A gas can be taken from A to B via two different paths ACB and ADB.

When path ACB is used, \(60\) J of heat flows into the system and \(30\) J of work is done by the system. If path ADB is used, work done by the system is \(10\) J.
The heat flow into the system in path ADB is

According to the first law of thermodynamics, the change in internal energy is given by:
\(\Delta U = Q - W\)
where \(Q\) is the heat added to the system and \(W\) is the work done by the system.

For path ACB:
Heat added, \(Q_{ACB} = 60\) J
Work done, \(W_{ACB} = 30\) J
\(\Delta U = 60 - 30 = 30\) J

Since internal energy is a state function, the change in internal energy \(\Delta U\) between states A and B is the same for any path. Therefore, for path ADB, \(\Delta U = 30\) J.

For path ADB:
Work done, \(W_{ADB} = 10\) J
Using the first law of thermodynamics again:
\(\Delta U = Q_{ADB} - W_{ADB}\)
\(30 = Q_{ADB} - 10\)
\(Q_{ADB} = 30 + 10 = 40\) J

The heat flow into the system in path ADB is \(40\) J.

Answer: \(40\) J