understand that the scale of thermodynamic temperature does not depend on the property of any particular substance

Temperature Scales – Cambridge A‑Level Physics 9702

Learning objectives

• Define thermal equilibrium and state the direction of spontaneous heat flow.
• Recall the relationship between kelvin (K) and degree Celsius (°C) and use it to convert temperatures.
• Explain why the kelvin (thermodynamic) scale is independent of any particular substance.
• Carry out simple temperature‑conversion calculations (AO2).

1. Thermal equilibrium

Thermal equilibrium is the state in which two or more bodies in contact exchange no net heat; their temperatures are equal. Heat always flows spontaneously from the hotter body to the colder body until thermal equilibrium is reached.

2. Thermodynamic temperature

• Definition: Thermodynamic temperature T is a fundamental property that characterises the thermal state of a system. It is defined by the second law of thermodynamics and does not refer to any specific material.
• In the SI system the unit of thermodynamic temperature is the kelvin (K).

3. The Kelvin scale

• Zero point – absolute zero: 0 K is the lowest possible temperature, the point at which a perfect crystal would have minimum internal energy and all molecular motion would cease.
• Unit size: Since the 2019 re‑definition of the SI, the kelvin is defined by fixing the numerical value of the Boltzmann constant:
  

  kB = 1.380 649 × 10⁻²³ J K⁻¹.
  

  This definition ties the size of the kelvin to a universal constant, not to any substance.

• Independence from any substance: The scale is based on universal physical laws (the third law of thermodynamics) and a fixed constant, so it applies equally to gases, liquids, solids, plasmas, or exotic states of matter. Historical reference points such as the triple point of water (273.16 K) are used only for convenience, not to define the scale.

4. Converting between kelvin and degree Celsius

Worked example 1 – Converting from kelvin

Problem: Convert 298 K to degrees Celsius and degrees Fahrenheit.

1. K → °C using the boxed formula:
   

   θ = 298 – 273.15 = 24.85 °C

2. °C → °F (optional third‑party scale):
   

   °F = (9/5) θ + 32 = (9/5) × 24.85 + 32 ≈ 76.7 °F

Result: 298 K = 24.85 °C ≈ 77 °F (rounded to the nearest whole degree for typical exam use).

Worked example 2 – Converting from Celsius

Problem: A laboratory sample is at 25 °C. Find its temperature in kelvin and in degrees Fahrenheit.

1. °C → K:
   

   T = 25 + 273.15 = 298.15 K

2. °C → °F**:
   

   °F = (9/5) × 25 + 32 = 45 + 32 = 77 °F

Result: 25 °C = 298.15 K = 77 °F.

5. Optional: Other common temperature scales (extra material)

6. Summary

1. The kelvin is the SI base unit for temperature, defined by fixing the Boltzmann constant.
2. Absolute zero (0 K) is the universal lower limit of temperature.
3. Conversion between kelvin and degree Celsius is linear: T(K) = θ(°C) + 273.15.
4. Thermal equilibrium is reached when no net heat flows between bodies; heat always moves from hot to cold.
5. The Kelvin scale’s definition relies on universal constants, not on any particular material, making it suitable for all physical systems.
6. Practice converting both ways (K ↔ °C) and, where required, to/from Fahrenheit.