A hypothetical thermometer reads 5^{circ} C i | vedclass

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(C) The general formula for temperature conversion between any two scales is: $\frac{X - 	ext{LFP}_X}{	ext{UFP}_X - 	ext{LFP}_X} = \frac{Y - 	ext{

Vedclass · Accepted Answer

$122$

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(C) The general formula for temperature conversion between any two scales is: $\frac{X - 	ext{LFP}_X}{	ext{UFP}_X - 	ext{LFP}_X} = \frac{Y - 	ext{LFP}_Y}{	ext{UFP}_Y - 	ext{LFP}_Y}$.For the faulty thermometer $(X)$: $	ext{LFP}_X = 5^{\circ} C$,$	ext{UFP}_X = 99^{\circ} C$,and reading $X = 52^{\circ} C$.For the Fahrenheit scale $(Y)$: $	ext{LFP}_Y = 32^{\circ} F$,$	ext{UFP}_Y = 212^{\circ} F$.Substituting the values: $\frac{52 - 5}{99 - 5} = \frac{T_F - 32}{212 - 32}$.$\frac{47}{94} = \frac{T_F - 32}{180}$.$0.5 = \frac{T_F - 32}{180}$.$T_F - 32 = 90$.$T_F = 122^{\circ} F$.

$A$ hypothetical thermometer reads $5^{\circ} C$ in melting ice and $99^{\circ} C$ at the steam point. What is the correct temperature in $^{\circ} F$ when this faulty thermometer reads $52^{\circ} C$?

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