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
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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:
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
Was this answer helpful ?
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