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
The correct answer is: D
Explanation
100oC ... 90oS\(\theta\)oC ... 60oS
0oC ... 30oS
\(\frac{\theta - 0}{100 - 0}\) = \(\frac{60 - (-30)}{90 - (-30)}\)
\(\theta\) = 75oC