Thermodynamics: An Engineering Approach 8th Edition

Published by McGraw-Hill Education
ISBN 10: 0-07339-817-9
ISBN 13: 978-0-07339-817-4

Chapter 2 - Energy, Energy Transfer, and General Energy Analysis - Problems - Page 105: 2-105

Answer

$T_{s}=276.9K$

Work Step by Step

As the problem says: $Q_{absorbed}=Q_{rad}$ $Q_{absorbed}=\alpha*Q_{solar}=0.2*1000\frac{W}{m^2}=200\frac{W}{m^2}$ $Q_{rad}=\varepsilon \sigma A\Delta T^4$ $\frac{Q_{rad}}{A}=\varepsilon \sigma \Delta T^4=200\frac{W}{m^2}$ Knowing that $T_{space}=0K$ $200\frac{W}{m^2}=0.6*5.67*10^-8\frac{W}{m^2K^4}*(T_{s}^4-(0K)^4)$ Solving for $T_{s}$: $T_{s}=\sqrt[4] {\frac{200\frac{W}{m^2}}{0.6*5.67*10^-8\frac{W}{m^2K^4}}}$ $T_{s}=276.9K$
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