By the end of this lesson you will be able to:
$\displaystyle \frac{T}{\text{K}} = \frac{\theta}{^\circ\text{C}} + 273$.The kelvin scale is an absolute temperature scale – 0 K is the theoretical point at which all molecular motion ceases. The Celsius scale is a relative scale anchored to the freezing and boiling points of water at 1 atm.
The conversion between the two scales is linear and can be written as:
$$T\ (\text{K}) = \theta\ (\text{°C}) + 273.15$$For most A‑Level work the approximation 273 is acceptable, but the more precise value is 273.15 K.
| From | To | Formula | Example |
|---|---|---|---|
| °C | K | $T = \theta + 273.15$ | $25^\circ\text{C} \rightarrow 298.15\ \text{K}$ |
| K | °C | $\theta = T - 273.15$ | $310\ \text{K} \rightarrow 36.85^\circ\text{C}$ |
Convert 0 °C to kelvin.
Using $T = \theta + 273.15$:
$$T = 0 + 273.15 = 273.15\ \text{K}$$Convert 500 K to degrees Celsius.
Using $\theta = T - 273.15$:
$$\theta = 500 - 273.15 = 226.85^\circ\text{C}$$Find the temperature in kelvin of a laboratory oven set at 200 °C.
$$T = 200 + 273.15 = 473.15\ \text{K}$$The conversion between kelvin and degrees Celsius is straightforward because the two scales are offset by a constant value:
$$\boxed{T\ (\text{K}) = \theta\ (\text{°C}) + 273.15}$$Remember to keep track of units, use the correct sign, and apply the precise offset when required.