Requirement of Temperature Scale:
The temperature scale on which temperature of the system can be read is required to assign the numerical values to the thermal state of the system. It requires the selection of basic unit and reference state.
Establish a correlation between Centigrade and Fahrenheit temperature scales.
Sol: Let the temperature't' be linear function of property x. (x can be length, resistance volume, pressure etc.) Then by using equation of Line;
t = A.x + B ...(i)
At Ice Point for Centigrade scale t = 0°, then
0 = A.xi +B ...(ii)
At steam point for centigrade scale t = 100°, then
100 =A.x S + B ...(iii)
From equation (iii) and (ii), we get
a = 100/(xs - xi ) and b = -100xi/(xs - xi) Finally equation becomes in centigrade scale is;
t_{0} C = 100x/(xs - xi ) -100xi/(xs - xi)
t_{0} C = [(x - xi )/ (xs - xi )]100 ...(iv)
Likewise if Fahrenheit scale is used, then
At Ice Point for Fahrenheit scale t = 32°, then
32 = A.xi + B ...(v)
At steam point for Fahrenheit scale t = 212°, then
212 =A.xS + B ...(vi)
From equation (v) and (vi), we get
a = 180/(xs - xi ) and b = 32 - 180xi/(xs - xi) Finally general equation becomes in Fahrenheit scale is;
t_{0} F = 180x/(xs - xi ) + 32 - 180xi/(xs - xi)
t_{0} F = [(x - xi )/ (xs - xi )]180 + 32 ...(vii)
Likewise if Rankine scale is used, then
At Ice Point for Rankine scale t = 491.67°, then
491.67 = A.xi + B ...(viii)
At steam point Rankine scale t = 671.67°, then
671.67 = A.xS + B ...(ix)
From equation (viii) and (ix), we get
a = 180/(xs - xi ) and b = 491.67 - 180xi/(xs - xi) Finally equation becomes in Rankine scale is;
t_{0} R = 180×/(xs - xi ) + 491.67 - 180xi/(xs - xi)
t_{0} R = [(x - xi )/ (xs - xi )] 180 + 491.67 ...(x)
Likewise if Kelvin scale is used, then
At Ice Point for Kelvin scale t = 273.15°, then
273.15 = A.xi + B ...(xi)
At steam point Kelvin scale t = 373.15°, then
373.15 = A.xS + B ...(xii)
From equation (xi) and (xii), we get
a = 100/(xs - xi ) and b = 273.15 - 100xi/(xs - xi) Finally equation becomes in Kelvin scale is;
t_{0} K = 100x/(xs - xi ) + 273.15 - 100xi/(xs - xi)
t_{0} K = [(x - xi )/ (xs - xi )] 100 + 273.15 ...(xiii)
Now compare between above four scales:
(x - xi )/ (xs - xi ) = C/100 ...(A)
= (F-32)/180 ...(B)
= (R-491.67)/180 ...(C)
= (K - 273.15)/100 ...(D)
Now joining all 4 values we get following relation
K = C + 273.15
C = 5/9[F - 32]
= 5/9[R - 491.67] F = R - 459.67
= 1.8C + 32