First slide

Thermometry

Question

On a hypothetical scale X, the ice point is 40⁰ and the steam point is 120⁰. For another scale Y the ice point and steam points are –30⁰ and 130⁰ respectively. If X-reads 50⁰ The reading of Y is

Easy

A

-5⁰
B

-8⁰
C

-10⁰
D

-12⁰

Solution

Scale    LFP    UFP    Temp
X          40⁰      120⁰     50
Y        -30⁰       130⁰     ?
we have $\large \left ( \frac {x-LFP}{UFP-LFP} \right )_x=\left ( \frac {y-LFP}{UFP-LFP} \right )_y$
$\large \frac {50-40}{120-40}=\frac {y+30}{130+30}\Rightarrow y = -10^0$

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