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At a certain temperature, T (K), during a process, 500 J is absorbed by the system and work of 200 J is done by the system. Then change in internal energy of the system is:
A
700 J
B
300 J
C
400 J
D
500 J
Detailed Solution
Heat absorbed by the system, q = +500 J
Work done by the system, w = $-200$ J
$\Delta U = q + w = 500 - 200 = 300$ J