B.1.4—Temperature scale changes

Same size of change

Because the Celsius and Kelvin scales have the same interval size, a temperature change has the same numerical value in both scales:

$\Delta T(\mathrm K)=\Delta\theta(^{\circ}\mathrm C)$

Read a change, not an absolute value

If a sample falls from +10 °C to −10 °C, then

$\Delta\theta=-10-10=-20^{\circ}\mathrm C$

The same change is −20 K. The zero point shifts, but the spacing between adjacent temperatures does not.

Common trap

Do not add 273.15 when converting a temperature difference. Add 273.15 only when converting an absolute Celsius temperature to Kelvin.

How it is assessed

The evidence uses multiple-choice questions asking for a temperature change after expressing the endpoints in Celsius or kelvin.

Command terms

Calculate / Determine

What earns marks

For a temperature change, subtract initial from final. The numerical interval is identical in kelvin and Celsius, so do not add or subtract 273 when calculating ΔT. Include the sign if the process cools.

Watch for

Adding 273 to a temperature difference instead of using ΔT=final−initial.

Representative question

[Maximum number: 1]

The temperature of an object is changed from $\theta_{1}{ }^{\circ} \mathrm{C}$ to $\theta_{2}{ }^{\circ} \mathrm{C}$ . What is the change in temperature measured in kelvin?

$\left(\theta_{2}-\theta_{1}\right)$

$\left(\theta_{2}-\theta_{1}\right)+273$

$\left(\theta_{2}-\theta_{1}\right)-273$

$273-\left(\theta_{2}-\theta_{1}\right)$

Microscopic story

Matter contains moving particles. Temperature tracks average random kinetic energy, while internal energy also includes intermolecular potential energy. Phase changes alter particle behaviour at constant temperature.

Transfer story

A temperature difference gives the net direction of thermal energy transfer. Conduction transfers energy through local interactions, convection through moving fluids, and radiation through electromagnetic waves.

Equation map

$\rho=\frac{m}{V}$

$Q=mc\Delta T,\quad Q=mL$

$\frac{\Delta Q}{\Delta t}=\frac{kA\Delta T}{\Delta x}$

$L=\sigma AT^4$

$b=\frac{L}{4\pi d^2}$

$\lambda_{\max}T=2.9\times10^{-3}\,\mathrm{m\,K}$

Question strategy

Identify whether the question concerns a state property, a transfer mechanism or a rate.
Convert to SI units and use kelvin whenever an absolute temperature appears.
Check whether temperature changes, remains constant during a phase change, or enters a fourth-power/inverse-square relation.
State the physical reason, not only the numerical substitution.

B.1.4—Temperature scale changes

Compare Temperature Changes in K and °C

B.1.4 Exam Analysis

Assessment in practice

Synthesize B.1 Thermal Energy Transfers

Question 1