On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of wat

Question

On a new scale of temperature (which is linear) and called the W scale, the freezing and boiling points of water are $39^{\circ}	ext{W}$ and $239^{\circ}	ext{W}$ respectively. What will be the temperature on the new scale, corresponding to a temperature of $39^{\circ}	ext{C}$ on the Celsius scale?

Accepted Answer

Let $T_W$ be the temperature on the W scale and $T_C$ be the temperature on the Celsius scale.
For any two linear temperature scales, the ratio of the difference between any temperature and the lower fixed point (freezing point) to the fundamental interval (difference between boiling and freezing points) is constant.
$\frac{T_W - 	ext{LFP}_W}{	ext{UFP}_W - 	ext{LFP}_W} = \frac{T_C - 	ext{LFP}_C}{	ext{UFP}_C - 	ext{LFP}_C}$
Given:
$	ext{LFP}_W = 39^{\circ}	ext{W}$, $	ext{UFP}_W = 239^{\circ}	ext{W}$
$	ext{LFP}_C = 0^{\circ}	ext{C}$, $	ext{UFP}_C = 100^{\circ}	ext{C}$
$T_C = 39^{\circ}	ext{C}$
Substituting the values:
$\frac{T_W - 39}{239 - 39} = \frac{39 - 0}{100 - 0}$
$\frac{T_W - 39}{200} = \frac{39}{100}$
$T_W - 39 = 39 	imes \frac{200}{100}$
$T_W - 39 = 78$
$T_W = 78 + 39 = 117^{\circ}	ext{W}$

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