The Celsius and Fahrenheit Scales — Question 7

Question 7

Degree-hours measure excess temperature, not an absolute reading

Use numerical readings $F=\frac95C+32$ and $T/\mathrm K=C+273.15$; these conversions are exact. An oven temperature rises linearly from $20$ to $80^\circ\mathrm C$ in two hours. Define excess exposure relative to $20^\circ\mathrm C$ by $H_C=\int_0^{2\,\mathrm h}[C(t)-20]dt$. This is a mathematical exposure index, not a heat quantity.

1. Find $H_C$ and the corresponding Fahrenheit-degree-hour exposure relative to the same physical reference.

2. Find the erroneous value obtained by applying $F=1.8C+32$ directly to the excess-temperature function before integration. Explain why the extra term appears.

Question 7 - Solution

Degree-hours measure excess temperature, not an absolute reading

(a) Method and calculation

With $t$ expressed in hours, $C(t)-20=30t$.$$H_C=\int_0^2 30t\,dt=60\,{}^\circ\mathrm C\,\mathrm h.$$$$H_F=1.8H_C=108\,{}^\circ\mathrm F\,\mathrm h.$$

(b) Interpretation and further calculation

$$H_{\mathrm{wrong}}=\int_0^2[1.8(30t)+32]dt=108+64=172\,{}^\circ\mathrm F\,\mathrm h.$$$$H_{\mathrm{wrong}}-H_F=32(2)=64\,{}^\circ\mathrm F\,\mathrm h.$$A difference channel has no 32-degree intercept; the reference subtraction already removes it.

Physical check

The exposure index has temperature-times-time units; it cannot be identified with joules without an additional physical model.