Question
Answer the following:
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?

Answer

We know,
Relation between the Fahrenheit scale and Absolute scale :
$\frac{\text{T}_{\text{F}}-32}{180}=\frac{\text{T}_{\text{K}}-273}{100}\ ...(1)$
For another set of $T'_F$ and $T'_K$
$\frac{\text{T}_{\text{F}}-32}{180}=\frac{\text{T}_{\text{K}}-273}{100}\ ...(2)$
Subtracting Equation (1) and (2):
$\therefore\text{T}_{\text{F}}-\text{T}_\text{F}=1.8(\text{T}_\text{K}-\text{T}_\text{K})$
For, $\text{T}_\text{K}-\text{T}_\text{K}=1\text{K}$
$\text{T}'_\text{F}-\text{T}_\text{F}=1.8$
For the triple point temperature = 273.16K, the temperature on the new scale = 1.8 × 273.16 = 491.688 Units.

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