question_answer 20)

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

Answer:

\[q\text{ }=+\text{ }701\text{ }J\](Heat absorbed) \[w=-394\text{ }J\] (Work done by the system) \[\Delta U=q-w=701-394\] \[=307J\] i.e., Internal energy of system will increase by 307 J.