convert temperatures between kelvin and degrees Celsius and recall that T / K = θ / °C + 273.

Cambridge International AS & A Level Physics (9702) – Temperature Scales

Topic‑Map – Syllabus Coverage Overview

Learning Objectives

• Define thermal equilibrium and state the Zeroth Law of Thermodynamics.
• Convert temperatures accurately between kelvin (K) and degrees Celsius (°C).
• Recall and apply the exact relationship T (K) = θ (°C) + 273.15 (or equivalently θ (°C) = T (K) – 273.15).
• Explain the physical meaning of absolute zero and why the Kelvin scale has no negative values.
• Use specific heat capacity and latent heat formulas and recognise where temperature conversion is required.
• Propagate measurement uncertainties through temperature conversions.
• Understand why the Kelvin scale must be used in the ideal‑gas law.

14.1 Thermal Equilibrium & Temperature

• Thermal equilibrium: Two or more bodies are in thermal equilibrium when, after being in contact, no net heat flows between them.
• Zeroth Law of Thermodynamics: If body A is in thermal equilibrium with body B, and body B is in thermal equilibrium with body C, then A is in thermal equilibrium with C. → This law justifies the use of a single scalar quantity – temperature – to describe the thermal state of a body.
• Temperature: A measure of the average kinetic energy of the particles in a system. It is an intensive property, independent of the amount of material.

14.2 Temperature Scales

Why Two Scales?

• Celsius (°C): Defined by the freezing point (0 °C) and boiling point (100 °C) of water at 1 atm. The size of one degree Celsius is the same as one kelvin.
• Kelvin (K): An absolute thermodynamic scale. 0 K (absolute zero) is the temperature at which the translational kinetic energy of particles would be zero. Because it starts at a physical limit, the Kelvin scale has no negative values.

Derivation of the Linear Relationship

The size of a degree is identical in both scales, so the relationship between a temperature θ in °C and an absolute temperature T in K is purely an offset:

Δ is the temperature difference between the two zero‑points. By definition, 0 °C corresponds to 273.15 K, therefore Δ = 273.15 K. Hence:

Conversion Formulas

Worked Examples

1. Convert 0 °C to kelvin.
   T = 0 + 273.15 = 273.15 K
2. Convert 500 K to degrees Celsius.
   θ = 500 – 273.15 = 226.85 °C
3. Temperature of an oven set at 200 °C in kelvin.
   T = 200 + 273.15 = 473.15 K
4. Uncertainty propagation. A thermometer reads 25.0 °C ± 0.5 °C.
   Because the offset 273.15 K is exact (no uncertainty), the absolute uncertainty is unchanged: T = (25.0 ± 0.5) + 273.15 = 298.15 ± 0.5 K

Physical Meaning of 0 K (Absolute Zero)

• At 0 K the translational kinetic energy of particles would be zero (quantum zero‑point energy remains).
• No temperature on the Kelvin scale can be negative; a result T < 0 K indicates a calculation error.
• Absolute zero provides the reference point for the ideal‑gas law PV = nRT and for calculations of internal energy.

Common Mistakes & How to Avoid Them

• Using 273 instead of 273.15: The syllabus expects the exact value 273.15 K for any numerical answer.
• Confusing symbols: T always denotes kelvin, θ (or t) denotes degrees Celsius.
• Negative Kelvin values: Any result T < 0 K signals a mistake; the Kelvin scale has no negatives.
• Ignoring uncertainties: When converting a measured temperature, copy the absolute uncertainty unchanged because the offset is exact.

14.3 Specific Heat Capacity & Latent Heat

• Specific heat capacity (c): Amount of heat required to raise the temperature of 1 kg of a substance by 1 K (or 1 °C). Q = m c Δθ where Δθ may be expressed in °C or K (they are numerically identical).
• Latent heat (L): Heat required for a phase change at constant temperature. Q = m L
• Because Δθ and ΔT have the same magnitude, the same temperature conversion is used when the heat‑capacity formula involves kelvin.

Example – Heating Water

Calculate the energy needed to heat 0.500 kg of water from 20 °C to 80 °C. Specific heat capacity of water: c = 4180 J kg⁻¹ K⁻¹.

Δθ = 80 °C – 20 °C = 60 °C   (ΔT = 60 K)
Q = m c Δθ = 0.500 kg × 4180 J kg⁻¹ K⁻¹ × 60 K = 1.25 × 10⁵ J

15 Ideal Gases – Connection to Temperature Scales

Outline for Remaining A‑Level Topics (16‑25)

Practice Questions

1. Convert -40 °C to kelvin.
2. A gas is heated from 300 K to 450 K. What is the temperature change in degrees Celsius?
3. The temperature of a star is 5800 K. Express this temperature in degrees Celsius.
4. Explain why the statement “0 °C = 0 K” is false, using the conversion formula.
5. A temperature reading is 22.3 °C ± 0.2 °C. Give the equivalent kelvin value with its uncertainty.
6. Calculate the heat required to melt 150 g of ice at -10 °C. (c_ice = 2100 J kg⁻¹ K⁻¹, L_fusion = 3.34 × 10⁵ J kg⁻¹.)

Quick Reference – Other Common Temperature Scales (Optional)

Summary

The Kelvin and Celsius scales differ only by a constant offset of 273.15 K. The exact conversion is:

Key points to remember:

• Use 273.15 K for all examined calculations; 273 K is only a rough shortcut.
• 0 K is absolute zero – a physical limit; the Kelvin scale cannot be negative.
• When converting measured temperatures, propagate the same absolute uncertainty because the offset is exact.
• Keep symbols straight: T = kelvin, θ = degrees Celsius.
• Specific heat capacity and latent heat calculations require the same temperature conversion.
• The ideal‑gas law explicitly requires kelvin; using Celsius will give incorrect results.