A resistance thermometer reads 102 Ω in melting ice at 0 °C and 142 Ω in steam at 100 °C under the calibration pressure. In a bath it reads 112 Ω. Assume resistance is linear in Celsius temperature between these points. (a) Calculate bath temperature in °C and K, using T/K = θ/°C + 273.15. (b) Explain thermal equilibrium and one reason this two-point calibration may fail outside its range.

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25 °C = 298.15 K; equilibrium means no net heat transfer; resistance may be nonlinear outside calibration range. The full resistance increase over 100 °C is 142 − 102 = 40 Ω. The bath reading is 10 Ω above the ice-point reading, representing fraction 0.25 of the calibrated interval. Thus θ = (112 − 102)/(142 − 102) × 100 = 25 °C. The nonzero zero-degree resistance must be subtracted before forming the ratio. Convert the temperature scale by adding 273.15, giving T = 25 + 273.15 = 298.15 K. A temperature difference has the same numerical size in kelvin and degrees Celsius, but an absolute reading does not. Wait until thermometer and bath are in thermal equilibrium, meaning no net thermal energy transfer between them and equal temperatures. The linear resistance model is only established between its calibration points; nonlinearity, self-heating or changed pressure at the reference point can produce systematic errors.

Calibrating a resistance thermometer 2

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