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## Q. 2.2

A thermometer reads 32°F when placed in melting ice and 212°F when placed in boiling water. What will it read when the measured temperature is 290 K ?

## Verified Solution

Let                     t = ax + b
So that              32 = a$x_i$+ b

212 = a$x_s$+ b

Which gives,  $a=\frac{180}{x_s-x_i}, b=32-180\left[\frac{x_i}{x_s-x_i}\right]$

$t^{\circ} F =180\left[\frac{x-x_i}{x_s-x_i}\right]+32$

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

$\therefore$  t°F = 1.8 t°C + 32

Now  t°C = 290 – 273.15 = 16.85

$\therefore$  t°F = 1.8 × 16.85 + 32 = 62.33