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

Question

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

Aicharya · Accepted Answer

$\begin{aligned} & \frac{39-0}{100-0}=\frac{x-39}{239-39} \ & \Rightarrow \quad x=117^{\circ} \mathrm{W} \end{aligned}$ .

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

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