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## Q. 2.1

The ice and steam points on a temperature scale °M are 80°M and 400°M respectively. Correlate this with the Celsius scale. The °N reading on this scale is a certain number of degrees on a corresponding absolute temperature scale. Find this absolute temperature at °N.

## Verified Solution

Let                   t = ax + b
So that            80 = a$x_i$+ b
400 = a$x_s$+ b

Which gives, $a=\frac{320}{x_s-x_i}, b=80-320\left[\frac{x_i}{x_s-x_i}\right]$

$t^{\circ} M =320\left[\frac{x-x_i}{x_s-x_i}\right]+80$

Also, $t^{\circ} C =100\left[\frac{x-x_i}{x_s-x_i}\right]$

$\therefore$ t°M = 3.2 t°C + 80

Now $\frac{T_s}{T_i}=1.366$

And $T_s-T_i=400-80=320$

\begin{aligned}&T_i=\frac{320}{0.366}=874.3^{\circ} N \\\\&T_s=1194.3^{\circ} N\end{aligned}

T°N abs = (t°N – 80) + 874.3 = t°N + 794.3

At t°N = 0°N, we get

T°N abs = 0°N + 794.3 = 794.3