# Absolute Thermodynamic Temperature Scale or Kelvin Scale

We have       ɳ HE  = W net / Q1 = Q1 –Q2 / Q1 = 1 – Q2 / Q1

as                      t1 > t2

so                    t1 – t2 > 0 ( must be for heat flow)

We know reversible heat engine’s efficiency depends on source and sink temperatures.

Thus                ɳ rev = /1 (t1, t2) where f 1 is a function of t1 and t2

From(7.7)

1 –Q2 / Q1 = f1 ( t1 , t2)

or                                               Q2 / Q1 = 1 – F1 ( T1, T2)

or                     Q1 / Q2 = 1 / 1- f ( t1 , t2) = f2 ( t1,t2)

Let us now take three heat engines arranged as shown in Fig. 7.5

Q2 / Q3 = f3 ( t2 ,t3)

And                             Q1 / Q3 = f4 ( t1 , t3)

Again

Q1 / Q2 = Q1 / Q3 / Q2/Q3

= f4(t1,t3) / f3 (t2,t3)

or                              f2 ( t1,t2) = f4 ( t1,t3 / f3 ( t2 , t3)

Now t 1, t2, t3 are independent and t3 does not affect the operation of HE1 which is

operating between t1 and t2.

To satisfy equation (7.11), we must have

and                               f4 ( t1,t3) = ϕ (t1) / ϕ(t3)

and                                  f3(t2,t3) = ϕ( t1) / ϕ(t2) = Q1 / Q2

Since ϕ (t) is an arbitrary function, the simpliest possible way to define absolute

thermodynamic temperature Tis to let <j>(t) = T, proposed by Kelvin.

Thus from equation (7.12), we have

Q1 / Q2 = T1/T2

This scale is also known as Kelvin scale. It is independent of characteristic of any substance. In Kelvin scale, the triple point of water is taken as the standard reference point. For a Carnot engine operating between T and T tr (triple point of water), we have

Q /Q tr = T / T tr

T tr = 273.16 k (arbitrary assigned value)

T=273.16 Q / Q tr

If T changes, Q also changes. Thus Q plays an important role of thermodynamic property.

Fig. 7.7 shows a series of heat engines.

T1 / T2 = Q1 / Q2

T1-T2 / T2 = Q1 – Q2 / Q2

T1 –T2 / T2 = Q1 – Q2 / Q2

T1 –T2 = T2 / Q2 ( Q1 –Q2)

Similarly                                  T2 – T3 = T3 / Q3 ( Q2 –Q3)

= T2 / Q2 ( Q2 – Q3)

T3 – T4 = T2 / Q2 ( Q3 –Q4)

And so on.

Let                                           W1 = W2 = W3 =… = W

Q1 –Q2 = Q2 – Q3 = Q3 –Q4

Thus                                        T1 – T2 = ( T2 –T3) = ( T3 –T4) =….(7.16)

Also if it is assumed Q1 – Q2 = Q2 – Q3 = Q3 – Q4 = …

Then also                     W1 = W2 = W3 = … = W

Equation (7.16) shows equal temperature intervals usually steam point and ice point are taken as fixed reference temperatures and the difference between the two is arbitrary taken as 100°C.

TS – Ti =100 , [TS = steam point Ti = ice point]

ɳ rev = 1 – Ti / TS= 0.268 [ ɳ rev measured and found to be 0.268]

Ti / TS = 0.732

(7.17) =>          TS – 0.732 ts = 100

Ts  = 100 / 0.268 = 373.15 ( steam point)

Ti = 273.15 (ice point)

Triple point of water is at 0.01°C. For this in Kelvin scale 273.15 + 0.01 = 273.16 K.