A system undergoes a cyclic process in which it absorbs heat `Q_1` and gives away heat `Q_2`. If the efficiency of the process is `eta` and the work done is W, then

Options

A. `W=Q_1-Q_2`
B. `eta=W/Q_1`
C. `eta=Q_2/Q_1`
D. `eta=1-Q_2/Q_1`

Answer

To solve the problem, we need to analyze the cyclic process of the system in terms of heat absorbed, heat given away, work done, and efficiency.

### Step-by-Step Solution:

1. **Understanding the Efficiency of the Process:**
The efficiency (η) of a heat engine is defined as the ratio of the work done (W) by the engine to the heat absorbed (Q1) from the hot reservoir. Mathematically, this can be expressed as:
\[ \eta = \frac{W}{Q_1} \]

2. **Applying the First Law of Thermodynamics:**
The first law of thermodynamics states that the change in internal energy (ΔU) of a system is equal to the heat added to the system (Q) minus the work done by the system (W):
\[ \Delta U = Q - W \]
For a cyclic process, the internal energy change over one complete cycle is zero (ΔU = 0). Therefore, we can rewrite the equation as:
\[ 0 = Q - W \implies Q = W \]

3. **Calculating the Net Heat Transfer:**
In a cyclic process, the net heat transfer (Q_net) is the difference between the heat absorbed (Q1) and the heat rejected (Q2). Thus, we can express this as:
\[ Q_{\text{net}} = Q_1 - Q_2 \]
Since we established that for a cyclic process, Q_net is equal to the work done (W), we can write:
\[ W = Q_1 - Q_2 \]

4. **Identifying the Correct Options:**
From the above analysis, we can conclude:
- The efficiency is given by \( \eta = \frac{W}{Q_1} \).
- The work done is equal to the net heat transfer, which is \( W = Q_1 - Q_2 \).

Therefore, the correct options based on the analysis are:
- Option 1: Efficiency is \( \eta = \frac{W}{Q_1} \) (Correct)
- Option 2: Work done is \( W = Q_1 - Q_2 \) (Correct)
- Option 3: \( Q_{\text{net}} = W \) (Correct, as it follows from the first law of thermodynamics for cyclic processes)

### Summary of Correct Options:
Thus, the correct answers are options 1, 2, and 4.