The Celsius and Fahrenheit Scales — Question 5

Question 5

An interval specification must survive conversion at both ends

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. A component has measured temperature $T=(310.00\pm.20)\,\mathrm K$, with a hard bound. Its permitted Fahrenheit interval is $[97.5,98.7]^\circ\mathrm F$. Certification requires the entire measured interval to lie inside the permitted interval.

1. Convert the central reading and its full interval to Fahrenheit. Can the component be certified?

2. Keeping the same central temperature, determine the largest kelvin half-width that would pass. Identify the limiting specification edge.

Question 5 - Solution

An interval specification must survive conversion at both ends

(a) Method and calculation

$$C_0=36.85,\qquad F_0=98.33,\qquad\delta F=1.8(.20)=.36.$$$$F\in[97.97,98.69].$$Both endpoints lie within $[97.5,98.7]$, so certification passes under the stated error model.

(b) Interpretation and further calculation

$$\delta F_{\max}=\min(98.33-97.5,98.7-98.33)=.37.$$$$\delta T_{\max}=.37/1.8=.205556\,\mathrm K.$$The upper-temperature limit is the tighter edge; the measurement center is not at the specification midpoint.

Physical check

The existing upper endpoint is only $.01$ Fahrenheit degree below the specification edge.