Triple point of water on absolute scale A, \(T_1\) = 200 A
Triple point of water on absolute scale B, \(T_2\) = 350 B
Triple point of water on Kelvin scale, \(T_K\) = 273.15 K
The temperature 273.15 K on Kelvin scale is equivalent to 200 A on absolute scale A.
\(T_1 = T_K\)
200 A = 273.15 K
∴ A = \(\frac{273.15}{200}\)
The temperature 273.15 K on Kelvin scale is equivalent to 350 B on absolute scale B.
\(T_2 = T_K\)
350 B = 273.15
∴ B = \(\frac{273.15}{350}\)
\(T_A\) is triple point of water on scale A.
\(T_B\) is triple point of water on scale B.
∴ \(\frac{273.15}{200}\) x \(T_A\) = \(\frac{273.15}{350}\) x \(T_B\)
\(T_A = \frac{200}{350} T_B\)
Therefore, the ratio \(T_A : T_B\) is given as 4 : 7.
The electrical resistance in ohms of a certain thermometer varies with temperature according to the approximate law :
R = \(R_0\) [1 + α (T – \(T_0\))]
The resistance is 101.6 Ω at the triple-point of water 273.16 K, and 165.5 Ω at the normal melting point of lead (600.5 K). What is the temperature when the resistance is 123.4 Ω ?
Figures 9.20(a) and (b) refer to the steady flow of a (non-viscous) liquid. Which of the two figures is incorrect ? Why ?