AeroLab Reading room

Physics / A2 level

Temperature

Connect thermometry, thermal equilibrium and absolute temperature.

Learning objectives

The big idea

Bodies in thermal equilibrium have the same temperature and no net energy transfer by heating between them. A thermometer uses a physical property that changes reproducibly with temperature, such as electrical resistance.

The kelvin scale is an absolute scale. A temperature interval of one kelvin equals an interval of one degree Celsius, but the zero points differ. Gas-law calculations require absolute temperature. Heating can raise temperature or change state; specific heat capacity describes the former, while specific latent heat describes energy per mass for the latter.

1. A thermometer must interact with its surroundings

Temperature identifies thermal state and determines the direction of net energy transfer by heating. A thermometer changes until it approaches thermal equilibrium with the measured body. Its reading relies on a reproducible thermometric property, such as resistance or gas pressure. Response time and contact can matter: reading immediately or touching a sensitive part may give a value that does not represent the intended object.

2. Absolute values and intervals differ

Kelvin temperature equals Celsius temperature plus 273.15. Thus 20 °C is about 293 K, but a rise of 20 °C is a rise of 20 K. Gas laws require absolute temperature, whereas Q = mcΔT uses an interval. Doubling a Celsius number does not double absolute temperature; that mistake produces incorrect kinetic-energy and gas-pressure predictions.

3. Specific heat capacity explains different warming rates

For a fixed mass without a state change, energy supplied approximately equals mcΔT when losses and the container’s energy change can be neglected. A large specific heat capacity means more energy is needed for the same temperature rise. For 0.40 kg with c = 500 J/(kg K), a 6 K rise requires 1200 J. In an experiment, energy delivered by a heater may also warm the container and surroundings.

4. Latent heat changes arrangement

During an ideal phase change at fixed pressure, energy can change particle arrangement without raising temperature. Specific latent heat is energy per mass for that change, so Q = mL. Melting and vaporisation have different latent heats. A heating graph can therefore have sloping temperature-rise regions and approximately flat state-change regions. Real gradients may be blurred by impurities, changing pressure or imperfect thermal contact.

A method you can reuse

  1. Decide whether the process changes temperature, state or both.
  2. Convert absolute temperatures to kelvin where required.
  3. Choose specific heat capacity or the appropriate latent heat.
  4. Include separate energy contributions for a multi-stage process.
  5. Identify heat losses, container heating and measurement limitations.
Why does a thermometer reading in Celsius sometimes need no conversion in a heating calculation?

If only a temperature difference is needed, Celsius and kelvin intervals have the same size. Absolute temperatures in gas or radiation laws do require the kelvin scale.

Equations and relationships

T/K=θ/∘C+273.15T/\mathrm K=\theta/{}^\circ\mathrm C+273.15

Conversion from Celsius to thermodynamic temperature.

Q=mcΔTQ=mc\Delta T

Heating without a change of state.

Q=mLQ=mL

Energy for a change of state at fixed temperature.

Worked example

To the nearest kelvin, what is 27 °C?

Follow the solution

27 + 273.15 = 300.15 K, approximately 300 K.

Investigate and explain

Compare two teacher-approved thermometers in the same water bath after allowing both to reach equilibrium; discuss response time.

Use particle and heating models to separate temperature rise from state change. They illustrate mechanisms but do not replace calibration, material-property data or a real experiment’s heat-loss accounting.

Open the interactive investigation

Check your understanding

1. What is the size of a 5 °C temperature rise in kelvin?

  1. 5 K
  2. 278.15 K
  3. −268.15 K
Check the answer and reasoning

5 K

Temperature intervals have equal sizes on these two scales.

2. To the nearest kelvin, what is 27 °C?

  1. 27 K
  2. 300 K
  3. 246 K
Check the answer and reasoning

300 K

27 + 273.15 = 300.15 K, approximately 300 K.

3. Which temperature must be used in an ideal-gas equation?

  1. Temperature in Celsius without conversion
  2. A temperature difference alone
  3. Temperature in kelvin
Check the answer and reasoning

Temperature in kelvin

Absolute temperature is proportional to the relevant particle kinetic energy.