B.1 Concept Builder Part 3 · HL Physics Notebook

1. Start with predictions

7 minPairs

ATL Thinking: predict and justify. Social: compare with a partner. You will test every prediction later with a simulation.

1.1A metal bench and a wooden bench have been outside all night at the same temperature. You touch both. The metal feels colder because…

A. the metal is at a lower temperature B. the metal conducts energy away from your hand much faster C. the wood makes its own heat

You test it in stage 2.

1.2To warm a whole room with one heater, the best place for the heater is…

A. near the floor B. near the ceiling C. it does not matter

You test it in stage 3.

1.3A star keeps the same size, but its surface temperature (in kelvin) doubles. The power it radiates becomes…

A. 2 times bigger B. 4 times bigger C. 16 times bigger

You test it in stage 4.

1.4One star looks red and another looks blue-white. Which has the higher surface temperature?

A. the red star B. the blue-white star C. you cannot tell from the colour

You test it in stage 5.

1.5Explain the reasons for one of your predictions.

What a good answer looks like

A good reason uses an idea, for example: “The metal takes energy from my hand faster because…” or “A hotter star gives out light of shorter wavelength, so…”. You will see in the simulations which reasons hold.

2. Conduction: test prediction 1.1

18 minPairs → individual

ATL Self-management: change mm to m every time. Thinking: link the rate to k, A, ΔT and Δx.

Simulation A · Conduction through a slab one face is hotter than the other; the arrow shows the rate of energy flow ΔQ/Δt

area A 1.5 m²thickness Δx 4.0 mmtemperature difference ΔT 10 K

thermal conductivity k—

rate of energy flow ΔQ/Δt—

energy in 1 hour—

Learn it word for word

Conduction: energy passes from particle to particle by collisions: faster-vibrating particles at the hot end pass kinetic energy to their neighbours. In metals, free electrons also carry energy quickly, so metals are good conductors. ΔQ/Δt = kAΔT/Δx.

Using ΔQ/Δt = kAΔT/Δx

k from the data (W m⁻¹ K⁻¹).
A in m², Δx in metres: 4.0 mm = 0.0040 m.
ΔT is the difference between the two faces (the same in K or °C).
The answer is a power, in W. Energy = power × time.

2.1A glass window: A = 1.5 m², Δx = 4.0 mm, ΔT = 10 K across the glass (k = 0.80 W m⁻¹ K⁻¹). Rate of energy flow =

W

2.2Energy lost through this window in 1 hour =

J

2.3Set Δx = 8.0 mm (glass twice as thick). Rate of energy flow =

W

Find the mistake

A student calculates the rate for the window:

“ΔQ/Δt = 0.80 × 1.5 × 10/4 = 3 W.”

What is wrong?

The thickness must be in metres: 4.0 mm = 0.0040 m, so the rate is 3000 W. ΔT should be in kelvin, 283 K. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Copper (k = 400) and glass (k = 0.80) slabs have the same A, Δx and ΔT. Rate through copper ÷ rate through glass =

A. 500 B. 1 C. 0.002

Back check

2.bBack check (no simulation): a copper rod, A = 1.0 × 10⁻⁴ m², length 0.50 m, ends at 100 °C and 20 °C. Rate =

W

2.4Was prediction 1.1 right? Explain with conduction.

What a good answer looks like

B. Both benches are at the same temperature. Metal has a much larger thermal conductivity than wood (it has free electrons), so it conducts energy away from your hand much faster. Your skin cools quickly, so the metal feels colder. Wood is a poor conductor, so the skin where you touch it stays warm.

3. Convection: test prediction 1.2

10 minPairs

ATL Communication: explain with expand, density, rise and sink.

Simulation B · Convection in a room air near the heater warms, expands and becomes less dense

air next to the heater—

convection current—

the room—

Learn it word for word

Convection: a fluid (liquid or gas) near a hot object warms, expands and becomes less dense, so it rises; cooler, denser fluid sinks to take its place. This flow is a convection current. The fluid itself moves and carries the energy. Convection cannot happen in a solid or in a vacuum.

3.1Why does warm air rise?

it expands, so its density is lower than the cooler air around it heat always rises its molecules are lighter

3.2The freezing unit of an old fridge is at the top. Why?

air cooled at the top becomes denser and sinks, so a convection current cools the whole fridge cold rises it is easier to reach

Find the mistake

A student explains convection:

“The heat particles rise to the top of the room.”

What is wrong?

There are no heat particles: the air itself warms, becomes less dense and rises, carrying energy with it. Heat particles sink, not rise. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

In which can convection happen?

A. water and air B. a copper rod C. the vacuum of space

3.3Was prediction 1.2 right? Explain with density.

What a good answer looks like

A, near the floor. The air next to the heater warms, expands and becomes less dense, so it rises. Cooler, denser air sinks and flows along the floor to the heater. A convection current carries energy around the whole room. A heater near the ceiling only warms the air at the top, which is already the least dense, so it stays there and the floor stays cold.

4. Black-body power: test prediction 1.3

16 minIndividual

ATL Self-management: use kelvin and A = 4πR². Thinking: power laws (T⁴, R²).

Simulation C · Black-body radiation the intensity of the radiation at each wavelength; the dashed curve is the Sun (5800 K)

surface temperature T 5800 Kradius R 1.0 R☉

luminosity L = σ4πR²T⁴—

L ÷ L of the Sun—

peak wavelength λ_max—

Learn it word for word

A black body absorbs all the radiation that falls on it; it is also the best possible emitter. Stefan–Boltzmann law: the power radiated is L = σAT⁴, with T the surface temperature in kelvin. For a star, A = 4πR². Luminosity is the total power a star radiates.

4.1The Sun: R = 6.96 × 10⁸ m, T = 5800 K. Luminosity =

W

4.2Set T = 11 600 K and R = 1.0 R☉. L ÷ L of the Sun =

4.3A star has half the radius of the Sun and twice its temperature. L ÷ L of the Sun =

Find the mistake

A student finds the power from a hot plate at 300 °C, area 0.020 m²:

“L = 5.67 × 10⁻⁸ × 0.020 × 300⁴ = 9.2 W.”

What is wrong?

T must be in kelvin: 573 K, so L = 5.67 × 10⁻⁸ × 0.020 × 573⁴ = 122 W. The area should be squared. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Two stars have the same surface temperature. Star X has 3 times the radius of star Y. L(X) ÷ L(Y) =

A. 9 B. 3 C. 81

Back check

4.bBack check: a star has L = 1.0 × 10²⁸ W and T = 4000 K. Its radius =

m

4.4Was prediction 1.3 right? Explain with the law.

What a good answer looks like

C, 16 times. L = σ4πR²T⁴. With the same R, L is proportional to T⁴, and 2⁴ = 16. In the simulation, raising T from 5800 K to 11 600 K with R = 1.0 R☉ changes L from 1.0 to 16 L☉.

5. Wien’s law and stars: test prediction 1.4

19 minPairs → individual

ATL Thinking: link colour, peak wavelength and temperature. Communication: compare stars with L and b.

Simulation D · Stars approximate values for four real stars; brightness is the power per square metre arriving at Earth

peak wavelength λ_max—

luminosity L—

brightness at Earth b—

Learn it word for word

Wien’s law: λ_maxT = 2.9 × 10⁻³ m K. A hotter body has its peak at a shorter wavelength. Apparent brightness b = L/(4πd²): the power received per unit area at a distance d from the star.

5.1Peak wavelength for the Sun (T = 5800 K) =

m

5.2A star’s spectrum peaks at 290 nm. Its surface temperature =

K

5.3Brightness of the Sun at Earth (d = 1.5 × 10¹¹ m, L = 3.9 × 10²⁶ W) =

W m⁻²

5.4Use the simulation. Which star has by far the largest luminosity?

Betelgeuse Sirius A the Sun

Find the mistake

A student uses Wien’s law for a peak at 500 nm:

“T = 2.9 × 10⁻³/500 = 5.8 × 10⁻⁶ K.”

What is wrong?

The wavelength must be in metres: 500 nm = 5.00 × 10⁻⁷ m, so T = 5800 K. Wien’s law needs T in °C. Nothing is wrong.

Hinge question

Choose your answer. You will see at once if it is right, and why.

Betelgeuse is cooler than the Sun but far more luminous. This means Betelgeuse…

A. has a much larger radius B. is much closer to Earth C. is not a black body

What if? · stretch

5.wTwo stars have the same luminosity. Star A is 4 times further away than star B. b(A) ÷ b(B) =

5.5Was prediction 1.4 right? Explain with Wien’s law.

What a good answer looks like

B, the blue-white star. By Wien’s law λ_maxT is constant, so a higher temperature gives a shorter peak wavelength. Blue light has a shorter wavelength than red light. A blue-white star (such as Sirius A, about 9900 K, peak about 290 nm) is hotter than a red star (such as Betelgeuse, about 3600 K, peak about 810 nm).

6. Show it, then reflect

10 minIndividual

ATL Communication + Self-management: choose how to show your understanding, retrieve without notes, plan your next step.

Success criteria: your answer must…

name the three ways energy is transferred and say how each works with particles;
use the thickness in metres and the temperature in kelvin;
use L = σ4πR²T⁴ and λ_maxT = 2.9 × 10⁻³ m K correctly;
use one number from a simulation today.

Option A · Written

One paragraph: “How we know the temperature and radius of a star we cannot visit.”

Option B · Worked example

Pick a star in Simulation D. From its λ_max find T, then use L to find R. Check your R against the simulation.

Option C · Spoken

Explain to your partner in 1 minute why a vacuum flask keeps tea hot (conduction, convection and radiation). Your partner ticks the success criteria.

6.1Write your paragraph, your worked example or your notes.

What a good answer looks like

Model answer (Option A): A star is close to a black body. We measure its spectrum and find the peak wavelength; Wien’s law then gives the surface temperature, for example 2.9 × 10⁻³/(5.0 × 10⁻⁷) = 5800 K for the Sun. If we know the distance d, we measure the brightness b and find the luminosity from L = 4πd²b. Then L = σ4πR²T⁴ gives the radius: R = √(L/(4πσT⁴)). For the Sun this gives 7 × 10⁸ m. So two measurements, the peak wavelength and the brightness, plus the distance, give both T and R.

Exit ticket (no notes)

1A copper rod: A = 1.0 × 10⁻⁴ m², length 0.50 m, ΔT = 80 K, k = 400 W m⁻¹ K⁻¹. Rate of energy flow, in W =

2Convection cannot happen in…

a solid a liquid a gas

3A star’s spectrum peaks at 725 nm. Its surface temperature, in K =

4A star has R = 7.0 × 10⁸ m and T = 6000 K. Its luminosity, in W =

5A star of the same size as another has twice the surface temperature (in K). Its luminosity is…

16 times bigger 4 times bigger 2 times bigger

6Which idea from today is still the least clear to you?

What a good answer looks like

There is no single right answer. A good answer names one idea exactly, for example: “I am not sure when to change units in these laws.”

7Feedforward: Next time I will check ________ before I ________.

What a good answer looks like

A good feedforward names an action: “Next time I will write Δx in m, λ in m and T in K before I put any number into an equation.”

TOK · 2 minutes

Theory of knowledge: how do we know? There is no single right answer. Write two or three sentences.

TOKWe know the temperature of a star thousands of light years away from the colour of its light, using a law measured on Earth. What do we assume when we apply laws found in the laboratory to distant stars?

Points you could make

We assume the laws of physics are the same everywhere in the universe.
Stars are only approximately black bodies; the model gives useful but not exact temperatures.
Independent methods (spectral lines, distances from parallax) can check the results and so test the assumption.

Read the notes Part 2 Back to B.1

Conduction, convection and radiation

Learning goals: by the end you can…

1. Start with predictions

2. Conduction: test prediction 1.1

3. Convection: test prediction 1.2

4. Black-body power: test prediction 1.3

5. Wien’s law and stars: test prediction 1.4

6. Show it, then reflect

Exit ticket (no notes)