# The internal energy of a system changes because the system gains 180 J of heat and performs 338 J of work. In returning to its initial state, the system loses 111 J of heat. During this return process, what work is involved? If the work is done by the system, then give the work involved as a positive number. If the work is done on the system, then give the work involved as a negative number.

## To find out the work involved during the return process, we need to use the First Law of Thermodynamics. The First Law states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.

Mathematically, it can be expressed as:

ΔU = Q - W

Where:

ΔU is the change in internal energy

Q is the heat added to the system

W is the work done by the system

In this case, the initial change in internal energy (ΔU) is not given, but we are given the amount of heat gained (Q1 = 180 J) and the work performed (W1 = 338 J). We are also given the heat lost during the return process (Q2 = -111 J).

Since the system is returning to its initial state, the change in internal energy (ΔU) would be zero. We can write this as:

0 = Q2 - W2

So, to find the work involved (W2), we rearrange the equation:

W2 = Q2

Substituting the given value, we have:

W2 = -111 J

Therefore, the work involved during the return process is -111 J. The negative sign indicates that the work is done on the system.