The ice and steam points on a mercury in glass thermometer are found to be 90mm apart. What temperature is recorded in degree Celsius when the length of the mercury thread is 33.6mm above the ice point mark?
\(\frac{x - 0}{100 - 0} = \frac{33.6 - 0}{90 - 0}\)
\(\frac{x}{100} = \frac{33.6}{90}\)
\(90x = 3360 \implies x = \frac{3360}{90}\)
= 37.3°C
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