DP Physics SL B.1 Thermal Energy Transfers

DP Physics SL B.1 Thermal Energy Transfers - Full notes

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DP Physics SL | B.1 Thermal Energy Transfers

DP PHYSICS · THEME B: THE PARTICULATE NATURE OF MATTER

B.1

Thermal Energy Transfers

Full notes for SL · IB-style

Contents

1	The particle model of matter
2	Specific heat capacity
3	Changes of state and latent heat
4	Conduction and convection
5	Thermal radiation
6	Data-based questions
7	Exam technique and further practice
8	Practical work
9	Common misconceptions and links
10	Exam-style question with markscheme
11	Check your understanding
12	Summary and topic map

These are full teaching notes. Read each section in order, try the Think about it question before you read on, and do every Quick check before moving to the next section.

Before you start

Rate yourself on each statement now, and again after you've worked through the notes. Shade R (not yet), A (getting there) or G (confident).

I can…	R	A	G
describe the particle model of solids, liquids and gases and use density
relate temperature to average kinetic energy and use the Kelvin scale
use Q = mcΔT and Q = mL, including mixing, multi-stage problems and graphs
describe conduction and convection and calculate conduction rates
use Wien's law and the Stefan–Boltzmann law for black bodies
calculate apparent brightness and compare stars
describe experiments to measure c and L

There is no additional HL content in this subtopic. SL and HL students study the same material.

Guiding questions

How do macroscopic observations provide a model of the microscopic properties of a substance? How is energy transferred by conduction, convection and radiation? How can black-body radiation be used to describe stars?

How this topic is examined

B.1 is tested in Papers 1 and 2: specific heat and latent heat calculations, heating and cooling curves, conduction rates, black-body radiation graphs, Wien's law and Stefan–Boltzmann.

Always convert temperatures to kelvin in radiation and kinetic energy equations.

Command term	What the examiner wants
State	a brief answer with no explanation
Outline	a brief account or summary
Describe	a detailed account without reasons
Explain	a detailed account including causes, reasons or mechanisms
Calculate	a numerical answer with working and correct units
Determine	obtain the only possible answer, often from a graph or data
Show that	obtain the given result, showing every step (give one more significant figure)
Estimate	an approximate value using reasonable assumptions
Sketch	a graph or diagram showing the general shape and key features
Deduce / Suggest	reach a conclusion from given information; propose a reasonable explanation

How to use these notes

Each section teaches one part of the syllabus in full. You'll see these boxes throughout; each has one job:

Box	What it's for
Think about it	a question to get you thinking before a section
Key point	the one idea you must not forget
Common mistake	an error examiners see every year, and how to avoid it
In the real world	where the physics shows up outside the classroom
Try this	something quick you can do yourself to test the idea
Worked example	full solutions with IB-style marks (M1 method, A1 answer, R1 reasoning)
Quick check	short questions to test yourself at the end of each section

1 The particle model of matter

What things are made of, and how the particles move.

Think about it

Why can you compress a gas in a syringe but not a liquid?

Thermal physics explains large-scale observations (temperature, pressure, expansion, melting) using the behaviour of tiny particles. This section sets up the particle model and the idea of density.

1.1 Solids, liquids and gases

All matter is made of particles (atoms, molecules or ions) in constant random motion. In a solid, particles vibrate about fixed positions and are held by strong bonds. In a liquid, particles are close together but can slide past one another. In a gas, particles are far apart and move freely at high speeds, interacting only in collisions.

State	Arrangement	Motion	Compressibility
solid	regular, close-packed	vibrate about fixed points	very low
liquid	irregular, close	slide past each other	very low
gas	random, far apart	fast, random	high

Figure 1. Particle arrangements in a solid, a liquid and a gas.

Quick check

1. Why do gases fill their container?

Answers: 1. particles move freely and rapidly in all directions

1.2 Evidence for moving particles: Brownian motion

Smoke particles viewed under a microscope move in random, jerky paths. They are being struck unevenly by fast-moving, invisible air molecules. This Brownian motion (explained by Einstein in 1905) was key evidence that atoms and molecules exist.

Figure 2. The random path of a smoke particle in air (Brownian motion).

WORKED EXAMPLE 1 Explaining Brownian motion (standard)

Explain why the smoke particles move randomly.

Solution

Air molecules are moving randomly at high speedR1

Collisions on different sides are unequal at any instant, giving a changing resultant forceR1

Quick check

1. How would the motion change if the air were heated?

Answers: 1. more vigorous: air molecules move faster

1.3 Density

Density ρ = m/V, measured in kg m⁻³. Solids and liquids have similar densities because their particles are close together; gases are roughly 1000 times less dense because their particles are about ten times further apart in each direction.

Material	Density / kg m⁻³
air (20 °C)	1.2
water	1000
ice	920
aluminium	2700
iron	7900
mercury	13 600

WORKED EXAMPLE 2 Density (basic)

A 0.25 kg aluminium block has a volume of 9.26 × 10⁻⁵ m³. Calculate its density.

Solution

0.25/9.26 × 10⁻⁵M1

= 2700 kg m⁻³A1

WORKED EXAMPLE 3 Spacing estimate (exam level)

Water is about 1600 times denser than steam at 100 °C. Estimate how much further apart the molecules are in steam.

Solution

Volume per molecule 1600 times largerM1

Spacing ≈ 16001/3 ≈ 12 times largerA1

1.4 Temperature and the Kelvin scale

Temperature measures the average random kinetic energy of the particles. The Kelvin (absolute) scale starts at absolute zero, where the particles have minimum possible kinetic energy. T(K) = θ(°C) + 273. A temperature change of 1 K equals a change of 1 °C. Thermal equilibrium means no net energy flows between bodies because they are at the same temperature.

WORKED EXAMPLE 4 Kinetic energy (standard)

Calculate the average translational kinetic energy of gas molecules at 27 °C.

Solution

T = 300 K; E = 1.5 × 1.38 × 10⁻²³ × 300M1

= 6.2 × 10⁻²¹ JA1

Quick check

1. Convert −40 °C to kelvin.

2. By what factor does the average kinetic energy change from 300 K to 600 K?

Answers: 1. 233 K 2. it doubles

1.5 Internal energy

The internal energy of a substance is the total random kinetic energy plus intermolecular potential energy of all its particles. Heating can increase kinetic energy (temperature rises) or potential energy (a change of state at constant temperature).

Common mistake

Treating heat and temperature as the same thing. Temperature measures the average kinetic energy per particle; heat (thermal energy transferred) depends also on how much substance there is.

Quick check

1. Why does a bath of water at 40 °C have more internal energy than a cup at 80 °C?

Answers: 1. it contains far more particles

2 Specific heat capacity

How much energy changes temperature.

2.1 Defining specific heat capacity

The specific heat capacity c is the energy needed to raise the temperature of 1 kg of a substance by 1 K. Q = mcΔT. Water's high value (4200 J kg⁻¹ K⁻¹) means it heats and cools slowly.

Substance	c / J kg⁻¹ K⁻¹
water	4200
ice	2100
aluminium	900
iron	450
copper	390
lead	130

WORKED EXAMPLE 5 Kettle (standard)

Calculate the energy needed to heat 1.5 kg of water from 20 °C to 100 °C, and the time taken by a 2.0 kW kettle with no losses.

Solution

Q = 1.5 × 4200 × 80 = 5.04 × 10⁵ JM1 A1

t = 5.04 × 10⁵/2000 = 252 sA1

2.2 Heating graphs

For a constant power input P, the temperature rises at a rate ΔT/Δt = P/mc. A temperature–time graph is a straight line whose gradient gives c; losses make the graph curve and flatten at higher temperatures.

Figure 3. Temperature of 1.0 kg of water heated by a 200 W heater.

WORKED EXAMPLE 6 Gradient (exam level)

Using the 'no losses' line, find c for water.

Solution

Gradient = 28.6/600 = 0.0476 K s⁻¹M1

c = P/(m × gradient) = 200/0.0476 = 4200 J kg⁻¹ K⁻¹A1

2.3 Comparing materials

For the same energy supplied to equal masses, a material with a smaller specific heat capacity heats up more. On a temperature–energy graph, the gradient is 1/mc.

Figure 4. Temperature of 1.0 kg blocks of three metals as energy is supplied.

WORKED EXAMPLE 7 From the graph (standard)

Find the energy needed to raise the iron block from 20 °C to 60 °C.

Solution

Q = 1.0 × 450 × 40M1

= 18 kJA1

In the real world

Storage heaters use bricks with high heat capacity, heated overnight using cheaper electricity and releasing the energy slowly during the day.

2.4 Measuring specific heat capacity

Heat a block of known mass with an electrical heater; measure the energy supplied (VIt) and the temperature rise. c = VIt/(mΔT). Insulate the block, use oil in the thermometer hole for good contact, and continue recording after switching off to find the maximum temperature. Energy losses make the measured c too large.

Figure 5. Apparatus for measuring the specific heat capacity of a metal block.

WORKED EXAMPLE 8 Experiment (exam level)

A 1.0 kg aluminium block is heated by a 12 V, 4.0 A heater for 5.0 min and rises by 16 K. Calculate c and comment.

Solution

Energy = 12 × 4.0 × 300 = 14 400 JM1

c = 14 400/16 = 900 J kg⁻¹ K⁻¹; agrees with the data valueA1

Try this

Plot temperature against time for the heated block. Use the gradient of the straight section to find c.

2.5 Mixing problems

When hot and cold substances are mixed in an isolated system, energy lost by the hot one equals energy gained by the cold one until they reach the same final temperature.

Key point

Write 'energy lost by hot = energy gained by cold' and keep every ΔT positive.

WORKED EXAMPLE 9 Water mixing (standard)

0.20 kg of water at 80 °C is mixed with 0.30 kg of water at 20 °C. Find the final temperature.

Solution

0.20(80 − T) = 0.30(T − 20)M1

T = 44 °CA1

WORKED EXAMPLE 10 Metal in water (exam level)

A 0.10 kg copper block (c = 390) at 200 °C is placed in 0.40 kg of water at 20 °C. Find the final temperature.

Solution

0.10(390)(200 − T) = 0.40(4200)(T − 20)M1

T = 24.1 °CA1

2.6 Thermal capacity and climate

The heat capacity C = mc of an object (J K⁻¹) tells how much energy changes its temperature by 1 K. Oceans have an enormous heat capacity, so coastal climates have smaller temperature swings than inland ones.

WORKED EXAMPLE 11 Ocean layer (exam level)

Estimate the energy needed to warm the top 100 m of an ocean area of 1.0 km² by 1.0 K.

Solution

m = 1000 × 10⁶ × 100 = 1.0 × 10¹¹ kgM1

Q = 1.0 × 10¹¹ × 4200 × 1.0 = 4.2 × 10¹⁴ JA1

In the real world

The oceans have absorbed more than 90% of the extra energy trapped by greenhouse gases since the 1970s, which is why ocean warming is a key climate indicator.

3 Changes of state and latent heat

Energy without temperature change.

Think about it

Why does steam at 100 °C cause worse burns than water at 100 °C?

3.1 Phase changes

During melting, boiling, freezing and condensation, the temperature stays constant. Energy supplied goes into breaking (or released by forming) intermolecular bonds, changing potential energy, not kinetic energy.

Figure 6. Heating curve for ice to steam at constant power: flat sections are changes of state.

Quick check

1. Why is the boiling plateau longer than the melting plateau for water?

Answers: 1. the latent heat of vaporization (2.26 MJ kg⁻¹) is about 7 times the latent heat of fusion

3.2 Specific latent heat

The specific latent heat L is the energy needed to change the state of 1 kg of a substance without changing its temperature: Q = mL. For water, Lf = 3.34 × 10⁵ J kg⁻¹ (fusion) and Lv = 2.26 × 10⁶ J kg⁻¹ (vaporization).

WORKED EXAMPLE 12 Melting ice (standard)

Calculate the energy to melt 0.50 kg of ice at 0 °C.

Solution

0.50 × 3.34 × 10⁵M1

= 1.67 × 10⁵ JA1

WORKED EXAMPLE 13 Steam burns (exam level)

Compare the energy released when 10 g of steam at 100 °C and 10 g of water at 100 °C cool to 37 °C on skin.

Solution

Water: 0.010 × 4200 × 63 = 2.6 kJM1

Steam: 0.010 × 2.26 × 10⁶ + 2.6 kJ = 25 kJ (about 10 times more)A1

3.3 Cooling curves

A cooling curve for a liquid that freezes shows a horizontal section at the freezing point: latent heat is released as the bonds form, keeping the temperature constant until all the liquid has solidified.

Figure 7. Cooling curve for a wax that freezes at 60 °C.

WORKED EXAMPLE 14 Latent heat from a graph (exam level)

The wax (mass 0.050 kg) loses energy at 2.0 W while freezing for 8.0 minutes. Find its specific latent heat of fusion.

Solution

Q = 2.0 × 480 = 960 JM1

L = 960/0.050 = 1.9 × 10⁴ J kg⁻¹A1

3.4 Multi-stage heating

To take ice below 0 °C to steam, add the energy for each stage: warm the ice, melt it, warm the water, boil it, warm the steam.

WORKED EXAMPLE 15 Ice to water (exam level)

Calculate the energy to convert 0.20 kg of ice at −10 °C into water at 30 °C.

Solution

Warm ice: 0.20 × 2100 × 10 = 4200 J; melt: 0.20 × 3.34 × 10⁵ = 66 800 JM1

Warm water: 0.20 × 4200 × 30 = 25 200 JA1

Total = 9.62 × 10⁴ JA1

WORKED EXAMPLE 16 Ice in a drink (exam level)

0.050 kg of ice at 0 °C is added to 0.25 kg of water at 40 °C. Find the final temperature.

Solution

0.25(4200)(40 − T) = 0.050(3.34 × 10⁵) + 0.050(4200)TM1

T = 20.1 °CA1

3.5 Measuring latent heat

Latent heat of vaporization: an immersion heater keeps water boiling on a balance; record the mass lost in a measured time. Lv = Pt/Δm. Repeating at two powers and subtracting cancels heat losses.

WORKED EXAMPLE 17 Boiling (exam level)

A 120 W heater boils away 0.020 kg of water in 400 s. Find Lv and explain why it is too high.

Solution

L = 120 × 400/0.020 = 2.4 × 10⁶ J kg⁻¹M1 A1

Some energy is lost to the surroundings, so less evaporates than expectedR1

3.6 Refrigerators and heat pumps

A refrigerator uses a refrigerant that evaporates inside the fridge (absorbing latent heat from the food compartment) and condenses outside (releasing it into the room). A compressor does work to move energy from cold to hot. Heat pumps for homes use the same cycle, moving energy from cold outside air into the warm house.

Figure 8. Energy transfers in a refrigerator cycle.

Quick check

1. Why does the back of a fridge feel warm?

Answers: 1. the refrigerant condenses there, releasing latent heat

3.7 Evaporation and cooling

Evaporation occurs at the surface at any temperature. The fastest molecules escape, lowering the average kinetic energy of those remaining, so the liquid cools. Evaporation increases with temperature, surface area and airflow and decreases with humidity.

Quick check

1. Explain why sweating cools the body less effectively in humid air.

Answers: 1. less evaporation occurs when the air already contains much water vapour

4 Conduction and convection

Energy transfer through materials and fluids.

4.1 Conduction

Conduction transfers energy through a material from hotter to colder regions by collisions between particles and, in metals, by fast-moving free electrons. The rate of energy flow is ΔQ/Δt = kAΔT/Δx, where k is the thermal conductivity (W m⁻¹ K⁻¹).

Material	k / W m⁻¹ K⁻¹
copper	400
aluminium	240
glass	0.8
brick	0.6
water	0.6
wood	0.1
air	0.025
polystyrene foam	0.03

Figure 9. Conduction through a slab of area A and thickness Δx.

WORKED EXAMPLE 18 Window (exam level)

A glass window (k = 0.80) is 1.5 m × 1.0 m and 4.0 mm thick, with surface temperatures 20 °C and 5 °C. Calculate the rate of energy loss.

Solution

0.80 × 1.5 × 15/0.0040M1

= 4500 WA1

4.2 Temperature gradients

In steady state, the temperature falls linearly through a uniform slab. In a wall made of layers, the largest temperature drop occurs across the layer with the lowest conductivity per thickness (the best insulator).

Figure 10. Temperature through a wall of plaster and foam.

WORKED EXAMPLE 19 Double glazing (exam level)

Explain why double glazing reduces heat loss much more than doubling the glass thickness.

Solution

The trapped gas layer has very low kR1

Most of the temperature drop occurs across the gas, reducing the energy flowR1

In the real world

Double glazing traps a layer of air or argon (k ≈ 0.02 W m⁻¹ K⁻¹), roughly halving heat loss through windows compared with single glazing.

4.3 Convection

Convection transfers energy by bulk movement of a fluid. Heated fluid expands, becomes less dense and rises; cooler, denser fluid sinks to replace it, forming a convection current. It cannot occur in solids.

Figure 11. A convection current set up by a radiator.

WORKED EXAMPLE 20 Sea breeze (exam level)

Explain why a breeze blows from the sea to the land on a sunny day.

Solution

Land heats faster (lower heat capacity); air above it risesR1

Cooler, denser air from over the sea moves in to replace itR1

4.4 Reducing energy transfers at home

Loft insulation and cavity-wall foam trap air, reducing conduction and convection. Curtains and draught excluders reduce convection currents. Shiny foil behind radiators reflects infrared radiation.

WORKED EXAMPLE 21 U-values (exam level)

A wall has U-value 0.30 W m⁻² K⁻¹ and area 40 m². Find the energy lost in a day when the temperature difference is 15 K.

Solution

P = UAΔT = 0.30 × 40 × 15 = 180 WM1

E = 180 × 86 400 = 1.6 × 10⁷ JA1

5 Thermal radiation

Energy carried by electromagnetic waves.

5.1 Radiation and surfaces

All objects above absolute zero emit electromagnetic radiation, mainly infrared at everyday temperatures. Radiation needs no medium. Dark, matt surfaces are good absorbers and good emitters; light, shiny surfaces are poor absorbers and poor emitters.

Figure 12. Cooling of cans of hot water with different outer surfaces.

Quick check

1. Why are emergency blankets shiny?

Answers: 1. poor emitters, so they reduce energy loss by radiation

5.2 The Leslie cube

A hollow cube with four differently finished faces is filled with hot water. An infrared detector placed at the same distance from each face shows that matt black emits most and shiny silver least, at the same temperature.

Figure 13. Relative infrared emission from the faces of a Leslie cube.

WORKED EXAMPLE 22 Control variables (standard)

State two variables to keep constant when comparing the faces.

Solution

Distance from detector to faceA1

Temperature of the water (measure quickly)A1

5.3 Black-body radiation

A black body absorbs all radiation falling on it and is a perfect emitter. Its emission spectrum depends only on its temperature: a hotter body emits more at every wavelength and its peak shifts to shorter wavelengths. Stars are close to black bodies.

Figure 14. Black-body spectra at three temperatures.

Quick check

1. What happens to the colour of a heated metal as its temperature rises?

Answers: 1. red, then orange, then white as the peak moves to shorter wavelengths

5.4 Wien's displacement law

λmaxT = 2.9 × 10⁻³ m K. This gives the surface temperature of a star from its peak wavelength.

WORKED EXAMPLE 23 Sun (standard)

The Sun's spectrum peaks at 500 nm. Estimate its surface temperature.

Solution

T = 2.9 × 10⁻³/5.00 × 10⁻⁷M1

= 5800 KA1

WORKED EXAMPLE 24 Human body (standard)

Find the peak wavelength emitted by skin at 310 K.

Solution

λ = 2.9 × 10⁻³/310M1

= 9.4 μm (infrared)A1

5.5 The Stefan–Boltzmann law

The total power (luminosity) radiated by a black body is L = σAT⁴, with σ = 5.67 × 10⁻⁸ W m⁻² K⁻⁴. Doubling the kelvin temperature increases the power 16 times. A real surface radiates P = eσAT⁴, with emissivity e < 1.

Figure 15. Power radiated per square metre of a black body against temperature.

WORKED EXAMPLE 25 Star (exam level)

A star has radius 7.0 × 10⁸ m and surface temperature 5800 K. Calculate its luminosity.

Solution

A = 4πr² = 6.16 × 10¹⁸ m²M1

L = 5.67 × 10⁻⁸ × 6.16 × 10¹⁸ × 5800⁴ = 3.9 × 10²⁶ WA1

WORKED EXAMPLE 26 Net radiation (exam level)

A person (area 1.8 m², skin 307 K, e = 0.97) stands in a room at 293 K. Estimate the net power radiated.

Solution

P = eσA(T₁⁴ − T₂⁴)M1

= 0.97 × 5.67 × 10⁻⁸ × 1.8 × (307⁴ − 293⁴) = 150 WA1

5.6 Apparent brightness

Apparent brightness b = L/4πd² is the power received per unit area at distance d from a source. It obeys the inverse-square law.

WORKED EXAMPLE 27 Solar constant (standard)

Calculate the intensity of sunlight at Earth (d = 1.5 × 10¹¹ m, L = 3.8 × 10²⁶ W).

Solution

b = 3.8 × 10²⁶/(4π(1.5 × 10¹¹)²)M1

= 1.3 × 10³ W m⁻²A1

WORKED EXAMPLE 28 Distance (exam level)

A star with the Sun's luminosity has apparent brightness 1.0 × 10⁻¹⁰ W m⁻². Find its distance.

Solution

d = √(L/4πb)M1

= 5.5 × 10¹⁷ m (about 58 ly)A1

5.7 Comparing stars

Combining L = 4πR²σT⁴ with Wien's law lets astronomers compare stellar radii. Stars of the same luminosity but lower temperature must be larger.

WORKED EXAMPLE 29 Radius ratio (exam level)

Star X has the same luminosity as the Sun but half its surface temperature. Find its radius in solar radii.

Solution

R²T⁴ constant: R² = 2⁴M1

R = 4 R☉A1

6 Data-based questions

Analysing thermal data.

6.1 Latent heat from two powers

Heat losses can be eliminated by boiling water at two different heater powers. If losses are the same rate h in both, P₁ = L(Δm₁/Δt) + h and P₂ = L(Δm₂/Δt) + h; subtracting removes h.

Heater power / W	Mass boiled in 300 s / g
60	6.2
120	14.2

WORKED EXAMPLE 30 Two powers (exam level)

Use the data to find Lv and the rate of energy loss.

Solution

60 = L(6.2 × 10⁻³/300) + h; 120 = L(14.2 × 10⁻³/300) + hM1

Subtract: 60 = L(8.0 × 10⁻³/300): L = 2.25 × 10⁶ J kg⁻¹A1

h = 60 − 46.5 = 13.5 WA1

6.2 Specific heat from a graph of data

Plotting energy supplied against temperature rise gives a straight line of gradient mc. A non-zero intercept indicates energy absorbed by the container or lost at the start.

Figure 16. Energy supplied against temperature rise for a 1.0 kg block.

WORKED EXAMPLE 31 Gradient (exam level)

Find c for the block.

Solution

Gradient = (23.5 − 1.0)/25 = 0.90 kJ K⁻¹M1

c = 900 J kg⁻¹ K⁻¹ (aluminium)A1

6.3 Stellar data

Questions often give a table of stars with peak wavelength and luminosity and ask for temperatures and radii.

Star	λmax / nm	L / L☉
Sun	500	1
Sirius A	290	25
Barnard's Star	910	0.0035

WORKED EXAMPLE 32 Sirius (exam level)

Find the surface temperature of Sirius A and its radius in solar radii (T☉ = 5800 K).

Solution

T = 2.9 × 10⁻³/2.9 × 10⁻⁷ = 10 000 KA1

R/R☉ = √(L/L☉)(T☉/T)² = 5 × 0.336 = 1.7M1 A1

7 Exam technique and further practice

Putting thermal physics together.

7.1 Rate of cooling from a graph

The rate of cooling at an instant is the gradient of the tangent to a temperature–time graph. Hotter objects cool faster because the temperature difference with the surroundings is larger, so the curve flattens over time.

Figure 17. Cooling of hot water, with a tangent drawn at 5 minutes.

WORKED EXAMPLE 33 Tangent (exam level)

The tangent at t = 5 min passes through (0, 76.0 °C) and (15 min, 26.6 °C). For 0.25 kg of water, find the rate of energy loss.

Solution

Gradient = −49.4/15 = −3.29 K min⁻¹ = −0.0549 K s⁻¹M1

P = mc × rate = 0.25 × 4200 × 0.0549 = 58 WA1

7.2 Method checklist

•Convert temperatures to kelvin for radiation laws and kinetic energy; °C differences are fine for Q = mcΔT.
•In mixing problems, write energy lost = energy gained with all ΔT positive.
•In multi-stage heating, list each stage and its formula.
•Use areas 4πR² for spheres in Stefan–Boltzmann problems.
•Convert mm, cm and minutes into SI units.

7.3 Multi-step problem: kettle and steam

WORKED EXAMPLE 34 Kettle (exam level)

A 2.4 kW kettle containing 1.2 kg of water at 15 °C is switched on and accidentally left on. Find (a) the time to reach boiling; (b) the mass boiled away in the next 3.0 minutes. Assume no losses.

Solution

(a) Q = 1.2 × 4200 × 85 = 4.28 × 10⁵ J; t = 178 sM1 A1

(b) 2400 × 180 = 4.32 × 10⁵ J; m = 4.32 × 10⁵/2.26 × 10⁶ = 0.19 kgM1 A1

7.4 Multi-step problem: star

WORKED EXAMPLE 35 Betelgeuse (exam level)

Betelgeuse has peak wavelength 850 nm and luminosity 4.5 × 10³¹ W. Find its surface temperature and radius.

Solution

T = 2.9 × 10⁻³/8.5 × 10⁻⁷ = 3400 KA1

R = √(L/4πσT⁴) = 6.8 × 10¹¹ m (about 1000 solar radii)M1 A1

7.5 Graded practice

WORKED EXAMPLE 36 Lead shot (standard)

Lead shot (c = 130) falls 1.5 m inside a tube 20 times. Assuming all energy heats the lead, find its temperature rise.

Solution

20 × g × 1.5 = 130ΔTM1

ΔT = 2.3 K (independent of mass)A1

WORKED EXAMPLE 37 Freezer (exam level)

A freezer removes energy at 80 W. How long does it take to freeze 0.50 kg of water already at 0 °C?

Solution

Q = 0.50 × 3.34 × 10⁵ = 1.67 × 10⁵ JM1

t = 2090 s = 35 minA1

WORKED EXAMPLE 38 Conduction through a pan (exam level)

A copper pan base (k = 400) is 3.0 mm thick with area 0.020 m². Water boils off at 1.0 g s⁻¹. Find the temperature difference across the base.

Solution

Power = 0.0010 × 2.26 × 10⁶ = 2260 WM1

ΔT = PΔx/kA = 2260 × 0.003/(400 × 0.020) = 0.85 KA1

7.6 Comparing the three transfer mechanisms

Feature	Conduction	Convection	Radiation
medium needed?	yes (mainly solids)	yes (fluids only)	no
mechanism	particle collisions, free electrons	bulk fluid flow due to density differences	electromagnetic waves (mainly infrared)
depends on	k, A, ΔT/Δx	fluid density changes, shape of space	T⁴, surface area, emissivity
reduced by	insulators, trapped air	trapping air, lids, barriers	shiny surfaces, low emissivity

WORKED EXAMPLE 39 Vacuum flask (exam level)

Explain how a vacuum flask reduces energy transfer by each mechanism.

Solution

Vacuum between walls stops conduction and convectionR1

Silvered walls reduce radiation; stopper reduces convection and evaporation from the topR1

7.7 Common exam errors

•Using °C in Wien's or Stefan–Boltzmann laws.
•Forgetting the latent heat stage in ice or steam problems.
•Writing that temperature rises during melting.
•Using the radius instead of the surface area for a star's luminosity.
•Confusing conduction (through solids) with convection (fluid movement).

8 Practical work

Investigating thermal properties.

8.1 Cooling curves and insulation

Fill identical cans with hot water, wrap them in different insulators and record temperatures every minute. Compare initial cooling rates (gradients) and times to fall by a fixed amount.

Try this

Compare the cooling of hot water in a mug with and without a lid. Which energy transfer does the lid mostly reduce?

Quick check

1. Name two variables that must be controlled.

Answers: 1. starting temperature and volume of water (and room temperature)

8.2 Measuring thermal conductivity (qualitative)

Rods of different metals with wax beads at equal intervals are heated at one end; the order in which the beads fall shows relative conductivities. Quantitative methods measure temperature gradient and energy flow in steady state.

Quick check

1. Why must the rods be the same length and diameter?

Answers: 1. for a fair comparison (A and Δx controlled)

9 Common misconceptions and links

Students often think…	Actually…
heat and temperature are the same	temperature measures average KE; heat is energy transferred
temperature rises during melting	it stays constant; energy increases potential energy
cold flows into objects	energy flows from hot to cold
only hot objects emit radiation	all objects above 0 K emit
Wien's law uses °C	temperature must be in kelvin

How B.1 connects to the rest of the course

Topic	Connection
B.2 Greenhouse effect	black-body radiation and Earth's energy balance
B.3 Gas laws	kinetic theory and temperature
B.4 Thermodynamics (HL)	internal energy and the first law
E.5 Fusion and stars	luminosity, HR diagrams
A.3 Energy	conservation of energy and power

10 Exam-style question with markscheme

Try this under exam conditions (about 18 minutes) before reading the markscheme.

A student heats 0.30 kg of water in an insulated container using a 50 W immersion heater. The water starts at 18 °C.

(a) Calculate the time to reach 30 °C, assuming no losses. [2]

(b) The actual time was 340 s. Calculate the percentage of energy lost. [2]

(d) The Sun's spectrum peaks at 500 nm. Calculate its surface temperature. [2]

(e) Explain why the Sun's apparent brightness at Mars is less than at Earth. [2]

Markscheme

Part	Answer	Marks
(a)	Q = 0.30 × 4200 × 12 = 15 120 J	M1
	t = 302 s	A1
(b)	energy supplied = 17 000 J	M1
	lost = 1880 J = 11%	A1
(c)	0.020 × 2.26 × 10⁶	M1
	4.5 × 10⁴ J	A1
(d)	T = 2.9 × 10⁻³/5.0 × 10⁻⁷	M1
	5800 K	A1
(e)	Mars is further away	R1
	b = L/4πd²: intensity spreads over a larger area	R1

Examiner's comments. In (b), several candidates subtracted times rather than energies. In (d), a few used the wavelength in nm.

11 Check your understanding

Section A: multiple choice

1. The specific heat capacity of a substance is the energy to raise

A. 1 kg by 1 K B. 1 g by 1 K C. 1 kg by 100 K D. 1 mol by 1 K

2. During boiling the temperature of water

A. rises B. falls C. stays constant D. oscillates

3. Absolute zero is

A. 0 °C B. −273 °C C. −100 °C D. 273 °C

4. Energy transfer through a vacuum occurs only by

A. conduction B. convection C. evaporation D. radiation

5. If a black body's kelvin temperature doubles, its emitted power

A. doubles B. increases 16 times C. quadruples D. halves

6. Wien's law relates peak wavelength to

A. temperature B. luminosity C. distance D. radius

7. Apparent brightness varies with distance as

A. 1/d B. d² C. 1/d³ D. 1/d²

8. Metals conduct well mainly because of

A. large atoms B. high density C. free electrons D. convection

Section B: structured

9. Calculate the final temperature when 0.050 kg of ice at 0 °C is added to 0.25 kg of water at 40 °C. [4]

10. A brick wall 0.20 m thick has area 12 m² and k = 0.60 W m⁻¹ K⁻¹. Find the rate of energy loss for a 15 K difference. [2]

11. Explain, in terms of particles, why temperature stays constant during melting. [3]

12. Star A has twice the radius and the same temperature as star B. Compare their luminosities. [2]

Answers

1. A 2. C 3. B 4. D 5. B 6. A 7. D 8. C

9. Energy to melt 16 700 J; 0.25(4200)(40 − T) = 16 700 + 0.050(4200)T: T = 20.1 °C.

10. 0.60 × 12 × 15/0.20 = 540 W.

11. Energy breaks intermolecular bonds, increasing potential energy; average KE unchanged.

12. L ∝ R²: LA = 4LB.

12 Summary and topic map

Idea	Remember
Particle model	solid, liquid, gas; density
Temperature	Ek = 3/2 kB T; kelvin
Heating	Q = mcΔT; Q = mL
Conduction	ΔQ/Δt = kAΔT/Δx
Radiation	λmax T = 2.9 × 10⁻³ m K; L = σAT⁴
Brightness	b = L/4πd²

Before the exam, make sure you…

•convert to kelvin for radiation laws
•add all stages in heating curve problems
•state 'energy lost = energy gained' in mixing
•use areas 4πR² for stars

Thermal Energy Transfers on one page

Cover this page with a sheet of paper and try to rebuild it from memory. Whatever you can't recall is what to revise next.

Figure 18. Topic map for Thermal Energy Transfers.

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