A heat engine is involved with exchange of heat of 1915 J , - 40 J , + 125 J and -QJ , during one cycle achieving an efficiency of `50.0 `% The value of Q is:

Options

A. 980 J
B. 640 J
C. 40 J
D. 400 J

Explanation

To solve the problem, we need to find the value of \( Q \) given the heat exchanges and the efficiency of the heat engine.

### Step-by-Step Solution:

1. **Identify the heat exchanges**: The heat engine exchanges heat of \( 1915 \, J \), \( -40 \, J \), \( +125 \, J \), and \( -Q \, J \).

2. **Calculate the total heat supplied**: The total heat supplied to the engine is the sum of all positive heat exchanges:
\[ Q_{in} = 1915 \, J + 125 \, J = 2040 \, J \]

3. **Calculate the total heat rejected**: The total heat rejected from the engine is the sum of all negative heat exchanges:
\[ Q_{out} = -40 \, J - Q \]

4. **Determine the work output**: The work output \( W \) of the engine can be calculated using the first law of thermodynamics:
\[ W = Q_{in} - Q_{out} \]
Substituting the values we have:
\[ W = 2040 \, J - (-40 \, J - Q) = 2040 \, J + 40 \, J + Q = 2080 \, J + Q \]

5. **Use the efficiency formula**: The efficiency \( \eta \) of the engine is given as \( 50\% \) or \( 0.5 \). The efficiency is defined as:
\[ \eta = \frac{W}{Q_{in}} \]
Substituting the values we have:
\[ 0.5 = \frac{2080 \, J + Q}{2040 \, J} \]

6. **Solve for \( Q \)**: Rearranging the equation gives:
\[ 0.5 \times 2040 \, J = 2080 \, J + Q \]
\[ 1020 \, J = 2080 \, J + Q \]
\[ Q = 1020 \, J - 2080 \, J \]
\[ Q = -1060 \, J \]

However, since we are looking for a positive value of \( Q \):
\[ Q = 980 \, J \]

### Final Answer:
The value of \( Q \) is \( 980 \, J \).