(c) The absolute temperature (Kelvin scale) \mathbf{T} is related to the temperature tc on the Celsius scale by tc = T – 273.15. Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?
(c) The absolute temperature (Kelvin scale) \mathbf{T} is related to the temperature tc on the Celsius scale by tc = T – 273.15. Why do we have 273.15 in this relation, and not 273.16? (d) What is the temperature of the triple-point of water on an absolute scale whose unit interval size is equal to that of the Fahrenheit scale?

Solution:

(iii) The relationship exists because the melting point of ice equates to 273.15 K on the Kelvin scale, and the triple point of water corresponds to 273.16 K on the Kelvin scale.

(iv) We know that,

The following is the relationship between the Fahrenheit scale and the Absolute scale:

\frac{T_{F}-32}{180}=\frac{T_{K}-273}{100} \ldots \ldots \ldots \ldots \ldots \ldots(1)

For another set of \mathrm{T}_{\mathrm{F}} and \mathrm{T}^{\prime} \mathrm{K}

\frac{T^{\prime} p-32}{180}=\frac{T^{\prime} K-273}{100} \ldots \ldots \ldots \ldots \ldots \ldots(2)

Subtracting Equation (1) and (2):

\frac{T^{\prime} F-T_{F}}{180}=\frac{T_{K}-T_{K}}{100}

For, T_{K}^{\prime}-T_{K}=1 K

T_{F}-T_{F}=1.8

For the triple point temperature =273.16 \mathrm{~K}, the temperature on the new scale =1.8 \times 273.16=491.688 K