Essential University Physics: Volume 1 (4th Edition) Clone

Published by Pearson
ISBN 10: 0-134-98855-8
ISBN 13: 978-0-13498-855-9

Chapter 16 - Section 16.3 - Heat Transfer - Example - Page 304: 16.5

Answer

$5.8 \times 10^3 \ K$

Work Step by Step

We simplify equation 16.9 to find: $ P = (\frac{P}{4 \pi R^2 \sigma})^{1/4}$ $ P = (\frac{3.9 \times 10^{26}}{4 \pi (7 \times 10^8)^2(5.7 \times 10^{-8})})^{1/4}= 5.8 \times 10^3 \ K$
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