × 10¹ ≈ 115 W
Note on exact quantities: ε = 1 (ideal black body, exact by definition) and the integer 4 in 4πR² or the exponent 4 in T⁴ are mathematical constants — they do not limit significant figures. The answer is reported to 3 significant figures, governed by the given area (4.0 × 10⁻⁴ has 2 sig figs) and temperature (1.50 × 10³ has 3 sig figs), so 2 sig figs controls: P_net ≈ 1.1 × 10² W. - 7
Final answer
P_net ≈ 1.1 × 10² W (two significant figures, governed by the area measurement 4.0 × 10⁻⁴ m²).
- 8
Common trap
Using Celsius instead of kelvin: if you mistakenly use 1500 °C as T without converting, the answer balloons because (1773)⁴ ≫ (1500)⁴. Always confirm T is in kelvin before raising to the fourth power.
- 9
Similar NEET-style question
A spherical black body of radius 0.10 m is at 800 K in surroundings at 300 K. Find the net rate of radiative heat loss. (Hint: compute A = 4πr², then apply P_net = σA(T⁴ − T₀⁴).)
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What to remember before solving Heat Transfer questions
Key Fact
Modes of heat transfer
Conduction: heat passes between neighbouring parts of a body through direct contact; in steady state the rate H = KA(TC − TD)/L, K being the thermal conductivity. Convection: bulk transport of different parts of a fluid (natural or forced). Radiation: electromagnetic waves; no medium needed; Stefan-Boltzmann law: a perfect radiator emits H = σAT⁴, other bodies a fraction of it set by their emissivity.
-- NCERT Class 11 Physics, Ch. 10, p. 214Formula
Stefan-Boltzmann law
Power radiated by a black body: P = σ A T⁴, where σ = 5.67 × 10⁻⁸ W/m²/K⁴. For grey body: P = ε σ A T⁴ with emissivity ε ≤ 1. Net power emitted to surroundings at T_s: P = ε σ A (T⁴ - T_s⁴).
-- NCERT Class 11 Physics, Ch. 10, p. 219Which Heat Transfer formulas do you need for NEET?
1 formula — click to collapse
Stefan-Boltzmann radiation law
Radiation power from a body. Black body epsilon=1; net to surroundings P = sigma*epsilon*A*(T^4 - T_s^4).
| Symbol | Quantity | SI Unit |
|---|---|---|
| sigma | Stefan-Boltzmann = 5.67e-8 | W/m^2/K^4 |
| epsilon | emissivity (0-1) | - |
| A | surface area | m^2 |
| T | absolute temp | K |
Valid when
- Body in radiative equilibrium
- T in kelvins
More in Properties of Bulk Matter: 5 exam traps and mistakes · 11 formulas · 5 question patterns from its other lessons.
Heat Transfer questions from past NEET papers
2 questions from NEET 2021, 2026. Answers verified against NTA official keys. — click to collapse
NEET 2026
1[L⁻¹T⁻¹K⁻²]
2[L⁻¹K⁻²]
3[L⁻¹T⁻¹K⁻³]
4[L⁻¹T⁻¹K⁻⁴]
NTA Answer: Option 1(final)
NEET 2021
1(5/13)t
2(13/10)t
3(13/5)t
4(10/13)t
NTA Answer: Option 3(final)
All 17 past-paper questions from Properties of Bulk Matter →
Sources
NCERT refs: Class 11 Physics Chapter 10, p.219 | Class 11 Physics Chapter 10, p.218
Page numbers are the ones printed in the current NCERT textbook (2023 rationalised edition), unless marked pre-2023. The books are free at ncert.nic.in.
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