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Thermodynamic Temperature Scale (Kelvin)

Key idea: Learn what makes the Kelvin scale an absolute thermodynamic temperature scale and how to convert between Celsius and kelvin for A Level Physics.

  • GCE A-Level H2 Physics 2027
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Learning objectives

  • Use thermodynamic temperature and convert between Celsius and kelvin.

1. Definitions (Must Know)

A. Thermodynamic temperature scale (Kelvin scale)

  • The thermodynamic temperature scale (the Kelvin scale) is an absolute temperature scale that is independent of the behaviour of any particular thermometric substance.
  • The unit is the kelvin (K).
  • Absolute zero: the lowest possible temperature on the thermodynamic scale, defined as 0 K (about -273.15°C).

B. Celsius vs kelvin

  • Celsius temperature is written θ/°C.
  • Thermodynamic temperature is written T/K.

2. Key Ideas (What Earns Marks)

  • A thermodynamic temperature scale is absolute (it has an absolute zero) and is not tied to any single material property.
  • Convert between Celsius and kelvin:
    • T/K = θ/°C + 273.15
  • Temperature intervals are the same size:
    • Δ T/K = Δ θ/°C
  • Use kelvin whenever you substitute into thermodynamics/ideal gas equations (e.g. pV = NkT).

3. Detailed Explanations

A. Why “thermodynamic” (and not just “centigrade”)?

A centigrade thermometer assumes the chosen thermometric property varies linearly with temperature. In practice, different thermometric substances (e.g., mercury vs alcohol) do not expand in exactly the same way, so different thermometers can give slightly different readings for the same physical situation.

The thermodynamic scale aims to be substance-independent, so “T” means the same physical temperature regardless of which thermometer you use.

B. Absolute zero (what it means in exams)

  • Absolute zero is the zero point of the thermodynamic scale: T = 0 K.
  • It corresponds to θ ≈ -273.15°C on the Celsius scale.
  • You cannot have a negative thermodynamic temperature in this syllabus context: T starts from 0 K.

C. Conversion and “same size intervals”

Kelvin and Celsius temperature scalesAligned Kelvin and Celsius scales compare absolute zero, the ice point and the steam point. Equal vertical spacings show that one kelvin and one degree Celsius are equal temperature intervals.Thermodynamic temperature, T / KCelsius temperature, θ / °C373.15 K100.00 °C273.15 K0.00 °C0 K−273.15 °Csteam pointice pointabsolute zero100 K interval= 100 °C intervalT / K = θ / °C + 273.15
Kelvin and Celsius intervals have the same size, but their zeros differ by 273.15. Thermodynamic temperature starts at absolute zero and is written in kelvin, without a degree symbol.

The Kelvin scale is just the Celsius scale shifted by 273.15:

T/K = θ/°C + 273.15

Because it is a pure shift, a temperature difference of 1°C is the same size as a difference of 1 K: Δ T = Δ θ

4. Common Mistakes

  • Using θ in Celsius directly inside gas/thermodynamics equations (convert to kelvin first).
  • Writing the unit as “°K” (incorrect). The unit is K, not degree-K.
  • Mixing up temperature with temperature change:
    • you add 273.15 when converting a temperature,
    • you do not add 273.15 to a temperature change.
  • Treating Celsius as an absolute scale (it is not: 0°C is not “no thermal energy”).

5. Exam Tips

  • Write temperatures in Kelvin when using thermodynamic/ideal gas equations.
  • Use the approximate conversion T ≈ θ + 273 if the question only needs an estimate.
  • If you later use Δ U = Q + W, keep track of signs and label “work done on” vs “work done by” clearly.

6. Worked Examples

Modelled example 1

Convert Celsius to kelvin

Core

Problem

Convert θ = 25°C to kelvin.
Study the worked solution

  1. Use the scale offset

    Method

    Add 273.15 to the Celsius reading.

    Reason

    The kelvin and Celsius scales have equal intervals but different zero points.

    Working

    T/K = θ/°C + 273.15

  2. Calculate

    Method

    The thermodynamic temperature is 298.15 K.

    Reason

    The value is now measured from absolute zero.

    Working

    T = 25 + 273.15 = 298.15 K

  3. Round if appropriate

    Method

    To the nearest kelvin, T ≈ 298 K.

    Reason

    The requested precision can be matched without changing the conversion method.

    Working

    298.15 K ≈ 298 K

Guided practice 2

Convert kelvin to Celsius

About 3 min

Problem

Convert T = 310 K to Celsius.

Try this before viewing the solution

Unit: °C

Hints

Hint 1: reverse the offset
Rearrange T = θ + 273.15 for θ.
View solution step by step

  1. Reverse the conversion

    Method

    Subtract 273.15 from the kelvin reading.

    Reason

    This returns from the absolute scale to the shifted Celsius scale.

    Working

    θ/°C = T/K-273.15

  2. Evaluate

    Method

    The temperature is 36.85°C, or 36.9°C to one decimal place.

    Reason

    The numerical offset changes the reading, while the Celsius unit retains its degree sign.

    Working

    θ = 310-273.15 = 36.85°C ≈ 36.9°C

Common misconception 3

Temperature changes

Find and correct the mistake

Learner claim

A gas warms from 18°C to 31°C. A learner finds the change as 31-18 + 273.15 = 286.15 K. Diagnose the claim and find the correct Δ T.

Try this before viewing the solution

Unit: K

View solution step by step

  1. Separate reading from interval

    Method

    The 273.15 offset converts a temperature reading, not a temperature change.

    Reason

    Kelvin and Celsius intervals have the same size.

    Working

    Δ T/K = Δθ/°C

  2. Subtract the endpoints

    Method

    The gas warms by 13°C.

    Reason

    Temperature change is final reading minus initial reading.

    Working

    Δθ = 31-18 = 13°C

  3. Express the interval in kelvin

    Method

    The same interval is 13 K.

    Reason

    Converting both endpoints would add the same offset to each, which cancels in their difference.

    Working

    Δ T = 13 K

Examiner practice 4

Why kelvin matters in a gas-law calculation

4 marks

Examination question

At constant volume, a gas has pressure 100 kPa at 27°C. Estimate its pressure at 127°C. [4 marks]

Try this before viewing the solution

Unit: kPa

View solution step by step

  1. Convert both readings

    1 mark

    Method

    The temperatures are approximately 300 K and 400 K.

    Reason

    Gas-law proportionality uses thermodynamic temperature.

    Working

    T₁ = 27 + 273 = 300 K, T₂ = 127 + 273 = 400 K

  2. Apply the fixed-volume relation

    1 mark

    Method

    For a fixed amount of gas at constant volume, p/T is constant.

    Reason

    From pV = NkT, both N and V remain fixed.

    Working

    p₂/p₁ = T₂/T₁

  3. Calculate

    1 mark

    Method

    The final pressure is about 133 kPa.

    Reason

    The absolute temperature rises by a factor 400/300.

    Working

    p₂ = 100(400/300) ≈ 133 kPa

  4. Check the physical trend

    1 mark

    Method

    The pressure increases, but it does not become nearly five times larger.

    Reason

    The absolute temperature changes from 300 to 400 K; the ratio 127/27 would misuse Celsius zero.

    Working

    p₂/p₁ ≈ 1.33

Challenge 5

Constant pressure heating (volume change)

Minimal support

Independent transfer

An ideal gas has volume V₁ = 2.0 L at 20°C. It is heated at constant pressure to 80°C. Find the new volume.

Try this before viewing the solution

Unit: L

Hints

Hint 1: change the fixed condition
At constant pressure for a fixed amount of ideal gas, V/T is constant.
View solution step by step

  1. Convert to kelvin

    Method

    The initial and final temperatures are 293.15 K and 353.15 K.

    Reason

    Volume is proportional to absolute temperature, not the Celsius reading.

    Working

    T₁ = 293.15 K, T₂ = 353.15 K

  2. Apply the new constraint

    Method

    Use V₂/V₁ = T₂/T₁.

    Reason

    Pressure and gas amount are constant, so the ideal-gas equation reduces to V/T = constant.

    Working

    V₂/2.0 = 353.15/293.15

  3. Calculate and interpret

    Method

    The new volume is approximately 2.41 L.

    Reason

    Heating at constant pressure requires expansion, so a result greater than 2.0 L is physically consistent.

    Working

    V₂ = 2.0(353.15/293.15) ≈ 2.41 L

7. Mind Stretchers

Mind stretcher 1: Doubling absolute temperatureExtension

An object is at 20°C. Its thermodynamic temperature is doubled. Find the final temperature in °C.

Show Answer

Initial: T₁ = 20 + 273.15 = 293.15 K

Double it: T₂ = 2T₁ = 586.3 K

Convert back: θ₂ = 586.3 - 273.15 = 313.15°C ≈ 313°C

Mind stretcher 2: Why must T be in kelvin for gas laws?Extension

Explain why you must use thermodynamic temperature (kelvin) in equations like pV = NkT.

Show Answer

The constant k is defined so that pV is proportional to absolute temperature. If you use Celsius, the zero point is arbitrary (0°C is not absolute zero), so the proportionality would fail.

Using kelvin makes the relationship physically meaningful: T starts from 0 K at absolute zero.

Mind stretcher 3: Constant-volume gas thermometerExtension

A real route to an “absolute” scale is to use a gas at low pressure (close to ideal). At constant volume, pressure varies approximately linearly with thermodynamic temperature. Extrapolating the pressure–Celsius graph to zero pressure gives an intercept near -273.15°C (absolute zero).

Continue with the next resource in this course.

Course and syllabus information
Course
GCE A-Level H2 Physics
Edition
GCE A-Level H2 Physics 2027