The Celsius and Fahrenheit Scales — Question 4

Section 18-4 · Physics pages 2878–2879

Question 4 An affine display fault is distinct from changing temperature units Use numerical readings F = 9 5C + 32 and T/K = C + 273.15; these conversions are exact. A nominal Fahrenheit display reads 35.0 in a 0.0◦C reference and 215.0 in a 100.0◦C reference. Model its reading as D = aC + b exactly over this range. C D, F The faulty and standard mappings have the same slope and different intercepts. (a) Infer a, b and the true Celsius temperature when D = 125.0. (b) A technician applies the standard Fahrenheit-to-Celsius conversion directly to D. Find the systematic Celsius error and the corrected Fahrenheit reading for the unknown bath.
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Question 4 - Solution An affine display fault is distinct from changing temperature units C D, F (a) Method and calculation a = 215 −35 100 −0 = 1.8, b = 35. C = 125 −35 1.8 = 50.0. The gain is correct but the display has an offset of three Fahrenheit degrees. (b) Interpretation and further calculation Cnaive = 125 −32 1.8 = 51.6667, Cnaive −C = 1.6667. F = 1.8(50) + 32 = 122.0 = D −3.0. Unit conversion cannot repair an unmodeled calibration fault. Physical check Both anchor readings are needed to distinguish an offset fault from an incorrect gain under the affine model.

Question 4

An affine display fault is distinct from changing temperature units

Use numerical readings F=95C+32F=\frac95C+32 and T/K=C+273.15T/\mathrm K=C+273.15; these conversions are exact. A nominal Fahrenheit display reads 35.035.0 in a 0.0∘C0.0^\circ\mathrm C reference and 215.0215.0 in a 100.0∘C100.0^\circ\mathrm C reference. Model its reading as D=aC+bD=aC+b exactly over this range.

Diagram shown in the original SVG worksheet.

  1. Infer a,ba,b and the true Celsius temperature when D=125.0D=125.0.

  2. A technician applies the standard Fahrenheit-to-Celsius conversion directly to DD. Find the systematic Celsius error and the corrected Fahrenheit reading for the unknown bath.

Question 4 - Solution

An affine display fault is distinct from changing temperature units

Diagram shown in the original SVG worksheet.

(a) Method and calculation

a=215−35100−0=1.8,b=35.a=\frac{215-35}{100-0}=1.8,\qquad b=35.C=125−351.8=50.0.C=\frac{125-35}{1.8}=50.0.The gain is correct but the display has an offset of three Fahrenheit degrees.

(b) Interpretation and further calculation

Cnaive=125−321.8=51.6667,Cnaive−C=1.6667.C_{\mathrm{naive}}=\frac{125-32}{1.8}=51.6667,\quad C_{\mathrm{naive}}-C=1.6667.F=1.8(50)+32=122.0=D−3.0.F=1.8(50)+32=122.0=D-3.0.Unit conversion cannot repair an unmodeled calibration fault.

Physical check

Both anchor readings are needed to distinguish an offset fault from an incorrect gain under the affine model.