(a) The triple point of water has a unique value of 273.16 K. The triple point of water remains equal to 273.16 K for particular values of volume and pressure. The melting point of ice and boiling point of water depends on pressure and temperature and do not have constant values.

(b) The absolute zero or 0 K is the other fixed point on the Kelvin absolute scale.

The temperature of 273.16 K is the triple point of water. It is not the melting point of ice. The melting point of ice is equal to 0°C on the Celsius scale and 273.15 K on the Kelvin scale.

Hence, absolute temperature (Kelvin scale) T, is related to temperature ${\mathrm{t}}_{\mathrm{c}}$, on the Celsius scale as: ${\mathrm{t}}_{\mathrm{c}}$ = T – 273.15.

(d) If ${\mathrm{T}}_{\mathrm{F}}$ is the temperature on the Fahrenheit scale and ${\mathrm{T}}_{\mathrm{K}}$ is the temperature on the absolute scale, then:

$\frac{{\mathrm{T}}_{\mathrm{F}}-32}{180}=\frac{{\mathrm{T}}_{\mathrm{K}}-273.15}{100} ....\left(\mathrm{i}\right)$

If ${{\mathrm{T}}_{\mathrm{F}}}_1$ is the temperature on the Fahrenheit scale and ${{\mathrm{T}}_{\mathrm{K}}}_1$ is the temperature on the absolute scale, then:

$\frac{{\mathrm{T}}_{{\mathrm{F}}_1}-32}{180}=\frac{{\mathrm{T}}_{{\mathrm{K}}_1}-273.15}{100} ....\left(\mathrm{ii}\right)$

Given that:

${\mathrm{T}}_{{\mathrm{K}}_1}-{\mathrm{T}}_{\mathrm{K}}=1 \mathrm{K}$

Subtracting equation (i) from equation (ii),

$$\begin{gathered} \frac{{\mathrm{T}}_{{\mathrm{F}}_1}-{\mathrm{T}}_{\mathrm{F}}}{180}=\frac{{\mathrm{T}}_{{\mathrm{K}}_1}-{\mathrm{T}}_{\mathrm{K}}}{100}=\frac{1}{100} \\ {\mathrm{T}}_{{\mathrm{F}}_1}-{\mathrm{T}}_{\mathrm{F}}=\frac{1\times 180}{100}=\frac{9}{5} \end{gathered}$$

The triple point of water = 273.16 K

The triple point of water on the absolute scale = $273.16\times \frac{9}{5}=491.69$