Question -
A system undergoes a process in which system releases 200 J heat and work done by the surroundings is 300 J. Change in internal energy of the system is
Options:
A .  −100 J
B .  100 J
C .  −500 J
D .  500 J
Answer: Option B : B ΔU=q+w Since system releases heat, q will be negative. It is also given that work done by the surroundings is 300 J which is the same as the work done on the system. Hence, w has a +ve sign. Keeping this in mind, we get: = −200+300=100J
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