p19_067.pdf

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Chapter 19 - 19.67
67. For a cylinder of height h , the surface area is A c =2 πrh , and the area of a sphere is A o =4 πR 2 .The
net radiative heat transfer is given by Eq. 19-40.
A c + A o =1 . 8m 2 .
The emissivity ε =0 . 80 is given in the problem, and the Stefan-Boltzmann constant is found in
§
≈
19-11: σ =5 . 67
×
10 − 8 W / m 2
·
K 4 . The “environment” temperature is T env = 303 K, and the skin
temperature is T = 9 (102
−
32) + 273 = 312 K. Therefore,
P net = σεA T env −
T 4 =
−
86 W .
The corresponding sign convention is discussed in the textbook immediately after Eq. 19-40. We
conclude that heat is being lost by the body at a rate of roughly 90 W.
(b) Half the body surface area is roughly A =1 . 8 / 2=0 . 9m 2 .Now,with T env = 248 K, we find
|
P net |
= σεA T env −
T 4 ≈
230 W .
(c) Finally, with T env = 193 K (and still with A =0 . 9m 2 )weobtain
|
P net |
= 330 W.
(a) We estimate the surface area A of the body as that of a cylinder of height 1 . 8 m and radius
r =0 . 15 m plus that of a sphere of radius R =0 . 10 m. Thus, we have A

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