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 be the temperature on Fahrenheit scale and T_K be the temperature on absolute scale. Both the temperatures can be related as:

`(T_1 - 32)/180 = (T_K - 273.15)/100` ... (i)

Let T_F1 be the temperature on Fahrenheit scale and T_K1 be the temperature on absolute scale. Both the temperatures can be related as:

`(T_eta - 32)/180 = (T_(K1) - 273.15)/100`   ...(ii)

It is given that:

T_K1 – T_K = 1 K

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

`(T_(F1)- T_F)/180 =(T_(K1) - T_K)/100 = 1/100`

`T_(F1) - T_F = (1xx 180)/100 =  9/5`

Triple point of water = 273.16 K

∴Triple point of water on absolute scale = `273.16xx 9/5 = 491.69`

Answer 2

The Fahrenheit scale and Absolute scale are related as

`(T_F - 32)/180 = (T_K-273)/100`

For another set of temperature `T'_F and T'_K'`

`(T'_F - 32)/180 = (T'_K - 273)/100`

Subtracting i from ii

`(T'_F - T_F)/180 = (T'_k -T_K)/100`

`:. T'_F - T_F = 180/100 (T'_K - T_K)`

For  `T'_K - T_K = 1K`

`T'_F - T_F = 180/100`

:. For a temperature of triple point i.e 273.16 K the temperature on the new scale is

`= 273.16 xx 180/100 = 491.688`