19  Thermodynamics

19.1 Syllabus inquiry question

  • How does energy transfer relate to temperature and phase change?
Feynman Insight

From The Feynman Lectures on Physics, Vol I, Chapter 39:

Heat is not a substance; it is energy in transit. Temperature measures the average kinetic energy of particles, not how much heat an object contains.

19.2 Learning Objectives

  • Distinguish heat from temperature.
  • Apply \(Q = mc\Delta T\) and \(Q = mL\).
  • Describe conduction, convection, and radiation.
  • Calculate thermal conduction power.

19.3 Content

19.3.1 Heat and Temperature

Temperature is a measure of the average kinetic energy of particles in a substance.

Heat is energy transferred between objects due to a temperature difference.

Key Distinction
  • Temperature measures “how hot” (°C or K)
  • Heat measures “energy transferred” (J)

An object can have high temperature but low heat capacity (e.g., a spark) or low temperature but transfer a lot of heat (e.g., melting ice).

19.3.2 Interactive: Energy Bar Chart - Heating

Visualizing thermal energy changes:

19.3.3 Specific Heat Capacity

When heat is added to a substance (without changing phase), temperature rises:

\[Q = mc\Delta T\]

where: - \(Q\) = heat energy (J) - \(m\) = mass (kg) - \(c\) = specific heat capacity (J/kg·K) - \(\Delta T\) = temperature change (K or °C)

Specific heat capacity (\(c\)) is the energy needed to raise 1 kg by 1°C.

19.3.4 Common Specific Heat Capacities

Material \(c\) (J/kg·K) Notes
Water 4186 Very high—good coolant
Aluminium 900 Moderate
Copper 390 Low—heats quickly
Iron 450 Moderate
Ice 2090 About half of water

19.3.5 Latent Heat

During a phase change, temperature remains constant while heat is absorbed or released:

\[Q = mL\]

where: - \(L_f\) = latent heat of fusion (solid ↔︎ liquid) - \(L_v\) = latent heat of vaporization (liquid ↔︎ gas)

19.3.6 Interactive: Phase Change Energy

Energy changes during melting:

Key observation: Temperature stays at 0°C throughout melting—all energy goes into breaking molecular bonds.

19.3.7 Latent Heat Values for Water

Phase Change \(L\) (J/kg) Process
Fusion \(3.34 \times 10^5\) Melting/Freezing
Vaporization \(2.26 \times 10^6\) Boiling/Condensing

19.3.8 Heat Transfer Modes

Conduction: Transfer through direct molecular contact

  • Requires matter (works in solids, liquids, gases)
  • Faster in metals (free electrons)

Convection: Transfer by fluid motion

  • Requires a fluid (liquid or gas)
  • Driven by density differences

Radiation: Transfer by electromagnetic waves

  • No medium required
  • All objects emit thermal radiation

19.3.9 Thermal Conduction

For a slab of material:

\[P = \frac{kA\Delta T}{L}\]

where: - \(P\) = power (rate of heat transfer) (W) - \(k\) = thermal conductivity (W/m·K) - \(A\) = cross-sectional area (m²) - \(\Delta T\) = temperature difference (K) - \(L\) = thickness (m)

19.3.10 Thermal Conductivity Values

Material \(k\) (W/m·K) Classification
Copper 400 Excellent conductor
Aluminium 240 Good conductor
Glass 0.8 Poor conductor
Wood 0.15 Insulator
Air 0.025 Excellent insulator

19.4 Worked Examples

19.4.1 Example 1: Temperature change

Heat 0.50 kg of water from 20°C to 80°C. Use \(c = 4200\) J/kg·K.

Solution:

  1. Use \(Q = mc\Delta T\)

  2. \(\Delta T = 80 - 20 = 60\) °C (or 60 K)

  3. \(Q = 0.50 \times 4200 \times 60 = 126,000\) J

  4. Energy required is 126 kJ

19.4.2 Example 2: Phase change

Melt 0.20 kg of ice at 0°C. Use \(L_f = 3.34 \times 10^5\) J/kg.

Solution:

  1. Use \(Q = mL\)

  2. \(Q = 0.20 \times 3.34 \times 10^5 = 6.68 \times 10^4\) J

  3. Energy required is 66.8 kJ (or 67 kJ)

Note: The water stays at 0°C until all ice has melted.

19.4.3 Example 3: Conduction power

A wall has area 12 m², thickness 0.25 m, thermal conductivity \(k = 0.80\) W/m·K. Temperature difference is 15 K.

Solution:

  1. Use \(P = \frac{kA\Delta T}{L}\)

  2. \(P = \frac{0.80 \times 12 \times 15}{0.25}\)

  3. \(P = \frac{144}{0.25} = 576\) W

  4. Heat flows through the wall at 576 W

19.4.4 Example 4: Combined heating and phase change

Calculate the energy needed to heat 0.30 kg of ice from -10°C to water at 20°C.

Solution:

Step 1: Heat ice from -10°C to 0°C - \(Q_1 = mc_{ice}\Delta T = 0.30 \times 2090 \times 10 = 6270\) J

Step 2: Melt ice at 0°C - \(Q_2 = mL_f = 0.30 \times 3.34 \times 10^5 = 100,200\) J

Step 3: Heat water from 0°C to 20°C - \(Q_3 = mc_{water}\Delta T = 0.30 \times 4186 \times 20 = 25,116\) J

Total: \(Q = Q_1 + Q_2 + Q_3 = 6270 + 100,200 + 25,116 = 131,586\) J ≈ 132 kJ

19.5 Common Misconceptions

Common Misconceptions
  • Misconception: Temperature is the same as heat. Correction: Temperature measures average kinetic energy; heat is energy transfer due to temperature difference.

  • Misconception: Phase change changes temperature. Correction: During phase change, temperature stays constant—all energy goes to breaking/forming bonds.

  • Misconception: Metal objects are colder than wooden objects at room temperature. Correction: Both are at room temperature. Metal feels colder because it conducts heat away from your hand faster.

  • Misconception: Heat rises. Correction: Hot air rises (convection). Heat itself transfers in all directions by conduction and radiation.

19.6 Practice Questions

19.6.1 Easy (2 marks)

Define specific heat capacity and state its SI unit.

  • Energy required to raise 1 kg of substance by 1°C (or 1 K) (1)
  • Unit: J/kg·K (or J/kg·°C) (1)

Answer: Specific heat capacity is the amount of energy required to raise the temperature of 1 kg of a substance by 1 K. Unit: J/kg·K

19.6.2 Medium (4 marks)

How much heat is needed to raise 0.80 kg of copper from 20°C to 80°C? Use \(c = 390\) J/kg·K.

  • Use \(Q = mc\Delta T\) (1)
  • Correct temperature change: \(\Delta T = 60\) °C (1)
  • Calculation: \(Q = 0.80 \times 390 \times 60\) (1)
  • Correct answer: \(Q = 18,720\) J = 18.7 kJ (1)

Answer: 18.7 kJ (or 19 kJ)

19.6.3 Hard (5 marks)

A 1.5 kW heater warms 2.0 kg of water from 18°C to 60°C. Estimate the heating time, ignoring losses. Use \(c = 4200\) J/kg·K.

  • Temperature change: \(\Delta T = 42\) °C (1)
  • Energy calculation: \(Q = mc\Delta T = 2.0 \times 4200 \times 42 = 352,800\) J (2)
  • Time from \(t = Q/P = 352,800 / 1500 = 235\) s (1)
  • Convert to minutes: 3.9 min ≈ 4 min (1)

Answer: About 235 seconds (approximately 4 minutes)

19.7 Multiple Choice Questions

Test your understanding with these interactive questions:

19.8 Quick Quiz: Heat Transfer

Test yourself with this timed quiz covering the key concepts:

19.9 Extended Response Practice

Practice writing detailed physics explanations:

19.10 Summary

Key Takeaways
  • Heat is energy transfer; temperature is average kinetic energy
  • Heating without phase change: \(Q = mc\Delta T\)
  • During phase change: \(Q = mL\) (temperature constant)
  • Conduction power: \(P = kA\Delta T / L\)
  • Heat transfers by conduction, convection, and radiation

19.11 Self-Assessment

Check your understanding:

After studying this section, you should be able to:

19.12 Module 3 Complete

Congratulations on completing Module 3: Waves and Thermodynamics!

What you’ve learned
  • Wave properties: amplitude, wavelength, frequency, speed
  • Wave behaviour: reflection, refraction, interference, standing waves
  • Sound as a longitudinal pressure wave
  • Ray optics: reflection, refraction, lenses
  • Thermodynamics: heat, temperature, specific heat, latent heat