Label Conversion factor (Temperature)

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The discussion revolves around converting temperatures between two hypothetical scales, X and Y, with specified ice and steam points. The user attempts to solve the conversion using a given formula but finds their answer incorrect. Another participant suggests that since temperature scales are linear, a linear equation of the form y = mx + b could be more effective for determining the relationship between the two scales. They recommend using the known data points to establish the equations needed to solve for the slope and intercept. The conversation emphasizes the importance of correctly applying linear relationships in temperature conversions.
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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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