A liquid-in-glass thermometer has fixed points at 0°C and 100°C. If the length between these points is 20 cm, what is the temperature in Celsius and Fahrenheit when the liquid level is 4.5 cm above th

The fixed points of a liquid-in-glass thermometer are typically the ice point (freezing point of water) and the steam point (boiling point of water).

• Ice point: $0^\circ\text{C}$ or $32^\circ\text{F}$
• Steam point: $100^\circ\text{C}$ or $212^\circ\text{F}$

The total length between the fixed points (ice point to steam point) is 20 cm. This corresponds to a temperature range of $100^\circ\text{C} - 0^\circ\text{C} = 100^\circ\text{C}$.
The mercury level is 4.5 cm above the lower mark (ice point).
We can use the linear relationship between length and temperature:

$$\frac{T_C - T_{ice}}{T_{steam} - T_{ice}} = \frac{L_x - L_{ice}}{L_{steam} - L_{ice}}$$

Where $T_C$ is the temperature in Celsius, $L_x$ is the mercury level, $L_{ice}$ is the length at the ice point, and $L_{steam}$ is the length at the steam point.

$$\frac{T_C - 0^\circ\text{C}}{100^\circ\text{C} - 0^\circ\text{C}} = \frac{4.5 \text{ cm}}{20 \text{ cm}}$$ $$\frac{T_C}{100^\circ\text{C}} = \frac{4.5}{20}$$ $$T_C = \frac{4.5}{20} \times 100^\circ\text{C}$$ $$T_C = 0.225 \times 100^\circ\text{C}$$ $$T_C = 22.5^\circ\text{C}$$

The formula to convert Celsius to Fahrenheit is:

$$T_F = \frac{9}{5} T_C + 32$$

Substitute the calculated Celsius temperature:

$$T_F = \frac{9}{5} (22.5^\circ\text{C}) + 32$$ $$T_F = 9 \times 4.5 + 32$$ $$T_F = 40.5 + 32$$ $$T_F = 72.5^\circ\text{F}$$

The temperature on the Fahrenheit scale is $72.5^\circ\text{F}$.