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A Level · Cambridge (CIE) · 9702

Use thermodynamic and Celsius temperature scales

Convert between kelvin and degrees Celsius, explain absolute zero and distinguish the thermodynamic scale from sensor properties.

Temperature · Cambridge International AS & A Level Physics 9702 · 14.2

A Level; Cambridge International Physics 9702, 2025–2027 syllabus, Topic 14. These original explanations and calculations are study aids; full timed assessment and human subject-teacher review remain outstanding.

Before you begin

  • Use temperature differences, proportional reasoning, energy, mass and SI units.

Work on paper. Try each question before revealing help, and explain why your method works.

Quick readiness check

Which statement correctly converts Celsius temperature θ to kelvin T?

What you will learn

  • Understand that the thermodynamic scale does not depend on a particular substance
  • Convert between kelvin and degrees Celsius
  • Identify 0 K as absolute zero and the lowest possible temperature

The thermodynamic scale is independent of the thermometer material

A thermometer may use a changing property such as resistance or liquid volume, but the thermodynamic temperature scale does not depend on the property of any one substance. Different sensor materials can be calibrated against the same temperature scale.

Kelvin is the SI unit of thermodynamic temperature. The kelvin scale begins at absolute zero, the lowest possible temperature; 0 K is not the same as 0 °C.

Convert using the offset between the scales

The syllabus relation is T/K = θ/°C + 273.15. For a numerical conversion, T in kelvin equals θ in degrees Celsius plus 273.15; conversely θ in degrees Celsius equals T in kelvin minus 273.15.

A temperature interval has the same numerical size in kelvins and degrees Celsius: a rise of 5 K is a rise of 5 °C. The offset applies to absolute temperatures, not to temperature differences.

T/K = θ/°C + 273.15 · absolute zero = 0 K = −273.15 °C

Pause and explain

What is 0 K called?

Put the idea to work

Worked example

Convert 27.0 °C to kelvin and 300.0 K to degrees Celsius.

Show the worked solution
  1. For Celsius to kelvin, add 273.15.
  2. For kelvin to Celsius, subtract 273.15.
  3. Round both answers to one decimal place.

Answer 27.0 °C = 300.2 K; 300.0 K = 26.9 °C.

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From guided practice to a new situation

Write your answer and reasoning first. Use a hint only if you are stuck. The model solution is for self-checking; your written work is not automatically marked.

1 · Guided

Convert −18.0 °C to kelvin.

Give me a hint

Add 273.15 to the Celsius temperature, keeping the sign of −18.0.

Compare my reasoning
  1. T/K = θ/°C + 273.15.
  2. T = −18.0 + 273.15 K.
  3. Round to one decimal place.

T = 255.2 K.

Look for these in your work

  • Correct offset used
  • Negative Celsius value handled
  • Kelvin unit stated
2 · Independent

Convert 350 K to degrees Celsius.

Give me a hint

Subtract 273.15 from the kelvin value.

Compare my reasoning
  1. θ/°C = T/K − 273.15.
  2. θ = 350 − 273.15 °C.
  3. Round to one decimal place.

θ = 76.9 °C.

Look for these in your work

  • Conversion direction correct
  • Offset subtracted
  • Celsius unit stated
3 · Transfer

A metal-resistance thermometer and a liquid thermometer are both calibrated against the thermodynamic scale. Explain why their different sensing materials do not create different temperature scales.

Give me a hint

Distinguish each thermometer's temperature-dependent property from the scale used to assign temperature.

Compare my reasoning
  1. Resistance and liquid volume are sensor properties.
  2. Each property can be calibrated against thermodynamic temperature.
  3. The thermodynamic scale is independent of the particular substance used in the thermometer.

They use different signals but can be calibrated to the same substance-independent thermodynamic temperature scale.

Look for these in your work

  • Sensor signals distinguished
  • Calibration described
  • Scale independence stated

Common mistakes

  • Using 273.15 as a multiplier
  • Adding 273.15 when converting kelvin to Celsius
  • Applying the offset to a temperature change
Recall without your notes

What can you explain now?

Which statement correctly converts Celsius temperature θ to kelvin T?

Compare with the explanation

T/K = θ/°C + 273.15

The Celsius value is shifted by 273.15 to give absolute temperature.

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Make it stick

How do you convert an absolute temperature between Celsius and kelvin, and what is the lowest possible kelvin value?

If you needed the solution, close it and solve the problem again from a blank page. A correct answer is stronger when you can explain the reason for each step.

Keep learning Physics

Course scope and assessment

One hundred and sixty-two original lessons cover all 12 published subject-content outcomes in Topic 1 (sections 1.1–1.4), all nine outcomes in Topic 2 (section 2.1), all 13 outcomes in Topic 3 (sections 3.1–3.3), all 13 outcomes in Topic 4 (sections 4.1–4.3), all 11 outcomes in Topic 5 (sections 5.1–5.2), all 10 outcomes in Topic 6 (sections 6.1–6.2), all 16 outcomes in Topic 7 (sections 7.1–7.5), all 12 outcomes in Topic 8 (sections 8.1–8.4), all 15 outcomes in Topic 9 (sections 9.1–9.3), all 16 outcomes in Topic 10 (sections 10.1–10.3), all 18 outcomes in Topic 11 (sections 11.1–11.2), all seven outcomes in Topic 12 (sections 12.1–12.2), all 12 outcomes in Topic 13 (sections 13.1–13.4), all eight outcomes in Topic 14 (sections 14.1–14.3), and all nine outcomes in Topic 15 (sections 15.1–15.3), plus all four outcomes in Topic 16 (sections 16.1–16.2), and all ten outcomes in Topic 17 (sections 17.1–17.3), plus all 12 outcomes in Topic 18 (sections 18.1–18.5), all nine outcomes in Topic 19 (sections 19.1–19.3), all 19 outcomes in Topic 20 (sections 20.1–20.5), and all eight outcomes in Topic 21 (sections 21.1–21.2), and all 17 outcomes in Topic 22 (sections 22.1–22.4), and all 13 outcomes in Topic 23 (sections 23.1–23.2), and all 16 outcomes in Topic 24 (sections 24.1–24.3), and all 11 outcomes in Topic 25 (sections 25.1–25.3), of the 2025–2027 syllabus. Each lesson has guided, independent and application practice with explained answers. All 25 syllabus topics now have mapped teaching and practice; the 2028–2030 changes outside Topics 23–25 still need reconciliation. Online study cannot demonstrate supervised Paper 3 laboratory competence. Paper 5 planning and analysis, full timed cumulative assessment and human subject-teacher review remain outstanding.

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