Subtopics - Transmission of Heat (NEET)
Three modes of heat transfer: conduction through solids (Fourier's law, thermal conductivity K, thermal resistance R = d/KA, series and parallel rod combinations, ice growth on lakes); convection in fluids (natural vs forced, Newton's law of cooling for small temperature differences); and radiation (electromagnetic energy transfer, Stefan-Boltzmann law P = εσAT⁴, Wien's displacement law λ_max·T = b, Kirchhoff's law of emission and absorption, Prevost's theory of heat exchange).
1) Conduction and Thermal Conductivity
Defines conduction as heat transfer through particle vibration without bulk displacement, applicable to all states of matter but dominant in solids. Fourier's law: Q/t = KA(θ₁ - θ₂)/l where K is thermal conductivity in W/m·K, A is cross-sectional area, l is length, and (θ₁ − θ₂) is the temperature difference. Thermal resistance R = l/KA analogous to electrical resistance; allows series (R_total = R₁ + R₂ + …) and parallel (1/R_total = 1/R₁ + 1/R₂ + …) combinations. Equivalent K for series slabs of equal length: K_s = 2K₁K₂/(K₁ + K₂). Temperature gradient = −Δθ/Δx; in steady state, dQ/dt = −KA(dθ/dx). Wiedemann-Franz law: K/σT = constant for metals. Ice growth on lake: time t = (ρL/2Kθ)y² — time grows as square of thickness.
2) Convection and Newton's Law of Cooling
Convection transfers heat by bulk movement of fluid. Natural convection is driven by density differences (hot fluid rises, cool sinks); forced convection is mechanically driven. Newton's law of cooling: dT/dt = −k(T − T₀), valid ONLY when temperature difference (T − T₀) is small. Integrated form: T − T₀ = Ae^(−kt), giving an exponential decay curve. For numerical problems: (θ₁ − θ₂)/t = k[(θ₁ + θ₂)/2 − θ₀], where θ₀ is surroundings temperature and (θ₁+θ₂)/2 is the average body temperature. Greater temperature excess → greater rate of cooling. Body can never be cooled below surroundings temperature by radiation. Heating done from bottom (convection possible); cooling done from top.
3) Radiation, Stefan-Boltzmann Law and Wien's Displacement Law
Radiation transfers heat as electromagnetic waves (infrared, λ = 7.8×10⁻⁷ to 4×10⁻⁴ m) without requiring a medium; fastest mode (c = 3×10⁸ m/s). Every body above 0 K emits radiation. Absorptance a + reflectance r + transmittance t = 1. Black body: a = 1, r = t = 0; perfect reflector: r = 1. Stefan-Boltzmann law: E = σT⁴ for black body; for ordinary body e = εσT⁴. Net emission in surroundings at T₀: e = εσ(T⁴ − T₀⁴). Stefan's constant σ = 5.67×10⁻⁸ W/m²K⁴. Emissivity ε: black body ε = 1, perfect reflector ε = 0. Wien's displacement law: λ_max·T = b = 2.89×10⁻³ m·K; hotter body → shorter λ_max (bluer peak); cooler body → longer λ_max (redder peak). Kirchhoff's law: e_λ/a_λ = constant at given temperature; good absorber = good emitter at same wavelength. Prevost's theory: all bodies continuously emit and absorb; equilibrium when emission = absorption.
Transmission of Heat Download Notes & Weightage Plan
For each topic in the Transmission of Heat chapter below, you get (2) the exact resources to download and how to use them, and (3) a simple importance & time plan so NEET students know what to do first and what to revise last.
Conduction and Thermal Conductivity
Fourier's law of heat conduction, thermal conductivity K, thermal resistance analogy, series and parallel rod combinations, interface temperature, and ice growth analysis.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: R = l/KA for each rod; H = ΔT_total/R_total for series; 1/R_total = 1/R₁ + 1/R₂ for parallel. Series equivalent K = 2K₁K₂/(K₁+K₂) for two equal-length rods. Interface temperature: equate H through both rods of composite bar.
- High-risk Area: Confusing K (thermal conductivity, property of material) with K_s (equivalent conductivity of the composite system). Also: using the parallel-rod formula when the rods are actually in series. Read the problem geometry carefully — same temperature at both faces = parallel; heat must pass through one to reach the other = series.
- Best Practice Style: Before calculating anything: (1) draw the heat-flow circuit, (2) label each element with R = l/KA, (3) identify series or parallel, (4) compute R_total, (5) apply H = ΔT/R_total. This 5-step approach eliminates configuration errors.
Convection and Newton's Law of Cooling
Natural vs forced convection, Newton's law of cooling dT/dt = −k(T−T₀) valid only for small temperature differences, exponential cooling curve, practical formula for cooling intervals.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: Interval formula: (θ₁−θ₂)/t = k[(θ₁+θ₂)/2 − θ₀]. Use first interval to find k, then apply for second interval. Know: body never cools below surroundings; rate of cooling α (θ−θ₀).
- High-risk Area: Applying Newton's law to problems where temperature excess is large (e.g., body at 1000 K, surroundings at 300 K). Newton's law is a linear approximation valid only for small ΔT. For large temperature differences, Stefan's T⁴−T₀⁴ law must be used.
- Best Practice Style: For every Newton's law problem, verify: (1) is ΔT small? (2) identify θ₁, θ₂, θ₀, and t₁. Then apply interval formula. If the problem involves two successive cooling intervals use the ratio property: (θ₁−θ₂)/(θ₂−θ₃) gives relative time ratio.
Radiation, Stefan-Boltzmann Law and Wien's Displacement Law
Electromagnetic radiation from all bodies above 0 K, Stefan-Boltzmann (P = εσAT⁴), Wien's displacement (λ_max·T = b), Kirchhoff's law (good absorber = good emitter), Prevost's exchange theory.
1) Download Packs For This Topic (And How To Use Them)
Don't download everything and forget it. Use these like a small "attack kit": read → highlight → test → revise the same sheet again.
2) Importance, Weightage & Time Allocation (Practical)
Use this to avoid over-studying. This topic is usually low effort, quick return if your recall is clean.
- Scoring Focus: P ∝ T⁴ — doubling T multiplies P by 16. λ_max ∝ 1/T — hotter star has shorter peak wavelength (blue shift). Kirchhoff: e_λ/a_λ = E_λ for black body — good absorber at λ is good emitter at λ.
- High-risk Area: Two traps: (1) Wien's direction — students confuse hotter with longer wavelength. The correct answer is hotter = shorter λ = bluer peak. (2) Stefan's T⁴ — confusing doubling T with doubling power (students pick 2× instead of 16×).
- Best Practice Style: For Wien's law questions: remember 'hotter means shorter wavelength peak, cooler means longer'. Mnemonic: 'hot blue, cold red' (matches stellar colour). For Stefan ratio questions: always raise both temperatures to the 4th power before computing ratio — never shortcut.
Transmission of Heat Chapter NEET Traps & Common Mistakes (Topic-Wise)
Each subtopic below is of the Transmission of Heat chapter and shows what NEET students usually do wrong in NEET examination, a short example of the mistake, and how NEET frames the question to trick you with close options are given below.
Mistake Snapshot (What Students Do Wrong)
- Concluding radiation doubles when temperature doubles:: P = εσAT⁴ means power varies as fourth power of absolute temperature. If T doubles (T → 2T), new power = εσA(2T)⁴ = 16εσAT⁴. Power increases by factor 16, not 2. Students who recall 'fourth power' but then divide by 4 (the exponent) instead of computing 2⁴ = 16 systematically get the wrong answer. NEET always includes 2× and 4× as distractor options.
- Using temperature in Celsius instead of Kelvin in Stefan's law:: Stefan's law requires absolute temperature T in Kelvin. If a body is at 57°C, the correct value is T = 57 + 273 = 330 K. Using 57 in the formula gives a drastically underestimated power. This error is especially common when the problem gives temperatures in Celsius and provides a conversion factor in the options.
A black body at 300 K emits power P. What power does it emit at 600 K? Correct: P₂ = σ(600)⁴ = σ × (2×300)⁴ = 16σ(300)⁴ = 16P. Wrong (common): P₂ = 2P (linear), or P₂ = 4P (using exponent value not power). NEET always places 4P or 2P as attractive distractors. Answer = 16P.
How NEET Frames The Trap
NEET phrasing: 'The temperature of a black body is doubled. By what factor does its radiated power change?' The presence of '4' in Stefan's law makes students answer 4 — but 4 is the exponent, not the factor. The factor is 2⁴ = 16.
Q. A body radiates maximum at wavelength λ₀ at temperature T. If the temperature is increased to 3T, the radiated power increases by a factor of:
A. 3 B. 9 C. 27 D. 81
Trick: P ∝ T⁴. New power ratio = (3T)⁴/T⁴ = 3⁴ = 81. Option D is correct. Option B (9) comes from squaring the temperature ratio instead of raising to fourth power; option A (3) is linear error.
Mistake Snapshot (What Students Do Wrong)
- Claiming a hotter body has a longer peak wavelength:: Wien's law: λ_max × T = b = 2.89×10⁻³ m·K. Therefore λ_max = b/T — peak wavelength is inversely proportional to temperature. A hotter body has a SHORTER peak wavelength (shifts towards blue/UV). Students who think in terms of infrared radiation and 'heat = longer wavelength' reverse this and claim hotter stars are redder. The correct statement: hotter stars peak in blue-UV; cooler stars peak in red-infrared.
- Calculating the wrong temperature for peak wavelength by forgetting the unit of b:: Wien's constant b = 2.89×10⁻³ m·K (metres × kelvin). When λ_max is given in nm or μm, students must convert to metres before dividing. Using λ_max = 500 nm without converting to 500×10⁻⁹ m gives a temperature 10⁹ times too large. Always convert λ to metres before applying T = b/λ_max.
Star A has peak wavelength 400 nm; Star B has peak wavelength 700 nm. Which is hotter? T_A = (2.89×10⁻³)/(400×10⁻⁹) = 7225 K. T_B = (2.89×10⁻³)/(700×10⁻⁹) = 4129 K. Star A (shorter λ) is hotter. NEET trap: Star B (red, longer λ) seems hotter because we associate red/heat — wrong. Shorter peak wavelength = higher temperature.
How NEET Frames The Trap
NEET question: 'The sun appears yellow and a distant star appears blue. Which is at a higher temperature?' The intuitive answer is wrong — blue star is hotter (shorter peak wavelength). Yellow sun is intermediate temperature. The blue-hot connection is counter-intuitive for students who associate fire colour with everyday red/orange visible flames.
Q. The wavelength of maximum radiation emitted by a black body is 0.5 μm. If Wien's displacement constant b = 2.9×10⁻³ m·K, the temperature of the body is:
A. 5800 K B. 580 K C. 1450 K D. 14500 K
Trick: T = b/λ_max = (2.9×10⁻³)/(0.5×10⁻⁶) = (2.9×10⁻³)/(5×10⁻⁷) = 5800 K. Option A correct. Option B (580 K) is from forgetting the 10⁻⁶ conversion for μm and computing 2.9×10⁻³/5×10⁻³ = 580.
Mistake Snapshot (What Students Do Wrong)
- Applying Newton's law when temperature difference is large:: Newton's law dT/dt = −k(T−T₀) is a linear approximation valid only when (T − T₀) is small (typically ≤30°C). For large temperature differences, the correct relation is Stefan's fourth-power law: dT/dt ∝ (T⁴ − T₀⁴). Using Newton's formula to calculate cooling time for a body at 800 K placed in surroundings at 300 K is incorrect — the quartic formula must be used. NEET tests this by asking which law applies under specified conditions.
- Using the wrong form of the interval formula when periods differ:: The interval formula (θ₁−θ₂)/t = k[(θ₁+θ₂)/2 − θ₀] requires the same surroundings temperature θ₀ in both equations. Students sometimes change θ₀ between intervals or use the endpoint temperature instead of the mean for the middle term, leading to wrong values of k and wrong time predictions.
A liquid cools from 70°C to 60°C in 5 min with surroundings at 30°C. How long to cool from 60°C to 50°C? Interval 1: (70−60)/5 = k[(70+60)/2 − 30] → 2 = k[65−30] = 35k → k = 2/35. Interval 2: (60−50)/t = k[(60+50)/2 − 30] = (2/35)[55−30] = (2/35)(25) = 50/35 → t = 10/(50/35) = 7 min. Wrong answer (common error): t = 5 min (assuming same time because same 10°C drop — ignores smaller mean excess temperature).
How NEET Frames The Trap
NEET gives two equal temperature drops (e.g., 10°C each) and asks if the times are equal. Students say yes — wrong. Equal temperature drops in successive intervals take INCREASING times because the mean temperature excess decreases at each step.
Q. A body cools from 80°C to 70°C in 2 minutes and from 70°C to 62°C in 2 minutes. The temperature of the surroundings is:
A. 20°C B. 30°C C. 10°C D. 25°C
Trick: Apply the interval formula twice. Interval 1: (80−70)/2 = k[(80+70)/2 − θ₀] → 5 = k(75−θ₀). Interval 2: (70−62)/2 = k[(70+62)/2 − θ₀] → 4 = k(66−θ₀). Dividing: 5/4 = (75−θ₀)/(66−θ₀) → 5(66−θ₀) = 4(75−θ₀) → 330−5θ₀ = 300−4θ₀ → 30 = θ₀. Answer = 30°C, Option B.
Mistake Snapshot (What Students Do Wrong)
- Forgetting that heat flows from higher temperature to lower temperature (not from higher 'heat content'):: In the thermal-electrical analogy, temperature plays the role of electric potential. Heat flows from high T to low T just as current flows from high V to low V. However, students sometimes confuse the direction with 'higher specific heat → heat flows toward' or 'greater thermal mass → heat flows toward'. The direction of heat flow is determined exclusively by temperature, not by heat capacity.
- Applying series formula when rods share the same temperature boundary (parallel configuration):: Series combination: rods are end-to-end; heat must pass through all rods; same heat current H through each; temperatures at interfaces are intermediate. Parallel combination: both rods share the same two end temperatures; heat splits between them; total H = H₁ + H₂. If both ends of the system are at the same two temperatures (not one rod feeding the other), it is a parallel circuit — not series.
Two identical rods of length l, area A, conductivity K₁ and K₂ join two plates at 100°C and 0°C. Case 1 (Series): R_total = 2l/KA where K_eff = 2K₁K₂/(K₁+K₂); H = ΔT/R_total = K_eff × A × 100/2l. Case 2 (Parallel): 1/R_total = K₁A/l + K₂A/l; H_total = H₁ + H₂. Confusing the two cases gives wrong H and wrong junction temperature. Parallel always carries more heat than series for same rods.
How NEET Frames The Trap
NEET diagram shows two rods connecting the same hot and cold plates. Students default to series when they are actually parallel. Key diagnostic: if both rods share the same two temperature endpoints without one feeding the other, they are in parallel.
Q. Two rods of same length and area but thermal conductivities K and 2K are connected in parallel between two walls at 100°C and 0°C. What is the effective thermal conductivity of the combination?
A. K B. 3K/2 C. 3K D. 2K
Trick: Parallel rods with same A and l: H_total = (KA×100/l) + (2KA×100/l) = 3KA×100/l. Effective: H = K_eff × (2A) × 100/l → K_eff × 2A = 3KA → K_eff = 3K/2. Option B is correct. Option D (2K) is wrong; students who average K₁ and K₂ directly get the wrong answer.