In Chapter 14 of Intermediate Physics for Medicine and Biology, Russ Hobbie and I discuss thermal radiation. If you’re a black body, the net power you radiate, wtot, is given by Eq. 14.41
wtot = S σSB (T4 – Ts4) , (14.41)
where S is the surface area, σSB is the Stefan-Boltzmann constant (5.67 × 10−8 W m−2 K−4), T is the absolute temperature of your body (about 310 K), and Ts is the temperature of your surroundings. The T4 term is the radiation you emit, and the Ts4 term is the radiation you absorb.
The fourth power that appears in this expression is annoying. It means we must use absolute temperature in kelvins (K); you get the wrong answer if you use temperature in degrees Celsius (°C). It also means the expression is nonlinear; wtot is not proportional to the temperature difference T – Ts.
On the absolute temperature scale, the difference between the temperature of your body (310 K) and the temperature of your surroundings (say, 293 K at 20 °C) is only about 5%. In this case, we simplify the expression for wtot by linearizing it. To see what I mean, try Homework Problem 14.32 in IPMB.
Section 14.9
Problem 32. Show that an approximation to Eq. 14.41 for small temperature differences is wtot = S Krad (T − Ts). Deduce the value of Krad at body temperature. Hint: Factor T4 – Ts4 = (T − Ts)(…). You should get Krad = 6.76 W m−2 K−1.
The constant Krad has the same units as a convection coefficient (see Homework Problem 51 in Chapter 3 of IPMB). Think of it as an effective convection coefficient for radiative heat loss. Once you determine Krad, you can use either the kelvin or Celsius temperature scales for T − Ts, so you can write its units as W m−2 °C−1.
Air and Water, by Mark Denny. |
Finally, our analysis implies that when the difference between the temperatures of the body and the surroundings is small, a body whose primary mechanism for getting rid of heat is radiation will cool exponentially following Newton’s law of cooling.
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