Ncert Exercises & Solutions

Q.7. In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

NEETprep Answer:

According to the first law of thermodynamics,
∆U = q + W (i)

Where,
∆U = change in internal energy for a process q
= heat
W = work Given,
q = + 701 J (Since heat is absorbed)
W = –394 J (Since work is done by the system)
Substituting the values in expression (i), we get
∆U = 701 J + (–394 J)
∆U = 307 J
Hence, the change in internal energy for the given process is 307 J.

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