Convert between Celsius and kelvin temperatures
Use the selected 273 offset in both directions and write the correct units.
Exam scope and assessment notes
Core and Extended · Physics 0972, 2026–2028, section 2.1.3. Explain pressure changes for a fixed gas mass at constant volume or constant temperature, and convert between kelvin and degrees Celsius using the selected +273 approximation. Numerical pV calculations and graphs belong to the labelled Supplement lessons. Written practice is self-checked. Human subject-teacher review, supervised practical experience and cumulative assessment remain pending. Cambridge states that IGCSE 0972 has the same subject content as IGCSE 0625, with 9–1 grading. This StudyForA pathway maps the 2026–2028 syllabus only. This 9–1 graded qualification is available only in a limited number of administrative zones. Confirm that your centre can enter you for this code and exam series with Cambridge International. Cambridge has published a separate syllabus for 2029. StudyForA has not reconciled that version yet; this page supports exams in 2026, 2027 and 2028 only.
Before you begin
- Identify absolute zero as the lowest-temperature reference.
- Add and subtract signed numbers.
Work on paper. Try each question before revealing help, and explain why your method works.
Quick readiness check
What is −10 + 273?
What you will learn
- Recall T in K = θ in °C + 273 for the selected syllabus.
- Convert positive and negative Celsius readings and kelvin readings consistently.
Add the offset for a Celsius reading
The selected syllabus uses T (K) = θ (°C) + 273. For 20°C, T = 293 K; for −10°C, T = 263 K. Kelvin temperatures use the symbol K without a degree sign. Negative Celsius temperatures can still correspond to positive kelvin temperatures, because the two scales have different zero points.
The school-level approximation places −273°C at 0 K and 0°C at 273 K. The exact thermodynamic offset is 273.15, but use the selected 273 when answering these syllabus-model questions. Do not alternate offsets halfway through a calculation or silently replace a supplied approximation with another precision.
Subtract the offset to return to Celsius
Rearrange the equation as θ (°C) = T (K) − 273. A reading of 310 K becomes 37°C, and 250 K becomes −23°C. Adding 273 to a kelvin reading would move the conversion in the wrong direction. Keep the input scale explicit before deciding which operation to use.
A proposed equilibrium temperature below 0 K is outside this absolute-temperature model. This does not make every negative Celsius reading impossible: −23°C corresponds to positive 250 K. The distinction is between the temperature's physical lower bound and the arbitrary zero of the Celsius scale.
Pause and explain
Convert 250 K to degrees Celsius using the selected approximation.
Worked example
Convert −10°C to kelvin and 310 K to degrees Celsius using the selected approximation.
Show the worked solution
- For Celsius to kelvin, add 273: −10 + 273 = 263.
- For kelvin to Celsius, subtract 273: 310 − 273 = 37.
- Use K without a degree sign and °C for the Celsius result.
Answer 263 K; 37°C.
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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.
Convert −10°C to kelvin and 310 K to degrees Celsius.
Give me a hint
Identify the input scale before adding or subtracting 273.
Compare my reasoning
- Celsius input needs +273: −10 + 273 = 263 K.
- Kelvin input needs −273: 310 − 273 = 37°C.
- Write each result with its appropriate scale symbol.
263 K; 37°C.
Look for these in your work
- Choose the direction correctly.
- Preserve the negative input sign.
Convert 25°C, −40°C and 373 K using the selected approximation.
Give me a hint
Use the same offset for all three conversions.
Compare my reasoning
- 25 + 273 = 298 K.
- −40 + 273 = 233 K.
- 373 − 273 = 100°C.
298 K; 233 K; 100°C.
Look for these in your work
- Convert both scales.
- Use K without a degree sign.
A conversion gives −280°C as −7 K. Check the arithmetic and whether the proposed equilibrium temperature fits the absolute-temperature model.
Give me a hint
Correct arithmetic can still reveal a physically inadmissible input.
Compare my reasoning
- Using the selected formula, −280 + 273 = −7.
- That kelvin result is below the model's 0 K lower bound.
- The proposed equilibrium temperature is inadmissible; a negative Celsius reading alone would not be sufficient to reject a different temperature.
The arithmetic gives −7 K, revealing an inadmissible equilibrium temperature in this model.
Look for these in your work
- Separate arithmetic from admissibility.
- Do not reject all negative Celsius readings.
Common mistakes
- Add for °C to K; subtract for K to °C.
- Kelvin uses K, not °K.
- Negative °C does not mean negative K.
What can you explain now?
What is 0°C in the selected kelvin conversion?
Compare with the explanation
273 K.
Add the selected offset: 0 + 273 = 273.
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Make it stick
Convert readings both ways, state the selected offset and distinguish negative Celsius from the absolute lower bound.
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