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Let us consider ${{T}_{f}}$ to be the temperature of water in Fahrenheit scale and ${{T}_{K}}$ to be the temperature on the absolute scale.

We can relate both the scales together by

$\dfrac{({{T}_{f}}-32)}{180}=\dfrac{({{T}_{K}}-273.15)}{100}$ …...................... (1)

Let us consider ${{T}_{f1}}$ to be the temperature of water in Fahrenheit scale and ${{T}_{K1}}$ to be the temperature on the absolute scale.

We can relate both the scales together by

$\dfrac{({{T}_{f1}}-32)}{180}=\dfrac{({{T}_{K1}}-273.15)}{100}$……………… (2)

From the question we can concur that

${{T}_{K1}}-{{T}_{K}}=1$K

On Subtracting equation 1 from equation 2 we get

$\dfrac{({{T}_{f1}}-{{T}_{f}})}{180}=\dfrac{({{T}_{K1}}-{{T}_{K}})}{100}$

$\Rightarrow \dfrac{({{T}_{f1}}-{{T}_{f}})}{180}=\dfrac{1}{100}$

$\Rightarrow ({{T}_{f1}}-{{T}_{f}})=\dfrac{9}{5}$

We know that the triple point of water is 273.16K

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

We need to analyze the requirement and the data that the question provides. In thermodynamics, the triple point of a substance is the temperature and pressure at which the three phases of that substance i.e., solid, liquid and gas coexist in thermodynamic equilibrium. In the phase diagram, the point where the sublimation curve, fusion curve and the vaporization curve meet, its temperature and pressure is known as the triple point. For example, the triple point of mercury occurs at a pressure of 0.2 mPa and the temperature of −38.83440$^{\circ }C$.