Calibrating a School Thermometer Before a Cooling Experiment
A BGCSE candidate is investigating how quickly hot water cools. She has a liquid-in-glass thermometer whose scale reads from -10 °C to 110 °C with 1 °C divisions, but the zero mark looks slightly above the true ice point. To check it, she places the bulb in a beaker of crushed ice and distilled water, stirs gently, and waits. After 4 minutes the reading stays steady at 1.5 °C. She then places the bulb in gently boiling distilled water at the school's altitude (about 1 000 m above sea level) and records 98.0 °C. She plans to record the water temperature every 30 s for 10 minutes, starting at about 80 °C, using the same thermometer. The room temperature is 24 °C.
1. Explain how the two fixed-point readings should be used to correct the cooling data, and state one assumption this correction relies on.
Read the worked answer
The thermometer has a zero error of +1.5 °C (it reads 1.5 °C at the true ice point) and a boiling-point error of about -2.0 °C (it reads 98.0 °C where pure water boils near 100 °C at this altitude). A simple linear correction can be made by assuming the error changes steadily between the two fixed points.
One workable method is to calculate a correction for each reading using the two-point calibration: take the true values (0 °C and approximately 100 °C) and the measured values (1.5 °C and 98.0 °C), then interpolate. For a reading R, the corrected value is roughly:
corrected = 0 + (R - 1.5) × (100 - 0)/(98.0 - 1.5)
So a reading of 80.0 °C corrects to (80.0 - 1.5) × 100/96.5 ≈ 81.3 °C. The assumption is that the thermometer's error varies linearly across the scale, which is reasonable for a uniform-bore capillary but should be stated as a limitation.
The candidate should also note that the true boiling point at her altitude is slightly below 100 °C, so the second fixed point is approximate unless she looks up the local value. Even so, the correction is small compared with the cooling she will observe, so the main conclusion (the shape of the cooling curve) is unaffected.
2. The candidate records temperatures every 30 s. Suggest how she should decide the uncertainty in each temperature reading and how that affects the cooling curve she plots.
Read the worked answer
Each reading has two contributions: the resolution of the scale (1 °C divisions, so a reading uncertainty of about ±0.5 °C) and the calibration uncertainty from the fixed-point check. The calibration correction itself carries uncertainty because the ice-point and steam-point readings were each taken to ±0.5 °C; combining these gives a calibration uncertainty of roughly ±0.7 °C. Adding in quadrature, the total uncertainty per corrected reading is about √(0.5² + 0.7²) ≈ 0.9 °C, so she should quote temperatures as, for example, (81.3 ± 0.9) °C.
When plotting the cooling curve, she should draw error bars of about ±1 °C on each point. If the smooth curve she draws passes through all the error bars, the trend is consistent with the data. If a point lies well outside its error bar, that point deserves checking - perhaps the thermometer was read too slowly, or the water was stirred unevenly.
The uncertainty also limits what she can claim. A temperature drop of 2 °C between two readings 30 s apart is comparable to the uncertainty, so she should not claim a definite change over such a short interval. The overall drop from 80 °C to near room temperature is far larger than the uncertainty, so the main conclusion is secure.