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.