ThermodynamicsHard
Question
The entropy change accompanying the transfer of 12,000 J of heat from a body A at 327°C to a body B at 127°C is
Options
A.−10.0 J/K
B.+10.0 J/K
C.−57.8 J/K
D.+57.8 J/K
Solution
$\Delta S = \Delta S_{A} + \Delta S_{B} = - \frac{12000}{600} + \frac{12000}{400} = + 10\text{ J/K}$
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