Essential University Physics: Volume 1 (4th Edition)

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

Chapter 16 - Exercises and Problems - Page 311: 69


1150 K

Work Step by Step

We know that the temperature is given by: $ \frac{P}{4 \pi e R^2 \sigma}$ We plug in the known values to find: $ T = (\frac{34,000}{.341425* 5.7* 10^{4}})^{1/4}*10^3$ $T=1150 K$
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