Answer
$T_3=\frac{m_1}{m_3}T_1+\frac{m_2}{m_3}T_2$
Work Step by Step
The energy balance for this system is:
$\Delta U=0$
$m_1c_v(T_3-T_1)+m_2c_v(T_3-T_2)=0$
$(m1+m2)T_3=m_1T_1+m_2T_2$
$T_3=\frac{m_1}{m_3}T_1+\frac{m_2}{m_3}T_2$
Thermodynamics: An Engineering Approach 8th Edition
by Cengel, Yunus; Boles, Michael
Published by
McGraw-Hill Education
ISBN 10:
0-07339-817-9
ISBN 13:
978-0-07339-817-4
Chapter 4 - Energy Analysis of Closed Systems - Problems - Page 209: 4-140
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