Understanding Cyclic Processes and Thermodynamics

A cyclic process in thermodynamics is one where the system returns to its initial state after undergoing a series of changes. A key property of a cyclic process is that the change in internal energy of the system is zero ($\Delta U = 0$).

The First Law of Thermodynamics relates the heat transfer ($Q$) to the work done ($W$) and the change in internal energy ($\Delta U$) for a process:

\(\Delta U = Q - W\)

For a cyclic process, since \(\Delta U = 0\), the First Law simplifies to:

\(0 = Q_{cycle} - W_{cycle}\)

This implies that the net heat transfer during the cycle is equal to the net work done during the cycle:

\(Q_{cycle} = W_{cycle}\)

Calculating Net Heat Transfer in the Cycle

The problem provides the heat transfers for different parts of the cyclic process: +20 kJ, -5 kJ, -10 kJ, and +15 kJ.

• A positive value for heat transfer means heat is added to the system.
• A negative value for heat transfer means heat is removed from the system.

To find the net heat transfer for the entire cycle ($Q_{cycle}$), we sum up the individual heat transfers:

\(Q_{cycle} = (+20 \text{ kJ}) + (-5 \text{ kJ}) + (-10 \text{ kJ}) + (+15 \text{ kJ})\)

\(Q_{cycle} = 20 - 5 - 10 + 15 \text{ kJ}\)

\(Q_{cycle} = (20 + 15) - (5 + 10) \text{ kJ}\)

\(Q_{cycle} = 35 - 15 \text{ kJ}\)

\(Q_{cycle} = 20 \text{ kJ}\)

Determining Net Work Done

As established by the First Law of Thermodynamics for a cyclic process, the net work done ($W_{cycle}$) is equal to the net heat transfer ($Q_{cycle}$).

\(W_{cycle} = Q_{cycle}\)

\(W_{cycle} = +20 \text{ kJ}\)

A positive value for work done means work is done by the system on its surroundings.

Therefore, the net work done for this cyclic process is +20 kJ.