Worked example: A copper coil has R0 = 20 Ω at T0 = 20°C (α = 0.0039/°C). (a) Find R at 100°C. (b) A thermistor's resistance halves for every 10°C rise near room temperature. If it is 10 kΩ at 25°C, estimate it at 45°C. Step 1: ΔT = 80°C. R = 20[1 + 0.0039 × 80] = 20 × 1.312 = 26.2 Ω — a 31% rise, significant for precision circuits. Step 2: Two 10°C steps: 10 kΩ → 5 kΩ → 2.5 kΩ at 45°C. The thermistor drops 75% over the same range where copper rose 31% — opposite sign and far more sensitive, which is why thermistors (not copper) sense temperature in thermostats. Step 3 (reality check): The linear formula is an approximation; over hundreds of degrees, higher-order terms matter, and near absolute zero metals deviate as impurity scattering takes over.