Answer
\[\underline{73{}^\circ \text{J}}\]
Work Step by Step
Calculate the conversion factor between the J scale and the H scale as follows:
\[\begin{array}{*{35}{l}}
\frac{\left( J-17 \right)}{\left( 97-17 \right)} & = & \frac{\left( H-0 \right)}{\left( 120-0 \right)} \\
\frac{\left( J-17 \right)}{80} & = & \frac{H}{120} \\
H & = & \left( \frac{120}{80} \right)\times \left( J-17 \right) \\
H & = & 1.5\left( J-17 \right) \\
\end{array}\]
Substitute \[84{}^\circ \] for H in the above equation as follows:
\[\begin{align}
& 84{}^\circ =1.5J-25.5 \\
& J=73{}^\circ
\end{align}\]
The boiling point on the Jekyll scale is \[\underline{73{}^\circ \text{J}}\].
