How Do You Convert Between Celsius and Kelvin?

For IB Physics SL and HL, convert Celsius to kelvin by adding $273.15$, and convert kelvin to Celsius by subtracting $273.15$.

Conversion Method

The Kelvin scale and Celsius scale have equal-sized intervals but different zero points. A temperature change of $1\text{ K}$ is therefore equal in size to a change of $1\,^{\circ}\text{C}$.

The Kelvin scale is an absolute temperature scale: its zero corresponds to the lowest possible thermodynamic temperature. Celsius instead places zero at water's freezing point under specified conditions. Because each kelvin and each Celsius degree represent the same interval, conversion changes the zero reference but does not rescale the measurement.

Here, $T$ represents temperature in kelvin and $\theta$ represents temperature in degrees Celsius.

For example, convert $25.0\,^{\circ}\text{C}$ to kelvin:

$$T = 25.0 + 273.15 = 298.15\text{ K}$$

Convert $300.0\text{ K}$ to Celsius:

$$\theta = 300.0 - 273.15 = 26.85\,^{\circ}\text{C}$$

In an ideal-gas calculation, a gas at $27.0\,^{\circ}\text{C}$ must be entered as $300.15\text{ K}$. If volume and amount of gas are constant, $P/T=\text{constant}$, so pressure is proportional to absolute temperature. Substituting $$ instead of would give a physically incorrect pressure ratio.

The Kelvin scale begins at absolute zero, so $0\text{ K} = -273.15\,^{\circ}\text{C}$. Kelvin temperatures are not written with a degree symbol: write $300\text{ K}$, not $300\,^{\circ}\text{K}$.

For differences, subtraction removes the offset: $(40+273.15)-(20+273.15)=20\text{ K}$. Therefore, a rise of $20\,^{\circ}\text{C}$ is also a rise of , although the final temperature readings differ.

The common misconception is that Celsius and Kelvin are related by multiplication. They are not: the scales differ by a fixed offset of $273.15$.

Exam Technique

This is part of B.1 Thermal energy transfers for both SL and HL. Convert temperatures to kelvin before using equations involving absolute temperature, such as the ideal gas law or Stefan and Boltzmann law. Show the conversion explicitly and use $\text{K}$ without a degree symbol.