Efficiency of cyclic process is `50%` if heat `Q_1=1915J` , `Q_2= -40J` , `Q_3= 125J`, then `Q_4` is unknown then find the value of `Q_4`.

Options

A. Option1 1080 J
B. Option2 -980 J
C. Option3 -1080J
D. Option4 -1280J

Answer

To solve the problem, we need to find the value of \( Q_4 \) given the efficiency of a cyclic process is \( 50\% \), and the values of \( Q_1 = 1915 \, J \), \( Q_2 = -40 \, J \), and \( Q_3 = 125 \, J \).

### Step-by-Step Solution:

1. **Understanding Efficiency**:
The efficiency \( \eta \) of a cyclic process is defined as:
\[ \eta = \frac{W}{Q_{supply}} \]
where \( W \) is the work done and \( Q_{supply} \) is the total heat supplied to the system.

2. **Identifying Heat Supplied and Rejected**:
In this case, the heat supplied \( Q_{supply} \) is the sum of \( Q_1 \) and \( Q_3 \):
\[ Q_{supply} = Q_1 + Q_3 = 1915 \, J + 125 \, J = 2040 \, J \]
The heat rejected \( Q_{rejected} \) is the sum of \( Q_2 \) and \( Q_4 \):
\[ Q_{rejected} = Q_2 + Q_4 = -40 \, J + Q_4 \]

3. **Setting Up the Efficiency Equation**:
Given that the efficiency is \( 50\% \) or \( \frac{1}{2} \):
\[ \frac{W}{Q_{supply}} = \frac{1}{2} \]
Rearranging gives:
\[ W = \frac{1}{2} Q_{supply} = \frac{1}{2} \times 2040 \, J = 1020 \, J \]

4. **Relating Work Done to Heat Supplied and Rejected**:
The work done can also be expressed as:
\[ W = Q_{supply} - Q_{rejected} \]
Substituting the expressions we have:
\[ 1020 \, J = 2040 \, J - (-40 \, J + Q_4) \]
Simplifying this gives:
\[ 1020 \, J = 2040 \, J + 40 \, J - Q_4 \]
\[ 1020 \, J = 2080 \, J - Q_4 \]
Rearranging to find \( Q_4 \):
\[ Q_4 = 2080 \, J - 1020 \, J = 1060 \, J \]

5. **Final Calculation**:
Since \( Q_4 \) is heat rejected, we need to express it as a negative value:
\[ Q_4 = -1060 \, J \]

### Conclusion:
The value of \( Q_4 \) is \( -1060 \, J \).