Two Carnot engines A and B operate in series such hat engine A absorbs heat at `T_(1)` and rejects heat to a sink at temperature T. Engine B absorbs half of the heat rejected by Engine A and rejects heat to the sink at `T_(3)` . When workdone in both the cases is equal, to value of T is :

Options

A. `(2)/(3)T_(1)+(3)/(2)T_(3)`
B. `1/3T_(1)+(2)/(3)T_(3)`
C. `(3)/(2)T_(1)+1/3T_(3)`
D. `(2)/(3)T_(1)+1/3T_(3)`

Answer

To solve the problem, we will analyze the two Carnot engines operating in series and derive the relationship between the temperatures involved.

### Step-by-Step Solution:

1. **Understanding the Carnot Engines**:
- Engine A absorbs heat \( Q \) from a source at temperature \( T_1 \) and rejects heat \( Q_0 \) to a sink at temperature \( T \).
- Engine B absorbs half of the heat rejected by Engine A, which is \( \frac{Q_0}{2} \), and rejects heat to a sink at temperature \( T_3 \).

2. **Work Done by Engine A**:
- The efficiency of Engine A (\( \eta_1 \)) is given by:
\[ \eta_1 = \frac{W_1}{Q} = 1 - \frac{T}{T_1} \]
- Rearranging gives:
\[ W_1 = Q \left(1 - \frac{T}{T_1}\right) = Q \frac{T_1 - T}{T_1} \]

3. **Work Done by Engine B**:
- The efficiency of Engine B (\( \eta_2 \)) is given by:
\[ \eta_2 = \frac{W_1}{\frac{Q_0}{2}} = 1 - \frac{T_3}{T} \]
- Rearranging gives:
\[ W_1 = \frac{Q_0}{2} \left(1 - \frac{T_3}{T}\right) \]

4. **Relating \( Q_0 \) and \( Q \)**:
- From the first engine, we know:
\[ Q_0 = Q - W_1 \]
- Substituting \( W_1 \) from Engine A into this equation gives:
\[ Q_0 = Q - Q \frac{T_1 - T}{T_1} = Q \frac{T}{T_1} \]

5. **Substituting \( Q_0 \) into Engine B's Work Done**:
- Substitute \( Q_0 \) into the expression for \( W_1 \) from Engine B:
\[ W_1 = \frac{Q \frac{T}{T_1}}{2} \left(1 - \frac{T_3}{T}\right) \]

6. **Setting the Work Done Equal**:
- Since the work done by both engines is equal, we set the two expressions for \( W_1 \) equal to each other:
\[ Q \frac{T_1 - T}{T_1} = \frac{Q \frac{T}{T_1}}{2} \left(1 - \frac{T_3}{T}\right) \]
- Cancel \( Q \) from both sides:
\[ \frac{T_1 - T}{T_1} = \frac{T}{2T_1} \left(1 - \frac{T_3}{T}\right) \]

7. **Cross-Multiplying and Simplifying**:
- Cross-multiply to eliminate the fractions:
\[ 2T_1(T_1 - T) = T(1 - \frac{T_3}{T}) \]
- This simplifies to:
\[ 2T_1^2 - 2T_1T = T - T_3 \]
- Rearranging gives:
\[ 2T_1^2 - 2T_1T - T + T_3 = 0 \]

8. **Finding the Value of \( T \)**:
- Rearranging the equation yields:
\[ 2T_1T - T + T_3 = 2T_1^2 \]
- Solving for \( T \) gives:
\[ T(2T_1 - 1) = 2T_1^2 - T_3 \]
- Finally, we get:
\[ T = \frac{2T_1 + T_3}{3} \]

### Final Answer:
\[ T = \frac{2T_1 + T_3}{3} \]