Label Conversion factor (Temperature)

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SUMMARY

The discussion focuses on converting temperatures between two hypothetical scales, X and Y, with defined ice and steam points. The ice point for scale X is 30° and the steam point is 150°, while for scale Y, the ice point is -15° and the steam point is 200°. Participants suggest using a linear relationship model, represented by the equation y = mx + b, to derive the conversion factors between the two scales based on the provided data points.

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  • Familiarity with temperature conversion concepts.
  • Basic algebraic manipulation skills.
  • Knowledge of the properties of linear relationships in mathematics.
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  • Research how to derive the slope (m) and y-intercept (b) from two data points.
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Students studying physics or mathematics, educators teaching temperature conversions, and anyone interested in understanding linear relationships in temperature scales.

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Homework Statement


On a hypothetical temperature scales X and Y, the ice and steam points of water are 30 and 150, and -15 and 200° respectively
a. 32 X° is equivalent to how many Y°
b. 100 °X is equivalent to how many K?

Homework Equations


Given value - (ice point)/ steam point = Unknown - (ice point)/ steam point - (ice point)


The Attempt at a Solution


32-(30)/150-(30) = Y-(-15)/200-(-15)
My answer is incorrect!
 
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kewlergel said:

Homework Statement


On a hypothetical temperature scales X and Y, the ice and steam points of water are 30 and 150, and -15 and 200° respectively
a. 32 X° is equivalent to how many Y°
b. 100 °X is equivalent to how many K?

Homework Equations


Given value - (ice point)/ steam point = Unknown - (ice point)/ steam point - (ice point)


The Attempt at a Solution


32-(30)/150-(30) = Y-(-15)/200-(-15)
My answer is incorrect!
Hi kewlergel, Welcome to Physics Forums.

I don't follow logic of your Relevant Equation; it could just be my thick-headedness :smile:

However, since temperature scales are linear you would expect there to be a linear relationship between the two scales: one of the form y = mx + b. You have data points for particular equivalent values from each scale, so why not set up the equations to solve for m and b?
 

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