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## Q. 2.14

A new scale N of temperature is devised in such a way that the freezing point of ice is 100°N and boiling point is 400°N. What is the temperature reading on this new scale when the temperature is 150°C? At what temperature both the Celsius and new temperature scale reading would be the same?

## Verified Solution

Let               t°N = ax + b

So that        100 = a$x_i$+ b

400 = a$x_s$ + b

\begin{aligned}a &=\frac{300}{x_s-x_i} \\b &=100-\frac{300 x_i}{x_s-x_i} \\t^{\circ} N &=\frac{300 x}{x_s-x_i}+100-\frac{300 x_i}{x_s-x_i} \\&=\frac{300\left(x-x_i\right)}{x_s-x_i}+100\end{aligned}

Also  $t^{\circ} C =100 \frac{100\left(x-x_i\right)}{x_s-x_i}$

$\therefore$  t°N = 3 t°C + 100

When         t°C = 150, then

t°N = 3 × 150 + 100 = 550

For              t°N = t°C

t°C = 3t°C + 100

t°C = – 50