PhysicsISC Class 11 · Thermal Properties of Matter

Heat Conduction, Newton's Law of Cooling and Radiation Laws

Updated 17 October 2026 · Fact-checked

In steady-state conduction, the rate of heat flow is H = KA(T₁ − T₂)/L. Newton's law of cooling states that the rate of cooling of a body is proportional to the temperature difference with its surroundings (for small differences). A blackbody radiates P = σAT⁴ (Stefan) and λ_m T = constant (Wien).

Heat flow H grows with area A and temperature difference, and falls as length L increases; a larger K means faster conduction.

Illustration drawn for QuestionBuddy.

Understand Conduction and Newton's Law of Cooling

Conduction. Heat flows through a solid without bulk movement of matter. In the steady state, the rate of flow through a slab or rod of cross-section A and length L, with end temperatures T₁ and T₂, is H = Q/t = KA(T₁ − T₂)/L. K is the thermal conductivity (W m⁻¹ K⁻¹). The temperature gradient is (T₁ − T₂)/L. Metals have large K, air and wool have small K.

Thermal resistance. Writing H = (T₁ − T₂)/R with R = L/(KA), conduction resembles Ohm's law. For slabs in series (same A) the total resistance is the sum, and the same H passes through each, which fixes the interface temperature. For slabs in parallel the heat flows add.

Searle's method. To find K of a good conductor like copper, a rod is heated at one end and cooled by flowing water at the other; the rise in the water's temperature gives the heat flow, and thermometers along the rod give the gradient. Lagging reduces side losses.

Convection and radiation. Convection involves the movement of the fluid itself. Radiation needs no medium. A blackbody absorbs all radiation; it radiates P = σAT⁴ (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴), with T in kelvin. Wien's displacement law: the wavelength of maximum intensity λ_m is inversely proportional to T, λ_m T = b (b = 2.9 × 10⁻³ m K).

Newton's law of cooling. dT/dt = −k(T − T₀). For a small drop in temperature, take the average temperature: (T₁ − T₂)/t = k[(T₁ + T₂)/2 − T₀]. The law holds when the temperature difference is small (about 30 K), mainly for forced convection.

Conduction and Newton's Law of Cooling: formulas and key facts

Conduction: $H = K A (T_{1} - T_{2}) / L ∣ R = L / (K A)$; Steady state; R adds in series.
Newton's law of cooling: $d T / d t = - k (T - T_{0}) ∣ (T_{1} - T_{2}) / t = k [(T_{1} + T_{2}) /2 - T_{0}]$; The second form is the average-temperature approximation.
Stefan's law: $P = σ A T^{4}$; Blackbody; σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴.
Wien's law: $λ_{m} T = b$; b = 2.9 × 10⁻³ m K.

Conduction and Newton's Law of Cooling in real life

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

A cold-box wall

Suppose a polystyrene box has total wall area 1.2 m², thickness 0.05 m and K = 0.04 W m⁻¹ K⁻¹. Its outside is at 30 °C and its inside at 0 °C.

Find the rate of heat flow and the mass of ice (L = 3.36 × 10⁵ J kg⁻¹) melted in one day.

Try it first, then show the solutionSolution

H = KAΔT/L = 0.04 × 1.2 × 30/0.05 = 28.8 W.
Heat in one day = 28.8 × 86 400 = 2.49 × 10⁶ J.
Mass melted = 2.49 × 10⁶/3.36 × 10⁵ = 7.4 kg.

Answer: 28.8 W; about 7.4 kg of ice per day.

Memory hook: A thicker wall or smaller K reduces the heat flow.

Tricks to remember Conduction and Newton's Law of Cooling

Short memory aids. Say each one aloud once, then use it on a question the same day.

Heat flow is like current
ΔT is the potential difference, R = L/KA is the resistance.
Use it when: Slabs in series or parallel.
Use the average temperature
In the cooling problems, replace T by the mean of the two readings.
Use it when: Newton's law numericals from two readings.

Conduction and Newton's Law of Cooling: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Unit of thermal conductivity · 1 mark · MCQBoard level

The SI unit of thermal conductivity is:
(a) W m⁻¹ K⁻¹
(b) J m⁻¹ K
(c) W m K⁻¹
(d) J K⁻¹

Try it first, then show the solutionSolution

K = HL/(AΔT), so the unit is W m/(m² K) = W m⁻¹ K⁻¹.

Answer: (a) W m⁻¹ K⁻¹

Example 2: Heat flow along a rod · 3 marks · Short answerBoard level

A copper rod 0.5 m long of cross-section 2 × 10⁻⁴ m² has its ends at 100 °C and 20 °C. Find the rate of heat flow in the steady state (K = 400 W m⁻¹ K⁻¹).

Try it first, then show the solutionSolution

H = KA(T₁ − T₂)/L. (1 mark)
= 400 × 2 × 10⁻⁴ × 80/0.5. (1 mark)
= 12.8 W. (1 mark)

Answer: 12.8 W

Example 3: Two slabs in series · 4 marks · Short answerHarder board level

Two slabs of equal area have thicknesses 0.1 m (K = 0.8 W m⁻¹ K⁻¹) and 0.2 m (K = 0.4 W m⁻¹ K⁻¹). The outer faces are at 100 °C and 20 °C. Find the heat flow per unit area and the temperature of the interface.

Try it first, then show the solutionSolution

R per unit area: 0.1/0.8 = 0.125 and 0.2/0.4 = 0.5; total 0.625 m² K W⁻¹. (1 mark)
H/A = 80/0.625 = 128 W m⁻². (1 mark)
Drop across the first slab = 128 × 0.125 = 16 K. (1 mark)
Interface temperature = 100 − 16 = 84 °C. (1 mark)

Answer: 128 W m⁻²; interface at 84 °C.

Example 4: Cooling of water · 4 marks · Short answerHarder board level

Hot water cools from 80 °C to 70 °C in 5 minutes in a room at 20 °C. Use Newton's law to find the time to cool from 70 °C to 60 °C.

Try it first, then show the solutionSolution

First interval: 10/5 = k(75 − 20) → k = 2/55 = 0.0364 min⁻¹. (2 marks)
Second interval: 10/t = k(65 − 20) = 0.0364 × 45 = 1.636. (1 mark)
t = 10/1.636 = 6.1 minutes (longer, since the difference from the room is smaller). (1 mark)

Answer: About 6.1 minutes.

Example 5: Rate of ice melting in a box · 5 marks · Long answerHOTS / competency

Suppose a cold box has walls of total area 1.2 m² and thickness 0.05 m made of material with K = 0.04 W m⁻¹ K⁻¹. It is kept at 0 °C inside while the outside is at 30 °C. How much heat enters per second, how much ice melts in 24 hours, and how would doubling the wall thickness change the melting rate?

Try it first, then show the solutionSolution

H = KAΔT/L = 0.04 × 1.2 × 30/0.05 = 28.8 W. (2 marks)
Heat in 24 h = 28.8 × 86 400 = 2.488 × 10⁶ J. (1 mark)
Mass = 2.488 × 10⁶/3.36 × 10⁵ = 7.4 kg. (1 mark)
With L doubled, H is halved: 14.4 W and about 3.7 kg per day. (1 mark)

Answer: 28.8 W; about 7.4 kg per day; halved by doubling the thickness.

Where students lose marks in Conduction and Newton's Law of Cooling

The slips examiners see most often, and what to write instead.

Mistake: Using °C in Stefan's law
Substituting 27 instead of 300 K.
Do this: Use absolute temperature in P = σAT⁴.
Usually costs: 2 marks.
Mistake: Adding conductivities in series
Adding K values for slabs in series.
Do this: Add the resistances L/(KA), not K.
Usually costs: 2 marks.
Mistake: Using T instead of T − T₀ in Newton's law
Taking the rate proportional to T.
Do this: Use the excess over the surroundings, with the average temperature.
Usually costs: 1 to 2 marks.

Exam tips

Draw the slab with temperatures labelled and the heat flow arrow.
Use thermal resistances for composite walls.
Write the average temperature clearly in cooling problems.
State that the law holds for small temperature differences.

Practice questions on Conduction and Newton's Law of Cooling

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQBoard level
Newton's law of cooling states that the rate of loss of heat is proportional to:
(a) the square of the temperature
(b) the temperature excess over the surroundings
(c) the fourth power of temperature
(d) the mass of the body only
Hint
It applies for a small difference of temperature.
Answer
(b) the temperature excess over the surroundings
Q2 · 2 marks · Short answerBoard level
A blackbody of surface area 0.5 m² is at 500 K. Find the power it radiates (σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴).
Hint
P = σAT⁴.
Answer
P = 5.67 × 10⁻⁸ × 0.5 × (500)⁴ = 5.67 × 10⁻⁸ × 0.5 × 6.25 × 10¹⁰ = 1772 W.
Q3 · 4 marks · Short answerHarder board level
A copper rod and a steel rod, each 0.5 m long and of equal area, are joined end to end. The free ends are at 100 °C and 0 °C. With K_Cu = 400 and K_steel = 50 W m⁻¹ K⁻¹, find the temperature of the joint when copper is at the hot end.
Hint
Use R = L/KA for each; same heat flow.
Answer
R ∝ 0.5/400 = 1.25 × 10⁻³ and 0.5/50 = 1.0 × 10⁻²; total 1.125 × 10⁻². Drop across copper = 100 × 1.25 × 10⁻³/1.125 × 10⁻² = 11.1 K, so the joint is at 88.9 °C.
Q4 · 3 marks · Short answerHarder board level
The wavelength of maximum emission of a star is 290 nm. Find its surface temperature (b = 2.9 × 10⁻³ m K). By what factor does the power per unit area change if the temperature doubles?
Hint
Use Wien's law then Stefan's law.
Answer
T = 2.9 × 10⁻³/2.9 × 10⁻⁷ = 10⁴ K. Power ∝ T⁴, so doubling T multiplies it by 16.
Q5 · 3 marks · Short answerHarder board level
A body is heated from 27 °C to 327 °C. By what factor does the power it radiates increase (consider only the body's own emission)?
Hint
Convert to kelvin first.
Answer
T₁ = 300 K, T₂ = 600 K. Ratio = (600/300)⁴ = 16.
Q6 · 4 marks · Case-basedHOTS / competency
Case study: Suppose a cup of tea at 90 °C cools to 70 °C in 4 minutes in a room at 30 °C.
(i) Find the cooling constant k by the average method. (1 mark)
(ii) Find the time to cool from 70 °C to 50 °C. (1 mark)
(iii) Find the rate of fall of temperature when the tea is at 70 °C. (1 mark)
(iv) What final temperature does the law predict? (1 mark)
Hint
Use (T₁ − T₂)/t = k(T_avg − T₀).
Answer
(i) 20/4 = k(80 − 30) → k = 0.1 min⁻¹. (ii) 20/t = 0.1 × (60 − 30) = 3, so t = 6.7 min. (iii) 0.1 × (70 − 30) = 4 K min⁻¹. (iv) 30 °C (the room temperature).

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write the conduction formula.Show answer
H = KA(T₁ − T₂)/L.
2.State Stefan's law.Show answer
P = σAT⁴ for a blackbody.
3.State Wien's law.Show answer
λ_m T = constant.
4.Which mode of heat transfer needs no medium?Show answer
Radiation.

Conduction and Newton's Law of Cooling: frequently asked questions

Why do metals feel colder than wood at the same temperature?

Metals conduct heat away from the hand quickly (high K), so the surface in contact stays near the temperature of the cold metal; wood conducts slowly.

Why does a thicker wall give better insulation?

H ∝ 1/L, so doubling the thickness halves the heat flow.

Is Newton's law exact?

No. It is a good approximation for small temperature differences. At large differences, radiation (∝T⁴) becomes important.