A2 · Practice questions · Temperature scales

One scale that needs no substance.

Six original Cambridge-style questions on the thermodynamic scale, absolute zero, converting between kelvin and degrees Celsius, and why temperature differences are the same on both.

Original questions All questions on this page are original work, written in the Cambridge AS & A Level style. They are not from past papers. They test the same concepts and skills the syllabus rewards.

Keep these straight

Only the zero differs between kelvin and Celsius.

Conversion: T / K = θ / °C + 273.15; a kelvin has the same size as a degree Celsius, so a temperature difference is the same number on both scales.
Absolute zero: 0 K = −273.15 °C, the temperature at which particles have their minimum (not zero) energy; it cannot be reached and kelvin temperatures are never negative.
Why thermodynamic: the scale does not depend on the property of any particular substance, so it avoids the disagreement between thermometers that are calibrated at the same fixed points.

Calculation

[2 marks]

Convert the normal human body temperature of 37.0 °C to kelvin, and convert the boiling point of liquid nitrogen, 77 K, to degrees Celsius.

T = 37.0 + 273.15 = 310 K (310.2 K). ✓
θ = 77 − 273.15 = −196 °C. ✓

Analysis

[3 marks]

A mercury-in-glass thermometer and a platinum resistance thermometer are both calibrated by marking 0 °C at the ice point and 100 °C at the steam point and dividing the interval into 100 equal parts. In the same bath of warm water their readings differ by 0.2 °C. Explain how both can be correctly calibrated and yet disagree, and explain how the thermodynamic scale avoids this problem.

Each thermometer assumes its own property (the expansion of mercury, the resistance of platinum) changes linearly with temperature between the fixed points. ✓
The two properties do not vary with temperature in exactly the same way, so although both agree at the fixed points they give slightly different values in between; neither is more correct than the other on its own terms. ✓
The thermodynamic scale is defined independently of the properties of any particular substance, so it provides one standard against which any thermometer can be checked. ✓

Calculation

[3 marks]

A sample of oil is warmed from 18.0 °C to a thermodynamic temperature of 318 K. Calculate the rise in temperature of the oil in kelvin and in degrees Celsius, and explain why the two values are the same.

Final temperature in Celsius: 318 − 273.15 = 44.9 °C, so Δθ = 44.9 − 18.0 = 26.9 °C (26.85). ✓
In kelvin: 18.0 °C = 291.2 K, so ΔT = 318 − 291.2 = 26.9 K. ✓
The Celsius scale is the kelvin scale shifted by 273.15 with the same size of unit, so the shift cancels in any difference and a change of 1 K equals a change of 1 °C. ✓

Analysis

[2 marks]

State what is meant by absolute zero, and explain why no temperature on the thermodynamic scale can be negative.

Absolute zero, 0 K (−273.15 °C), is the temperature at which the particles of a substance have their minimum possible energy; it cannot be reached. ✓
The thermodynamic scale takes its zero at this minimum-energy state, so there is no lower temperature to represent and every real temperature is above zero kelvin. ✓

Calculation

[3 marks]

A constant-volume gas thermometer contains a fixed mass of gas whose pressure is 1.00 × 10⁵ Pa at the ice point, 0 °C. For an ideal gas at constant volume the pressure is proportional to the thermodynamic temperature. Calculate the Celsius temperature of a liquid in which the pressure becomes 1.37 × 10⁵ Pa, and state the pressure the gas would exert at absolute zero.

p / T is constant, so T = 273.15 × (1.37 × 10⁵ / 1.00 × 10⁵) = 374 K. ✓
θ = 374 − 273.15 = 101 °C. ✓
At absolute zero p = 0, since p ∝ T on the thermodynamic scale and the molecules would have their minimum energy. ✓

Calculation

[4 marks]

(a) The surface of the Sun is at about 5800 K. Express this in degrees Celsius. (b) Liquid nitrogen boils at −196 °C. Express this in kelvin. (c) A cup of coffee cools by 25 °C. State the fall in its temperature in kelvin. (d) Explain why the answer to (c) is found without adding or subtracting 273.

(a) θ = 5800 − 273.15 = 5.53 × 10³ °C (5527 °C). ✓
(b) T = −196 + 273.15 = 77 K. ✓
(c) A fall of 25 K. ✓
(d) The 273.15 is only a shift of the zero; it appears in both the starting and finishing temperatures and cancels when they are subtracted, and one kelvin is the same size as one degree Celsius. ✓

Mark this once you have attempted all six and checked your working. It records a Practiced badge on the topic and adds a one-time bonus. Revealing the solutions alone does not count.