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?
Solution 1
Let TF be the temperature on Fahrenheit scale and TK be the temperature on absolute scale. Both the temperatures can be related as:
`(T_1 - 32)/180 = (T_K - 273.15)/100` ... (i)
Let TF1 be the temperature on Fahrenheit scale and TK1 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:
TK1 – TK = 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`
Solution 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`
