Topic: Thermodynamics
A system is undergoing a cycle that consists of three processes. During the first process, the work is 5 kJ and the heat is 23 kJ. In the second process no work takes place and the heat interaction is -50 kJ. The third process is adiabatic. The work in the third process is
- A. -32 kJ
- B. -23 kJ
- C. -22 kJ
- D. zero
Answer: A. -32 kJ
Explanation: According to the First Law of Thermodynamics for a complete cycle, the net heat transfer is equal to the net work transfer ($\Sigma Q = \Sigma W$). The total heat transfer is $Q_{net} = 23 - 50 + 0 = -27 \text{ kJ}$. The total work transfer is $W_{net} = 5 + 0 + W$. Equating the two gives $-27 = 5 + W$, solving for $W$ yields $W = -32 \text{ kJ}$.