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56. (a) The surface area of the cylinder is given by
22
2
2
2
2
2
11
1
1
2
2
2 (2.5 10 m)
2 (2.5 10 m)(5.0 10 m) 1.18 10 m
Arr
h
ππ
π
−−
−
−
=+
=
×
+
×
×
=
×
,
its temperature is
T
1
= 273 + 30 = 303 K, and the temperature of the environment is
T
env
=
273 + 50 = 323 K. From Eq. 1839 we have
()
(
)
44
2
2
4
4
e
n
v
0.85 1.18 10 m
(323K)
(303K)
1.4W.
PA
T
T
σε
−
=−
=
×
−=
(b) Let the new height of the cylinder be
h
2
. Since the volume
V
of the cylinder is fixed,
we must have
Vr
hr
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 Spring '08
 Any
 Physics

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