question_answer

A mercury thermometer is transferred from melting ice to a hot liquid. The mercury rises 0.9 of the distance between lower and upper fixed points. What is the temperature of the liquid in\[{}^\circ \mathbf{C}\]? What in \[{}^\circ \mathbf{F}\]?

Answer:

                \[{{t}_{C}}=\frac{{{l}_{t}}-{{l}_{0}}}{{{l}_{100}}-{{l}_{0}}}\times {{100}^{\circ }}C=\frac{0.9-0}{10-0}\times 100=\mathbf{9}{{\mathbf{0}}^{\mathbf{o}}}\mathbf{C}.\]                 \[{{t}_{F}}=\frac{9}{5}{{t}_{C}}+32=\frac{9}{5}\times 90+32=\mathbf{19}{{\mathbf{4}}^{\mathbf{o}}}\mathbf{F}.\]