All modules

Subject module

Heat, Thermodynamics & Waves

Thermal expansion and calorimetry, conduction and radiation, the laws of thermodynamics and its processes, heat engines and refrigerators, kinetic theory of gases, simple harmonic motion, and sound waves including standing waves, beats and the Doppler effect.

0/8
1Thermal expansion and calorimetry7 min read+15 XP

Solids and liquids expand on heating in proportion to the temperature rise; in a mixture, heat lost by hot bodies equals heat gained by cold ones, including latent heat for any change of state.

How it's asked: Relation α : β : γ = 1 : 2 : 3, thermal stress in a clamped rod, temperature scale conversion, anomalous expansion of water, and mixing ice with water to find the final temperature.

Why this is true

Heating increases the amplitude of atomic vibrations, and because the interatomic potential is asymmetric the average spacing grows, so length increases by ΔL = αLΔT. Area and volume involve two and three lengths, which is why β ≈ 2α and γ ≈ 3α for an isotropic solid. If the rod is clamped so it cannot expand, the walls must compress it by exactly the strain αΔT it would have had, producing a thermal stress YαΔT. During melting or boiling, the heat supplied breaks intermolecular bonds instead of raising the kinetic energy, so the temperature stays constant and Q = mL. Calorimetry is just energy conservation in an insulated system: always check first whether the available heat is enough to complete a change of state.

Key points

Key formulae

  • Linear expansion

    ΔL = αLΔT

  • Expansion coefficients

    β = 2α, γ = 3α

  • Thermal stress

    σ = YαΔT

    rod clamped at both ends

  • Sensible heat

    Q = mcΔT

  • Latent heat

    Q = mL

  • Temperature scales

    C/100 = (F − 32)/180 = (K − 273.15)/100

Fixed points of temperature scales

ScaleIce pointSteam pointDivisions
Celsius0 °C100 °C100
Fahrenheit32 °F212 °F180
Kelvin273.15 K373.15 K100

Common traps

  • Do not assume all the ice melts — compare heat available with heat needed to melt it first.
  • A temperature difference of 1 °C equals 1 K, but 1.8 °F — convert differences and readings differently.

Worked example

20 g of ice at 0 °C is dropped into 100 g of water at 40 °C in an insulated vessel of negligible heat capacity. Take c(water) = 1 cal g⁻¹ °C⁻¹ and L(fusion) = 80 cal g⁻¹. Find the final temperature.

High-yield

α : β : γ = 1 : 2 : 3; thermal stress YαΔT; water densest at 4 °C; temperature is constant during a change of state.

Quick check

Q1.The temperature at which the Celsius and Fahrenheit readings are equal is:

Q2.A metal rod is fixed between two rigid walls and heated by ΔT. The thermal stress developed depends on:

Other modules