Question

A system goes from state a to b along the path abc, when 20 kJ of heat flows into the system and the system does 7.6 kJ of work. If the system is returned back from b to a along the curved path bca due to a work of 5 kJ done on it, then

• A. 12.6 kJ of heat will be absorbed by the system
• B. 19.5 kJ of heat will be rejected by the system
• C. 7.5 kJ of heat will be absorbed by the system
• D. 17.5 kJ of heat will be rejected by the system

Hint:

When analyzing thermodynamic processes, always use the first law of thermodynamics to relate the heat, internal energy, and work done. Remember that the sign of work changes based on whether the work is done on the system or by the system.

Answer 1

The Correct Option is D

The first law of thermodynamics states: \[ \Delta Q = \Delta U + \Delta W \] Where:
- \( \Delta Q \) is the heat added to the system,
- \( \Delta U \) is the change in internal energy,
- \( \Delta W \) is the work done by the system.
In the process from a to b: \[ \Delta Q_{\text{ab}} = 20 \, \text{kJ}, \quad \Delta W_{\text{ab}} = 7.6 \, \text{kJ} \] So, \[ \Delta U_{\text{ab}} = \Delta Q_{\text{ab}} - \Delta W_{\text{ab}} = 20 - 7.6 = 12.4 \, \text{kJ} \] In the process from b to a, since the work is done on the system: \[ \Delta W_{\text{ba}} = -5 \, \text{kJ} \] Thus, the change in internal energy from b to a is: \[ \Delta U_{\text{ba}} = \Delta U_{\text{ab}} = 12.4 \, \text{kJ} \] The total heat added in the process b to a is: \[ \Delta Q_{\text{ba}} = \Delta U_{\text{ba}} - \Delta W_{\text{ba}} = 12.4 - (-5) = 12.4 + 5 = 17.5 \, \text{kJ} \] Since the system returns to its original state, this heat is rejected.
Thus, 17.5 kJ of heat is rejected by the system.