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IB Physics · Theme B.1 · by Ahmed Gomaa

Thermal Energy Field Guide

Watch particles in solids, liquids and gases, change between °C and kelvin, see where the energy goes while ice melts, mix hot and cold water, read a heating curve step by step, send energy through a wall, and find the temperature and brightness of a star from its light.

Hot, kinetic energy, warming Phase changes, results Cold, potential energy, particles Radiation, temperature profile
Part 1Particles and heatingSections 01–07

01Solid, liquid, gas: the particle picture

← Back to the notes: Solids, liquids and gases

In a solid the particles vibrate about fixed positions. In a liquid they are still close together but slide past each other. In a gas they are far apart and move fast in straight lines between collisions. From solid to gas the forces between the particles get weaker and their potential energy gets bigger.

arrangement—
motion—
forces and energy—

02Temperature is the average kinetic energy: use kelvin

← Back to the notes: Temperature

The two scales have the same step size, so a change of 1 °C is a change of 1 K. But only the kelvin scale starts at absolute zero, where the particles have their lowest possible kinetic energy. The average kinetic energy is proportional to T in kelvin: at 20 °C = 293 K it is 1.5 × 1.38 × 10⁻²³ × 293 = 6.07 × 10⁻²¹ J.

T in kelvin—
average kinetic energy—
Ēk compared with 20 °C—

03Internal energy: kinetic energy plus potential energy

← Back to the notes: Internal energy

0.10 kg of ice starts at −20 °C. The first 4200 J raise its temperature: the random kinetic energy of the molecules grows. Then the temperature stops at 0 °C while the next 33 400 J break bonds and melt the ice: the potential energy grows. Supply 20 kJ in total: the ice is at 0 °C and 47 g has melted. The internal energy is the sum of both bars.

state—
temperature—
energy going into—

04Specific heat capacity: same energy, different rises

← Back to the notes: Specific heat capacity

Each block has a mass of 1.0 kg and gets the same energy. ΔT = Q/(mc), so a small c gives a big temperature rise. 10 kJ warms water by only 2.38 K but copper by 26.0 K: water needs 10.9 times as much energy per kelvin.

ΔT of water—
ΔT of copper—
c(water) ÷ c(copper)—

05Mixing: the energy lost by the hot = the energy gained by the cold

← Back to the notes: Method of mixtures

Hot water is poured into 0.30 kg of water at 20 °C. Nothing escapes, so the two curves meet at one final temperature and the two bars are equal. 0.20 kg at 80 °C gives (0.20 × 80 + 0.30 × 20)/0.50 = 44 °C: the hot water loses 30 240 J and the cold water gains 30 240 J.

final temperature—
lost by the hot—
gained by the cold—

06Reading a heating curve, one part at a time

← Back to the notes: Phase changes and heating curves

This is the 150 g sample of worked example 5. Move the time and watch which equation fits each part. Every part uses Q = Pt. The first sloping part gives the power: 0.15 × 2000 × 40/100 = 120 W. On the flat part the sample melts at 55 °C: 120 × 400 = 48 000 J melts 0.15 kg, so L = 3.2 × 10⁵ J kg⁻¹.

part of the curve—
equation—
energy supplied so far—

07From ice to warm water: one step for each change

← Back to the notes: Specific latent heat

0.20 kg of ice at −10 °C becomes water at 40 °C. Do it in steps and add them: warm the ice (mcΔT, 4200 J), melt it (mL, 66 800 J), warm the water (mcΔT, 33 600 J). Total: 104 600 J. Use the right c for each phase, and never use mcΔT across 0 °C.

melting step—
warming steps—
total—
Part 2Transfer and radiationSections 08–12

08Conduction through a wall

← Back to the notes: Conduction

A wall of area 10 m² has 20 °C inside and 5 °C outside. The temperature falls steadily through the wall. A 20 cm brick wall (k = 0.60 W m⁻¹ K⁻¹) lets through 0.60 × 10 × 15/0.20 = 450 W: about 3.9 × 10⁷ J every day. Make it thicker and the rate falls; change the material and k changes the rate.

rate ΔQ/Δt—
temperature gradient—
energy in one day—

09Convection: the fluid carries the energy

← Back to the notes: Convection

Water at the bottom of the pan warms, expands and becomes less dense, so it rises above the flame. Cooler, denser water sinks at the sides to take its place. The loop is a convection current. Unlike conduction, the fluid itself moves; that is why convection cannot happen in a solid or in a vacuum.

warm water rises—
because of its—
energy carried by—

10Radiated power grows as T⁴

← Back to the notes: Thermal radiation and the Stefan–Boltzmann law

A black sphere of radius 0.10 m (A = 4πr² = 0.126 m²) at 600 K radiates 5.67 × 10⁻⁸ × 0.126 × 600⁴ = 923 W. The curve rises steeply: double T and P becomes 2⁴ = 16 times bigger. Always use kelvin: heating from 50 °C to 250 °C is 323 K → 523 K, a factor of 6.9, not 5⁴.

power P—
P ÷ P at 300 K—
P ÷ P at T/2—

11Wien’s law: the colour tells the temperature

← Back to the notes: Wien’s displacement law

The curve shows how much radiation a black body gives out at each wavelength. As T rises the peak moves to shorter wavelengths: λmaxT = 2.9 × 10⁻³ m K. The Sun (5800 K) peaks at 500 nm, in the visible. A 3000 K star peaks in the infrared and looks red; a 10 000 K star peaks in the ultraviolet and looks blue-white.

peak wavelength—
peak in the—
colour—

12Luminosity and apparent brightness

← Back to the notes: Stars: luminosity and apparent brightness

A star sends its power L out equally in all directions. At a distance d it is spread over a sphere of area 4πd², so b = L/(4πd²). The Sun gives 3.9 × 10²⁶ W; at 1 AU (1.5 × 10¹¹ m) that is 1.38 × 10³ W m⁻². Double the distance: the same rays cover 4 times the area, so b falls to a quarter.

apparent brightness b—
b ÷ b of the Sun at 1 AU—
area 4πd²—

13Practice and self-check

← Back to the notes: Practise next
Answers checked by calculation. Try each one before you open it.
1. Change 37 °C to kelvin and find the average kinetic energy of the molecules.
T = 37 + 273 = 310 K; Ēk = 1.5 × 1.38 × 10⁻²³ × 310 = 6.4 × 10⁻²¹ J.
2. Find the energy needed to heat 2.0 kg of water from 15 °C to 85 °C, and the time a 2.0 kW kettle takes (no losses).
Q = 2.0 × 4200 × 70 = 5.88 × 10⁵ J; t = Q/P = 5.88 × 10⁵/2000 = 294 s.
3. A 0.30 kg aluminium block (c = 900 J kg⁻¹ K⁻¹) at 150 °C is put into 0.50 kg of water at 20 °C. Find the final temperature.
0.30 × 900 × (150 − θ) = 0.50 × 4200 × (θ − 20) → θ = 82 500/2370 = 35 °C.
4. Find the energy needed to change 0.50 kg of ice at −15 °C into water at 30 °C.
0.50 × 2100 × 15 + 0.50 × 334 000 + 0.50 × 4200 × 30 = 15 750 + 167 000 + 63 000 = 2.5 × 10⁵ J.
5. 25 g of ice at 0 °C is dropped into 150 g of water at 30 °C. Find the final temperature.
Check: the water can give 0.150 × 4200 × 30 = 18 900 J; melting needs 0.025 × 334 000 = 8350 J, so all the ice melts. 18 900 − 8350 = 10 550 J warms 0.175 kg from 0 °C: θ = 10 550/(0.175 × 4200) = 14 °C.
6. Explain, in terms of molecules, why the temperature of boiling water stays at 100 °C although it is still heated.
The energy supplied increases the potential energy of the molecules (they are separated completely). Their average kinetic energy does not change, so the temperature stays constant.
7. A window has area 2.0 m² and glass 5.0 mm thick (k = 0.80 W m⁻¹ K⁻¹). The two faces differ by 8.0 K. Find the rate of energy flow.
ΔQ/Δt = 0.80 × 2.0 × 8.0/0.0050 = 2.6 × 10³ W.
8. A star’s spectrum peaks at 480 nm. Find its surface temperature.
T = 2.9 × 10⁻³/(480 × 10⁻⁹) = 6.0 × 10³ K.
9. A star has twice the Sun’s radius (R☉ = 6.96 × 10⁸ m) and a surface temperature of 5000 K. Find its luminosity.
L = σ4πR²T⁴ = 5.67 × 10⁻⁸ × 4π × (1.39 × 10⁹)² × 5000⁴ = 8.6 × 10²⁶ W.
10. Two stars have the same luminosity. Star A is 3 times as far away as star B. Find bA/bB.
b ∝ 1/d²: bA/bB = 1/3² = 0.11.

Before the exam, can you…

  • describe the particles in a solid, a liquid and a gas (arrangement, motion, forces)?
  • change °C to K, and say why Ēk = (3/2)kBT uses kelvin?
  • define internal energy, naming the molecules (atoms, particles) and both kinds of energy?
  • use Q = mcΔT and Q = Pt, and solve a mixing problem with lost = gained?
  • read every part of a heating curve and explain the flat parts with molecules?
  • define specific latent heat and use Q = mL, doing one step for each change?
  • check first whether all the ice melts?
  • use ΔQ/Δt = kAΔT/Δx with Δx in metres, and explain conduction and convection?
  • use L = σAT⁴ (T in kelvin, A = 4πR²) and λmaxT = 2.9 × 10⁻³ m K?
  • use b = L/(4πd²) to find a luminosity or a distance?