If a thermometer reads the freezing point of | vedclass

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(A) The temperature on any scale can be converted to another scale using the linear relationship: $\frac{X - LFP}{UFP - LFP} = 	ext{constant}$ Where

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$98$

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(A) The temperature on any scale can be converted to another scale using the linear relationship: $\frac{X - LFP}{UFP - LFP} = 	ext{constant}$ Where $X$ is the temperature reading,$LFP$ is the Lower Fixed Point (freezing point),and $UFP$ is the Upper Fixed Point (boiling point). For the Celsius scale: $LFP = 0^{\circ}C$,$UFP = 100^{\circ}C$. For the given thermometer: $LFP' = 20^{\circ}C$,$UFP' = 150^{\circ}C$. Equating the two: $\frac{X - 20}{150 - 20} = \frac{60 - 0}{100 - 0}$ $\frac{X - 20}{130} = \frac{60}{100}$ $\frac{X - 20}{130} = 0.6$ $X - 20 = 0.6 	imes 130$ $X - 20 = 78$ $X = 98^{\circ}C$.

If a thermometer reads the freezing point of water as $20^{\circ}C$ and the boiling point as $150^{\circ}C$,what will the thermometer read when the actual temperature is $60^{\circ}C$?

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