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?

In a process, `701 J` of heat is absorbed by a system and `394 J` of w

1.	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?
Answer» Correct Answer - q=+701 J w=-394 J, since work is done by the ststem `DeltaU=307 J` According to the first law of thermodynamics, `DeltaU=q+W(i)` Where, `DeltaU`= change in internal energy for a process q=heat W=work Given, q=+701 (Since heat is absorbed) W=-394 J (Since work is done by the system) Substituting the values in expression, (i), we get `DeltaU=701 J +(-394 J)` `DeltaU=307 J` Hence, the change in internal energy for the given process is 307 J.

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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?

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