On a new scale of temperature (which is linea | vedclass

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(D) The relationship between two linear temperature scales is given by the formula: $\frac{T_W - 	ext{Freezing point}_W}{	ext{Boiling point}_W - 	e

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$117$

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(D) The relationship between two linear temperature scales is given by the formula: $\frac{T_W - 	ext{Freezing point}_W}{	ext{Boiling point}_W - 	ext{Freezing point}_W} = \frac{T_C - 	ext{Freezing point}_C}{	ext{Boiling point}_C - 	ext{Freezing point}_C}$.Given for the $W$ scale: Freezing point = $39\,^{\circ}W$,Boiling point = $239\,^{\circ}W$.Given for the Celsius scale: Freezing point = $0\,^{\circ}C$,Boiling point = $100\,^{\circ}C$.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 = \frac{39}{100} 	imes 200$$T_W - 39 = 39 	imes 2 = 78$$T_W = 78 + 39 = 117\,^{\circ}W$.

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

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