JAMB - Physics (1991 - No. 13)
A thermometer with an arbitrary scale S, of equal division registers -30oS at the ice point and +90oS at the steam point. Calculate the Celsius temperature corresponding to 60oS
25.0oC
50.0oC
66.7oC
75.0oC
Explanation
100oC ... 90oS
\(\theta\)oC ... 60oS
0oC ... 30oS
\(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\)
\(\theta\) = 75oC
\(\theta\)oC ... 60oS
0oC ... 30oS
\(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\)
\(\theta\) = 75oC
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