Question

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 1

Let $T_F$ is the temperature on Farhenhiet scale.
$T_K$ is the temperature on absolute scale.

both the Temperature can be related as
$\frac{(T_F-32)}{180}=\frac{(T_K-273.15)}{100}$

it is given that, $\Delta{T_K}=1K$

so, $\frac{\Delta{T_F}}{180}=\frac{\Delta{T_K}}{100}$

or, $\frac{\Delta{T_F}}{9}=\frac{1}{100}$

or, $\Delta{T_F}=\frac{9}{5}$

triple point of water = 273.16K

triple point of water in farhenhiet = 273.16 × 9/5°F
= 491.688°F

Answer 2

Answer:

491.69°F

Explanation:

Let $T_{F}$ be the temperature on Fahrenheit scale and $T_{K1}$ be the temperature

on the absolute scale. Both the temperatures can be related as:

$\frac{T_{FI}-33}{180}= \frac{T_{KI}-273.15}{100}. . . . . (ii)$

It is given that:

$T_{KI}-T_{K} = 1K$

Subtracting equation (i) from equation (ii), we get:

$\dfrac{T_{FI}-T_{F}}{180} = \dfrac{T_{KI}-T_{K}}{100}=\dfrac{1}{100}$

$T_{KI}-T_{K} = \dfrac{1\times 180}{100} = \dfrac{9}{5}$

The triple point of water = 273.16 K

The triple point of water on absolute scale:

$\implies\:\:\:273.16\times \frac{9}{5}$

$\implies\:\:\: 491.69^{\circ}F$