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Q. 11.4 Answer the following :

(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?

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Let at a certain temperature the reading on Fahrenheit and Kelvin Scale be T and TK respectively

T_{F}-32=\frac{9}{5}(T_{K}-273)                                (i)

Let at another temperature the reading on Fahrenheit and Kelvin Scale be T' and T'K respectively

T'_{F}-32=\frac{9}{5}(T'_{K}-273)                                 (ii)

Subtracting equation (ii) from (i)

T_{F}-T'_{F}=\frac{9}{5}(T_{K}-T'_{K})

For T- T'K = 1 K, TF - T'F = 9/5

Therefore corresponding to 273.16 K the absolute scale whose unit interval size is equal to that of the Fahrenheit scale

\\T_{F}=\frac{9}{5}\times 273.16\\ T_{F}=491.688

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