Question

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

Answer 1

We know that: \(\dfrac{T_X-T_{FP}}{T_{BP}-T_{FP}}=\) constant

Given:
\(T_{FP}=39^{\circ}~W, T_{BP}=239^{\circ}~W, T_C=39^{\circ}~C\)

\(\dfrac{T_C-0}{100-0}=\dfrac{T_W-39}{239-39}\)

\(\dfrac{39}{100}=\dfrac{T_W-39}{200}\)

\(T_W=117^{\circ}~W\)
Final answer \(117^{\circ}~W\)